Re: Question on exact sequence

Re: Question on exact sequence

[George Janelidze]

The "curious discovery" is Exercise 6 at the end of Chapter VIII ("Abelian
Categories") of Mac Lane's "Categories for the Working Mathematician"...

However, I think it is an interesting question, and:

When for the standard snake lemma Michael says "...there is an exact
sequence
0 --> ker f --> ker g --> ker h --> cok f --> cok g --> cok h --> 0", what
does "there is" mean?

There are two well known answers:

ANSWER 1.  ker f --> ker g --> ker h   and   cok f --> cok g --> cok h are
the obvious induced morphisms and there exists a "connecting morphism" d :
ker h ---> cok f making the sequence above exact. Such a d is not unique:
for instance if d is such, then so is -d. However, since the snake lemma
holds in functor categories, the unnaturality of d does not make big
problems in concrete situations.

ANSWER 2. ker f --> ker g --> ker h   and   cok f --> cok g --> cok h are
the obvious induced morphisms as before, while THE "connecting morphism" d :
ker h ---> ker f is the composite of the zigzag

ker h ---> C <--- B ---> B' <---A' ---> cok f

(where the arrows are considered as internal relations). This "canonical
connecting morphism" d works even in the non-abelian case of Dominique Bourn
as I learned from my daughter Tamar who developed the "relative version".
Note also, that the desire to have such a canonical d (in the abelian case)
was a big original reason for developing what we call today "calculus of
relations" (at the beginning with great participation of Saunders himself).

And... in the "curious case = Exercise 6" the "canonical d" does not work!
For, consider the simplest case of the composite 0 ---> B ---> 0: the exact
ker-cok sequence will become

0 --> 0 --> 0 --> B --> B --> 0 --> 0 --> 0,

where B --> B must be an isomorphism, while it is easy to check that the
"canonical d" will become the relation opposite to the zero morphism B -->
B.

A possible conclusion is that the "master theorem" should involve some kind
of "d" as an extra structure.

To Steve's message: does Enrico really generalize the standard snake lemma
and the "curious case" simultaneously?

George

----- Original Message -----
From: "Michael Barr" <barr@math.mcgill.ca>
To: "Categories list" <categories@mta.ca>
Sent: Tuesday, November 10, 2009 12:57 AM
Subject: categories: Question on exact sequence


> I have recently discovered a curious fact about abelian categories.
> First, let me briefly describe the well-known snake lemma.  If we have a
> commutative diagram with exact rows (there are variations without the 0
> at the left end of the top and without the 0 at the right end of the
> bottom, but here is the strongest form)
>
>       0 ---> A ----> B ----> C ----> 0
>              |       |       |
>              |       |       |
>              |f      |g      |h
>              |       |       |
>              v       v       v
>       0 ---> A' ---> B' ---> C' ---> 0
>
> then there is an exact sequence
>   0 --> ker f --> ker g --> ker h --> cok f --> cok g --> cok h --> 0
>
> The curious discovery is that you have any pair of composable maps f: A
> --> B and h: B --> C and you form the diagram (with g = hf)
>                  1       f
>              A ----> A ----> B
>              |       |       |
>              |       |       |
>              |f      |g      |h
>              |       |       |
>              v       v       v
>              B ----> C ----> C
>                  h       1
> you get the same exact sequence.  So I would imagine that there must be
> a "master theorem" of which these are two cases.  Does anyone know what
> it says?  The connecting map here is just the inclusion of ker h into B
> followed by the projection on cok f.
>
> Michael
>
>
>
> [For admin and other information see: http://www.mta.ca/~cat-dist/ ]



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Re: Question on exact sequence

[Michael Barr]

I do appreciate the example since I wondered if the "connecting
homomorphism" could be induced by a composite of relations as in the snake
lemma.  I thought not and George has provided an example.  Since Tuesday,
we have had house guests so I really have not had time to absorb all the
replies, but when I have time, I plan to collect them all and try to see
if there is a satisfactory general answer of which the two instances I
described are special cases.  There is something going on here that I
don't quite comprehend (although maybe the answer is in the theorem Marco
mentioned.

Since my curious sequence was an exercise in CWM, it is surprising that
Saunders never raised the question in the form I did.  The conclusion
certainly looks like something out of the snake lemma, but I was unable to
formulate it as a cosequence.

Incidentally, the theorem on acyclic models, as it appears in my book,
can be described as a map induced by a composite of relations that, in
homology, becomes functional.

Michael

On Wed, 11 Nov 2009, George Janelidze wrote:

> The "curious discovery" is Exercise 6 at the end of Chapter VIII ("Abelian
> Categories") of Mac Lane's "Categories for the Working Mathematician"...
>
> However, I think it is an interesting question, and:
>
> When for the standard snake lemma Michael says "...there is an exact
> sequence
> 0 --> ker f --> ker g --> ker h --> cok f --> cok g --> cok h --> 0", what
> does "there is" mean?
>
> There are two well known answers:
>
> ANSWER 1.  ker f --> ker g --> ker h   and   cok f --> cok g --> cok h are
> the obvious induced morphisms and there exists a "connecting morphism" d :
> ker h ---> cok f making the sequence above exact. Such a d is not unique:
> for instance if d is such, then so is -d. However, since the snake lemma
> holds in functor categories, the unnaturality of d does not make big
> problems in concrete situations.
>
> ANSWER 2. ker f --> ker g --> ker h   and   cok f --> cok g --> cok h are
> the obvious induced morphisms as before, while THE "connecting morphism" d :
> ker h ---> ker f is the composite of the zigzag
>
> ker h ---> C <--- B ---> B' <---A' ---> cok f
>
> (where the arrows are considered as internal relations). This "canonical
> connecting morphism" d works even in the non-abelian case of Dominique Bourn
> as I learned from my daughter Tamar who developed the "relative version".
> Note also, that the desire to have such a canonical d (in the abelian case)
> was a big original reason for developing what we call today "calculus of
> relations" (at the beginning with great participation of Saunders himself).
>
> And... in the "curious case = Exercise 6" the "canonical d" does not work!
> For, consider the simplest case of the composite 0 ---> B ---> 0: the exact
> ker-cok sequence will become
>
> 0 --> 0 --> 0 --> B --> B --> 0 --> 0 --> 0,
>
> where B --> B must be an isomorphism, while it is easy to check that the
> "canonical d" will become the relation opposite to the zero morphism B -->
> B.
>
> A possible conclusion is that the "master theorem" should involve some kind
> of "d" as an extra structure.
>
> To Steve's message: does Enrico really generalize the standard snake lemma
> and the "curious case" simultaneously?
>
> George

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Re: Question on exact sequence

[George Janelidze]

I have further comments to Marco and Steve (maybe tomorrow...), but now I am
only answering

> Since my curious sequence was an exercise in CWM, it is surprising that
> Saunders never raised the question in the form I did.  The conclusion
> certainly looks like something out of the snake lemma, but I was unable to
> formulate it as a cosequence.

from Michael's message:

Dear Michael,

Does "formulate" mean "obtain/deduce"? Obtaining the curious sequence as a
consequence of the snake lemma is actually easy, and Saunders surely knew
it - which probably explains why did not he raise your question. Given your
f : A ---> B, h : B ---> C and g = hf, just apply the snake lemma to

    <1,f>         [f,-1]
A ---> A + B ---> B
 |              |              |
 | f            | g+1       | h
v             v             v
B ---> C + B ---> C
    <h,1>        [1,-h]

where + denotes the direct sum, <...> "uses" it as product, and [...] "uses"
it as coproduct (and use the fact that Ker(g) = Ker(g+1)).

However, this does not answer your original question of course.

George

----- Original Message -----
From: "Michael Barr" <barr@math.mcgill.ca>
To: "George Janelidze" <janelg@telkomsa.net>; <categories@mta.ca>
Sent: Thursday, November 12, 2009 2:41 PM
Subject: categories: Re: Question on exact sequence


> I do appreciate the example since I wondered if the "connecting
> homomorphism" could be induced by a composite of relations as in the snake
> lemma.  I thought not and George has provided an example.  Since Tuesday,
> we have had house guests so I really have not had time to absorb all the
> replies, but when I have time, I plan to collect them all and try to see
> if there is a satisfactory general answer of which the two instances I
> described are special cases.  There is something going on here that I
> don't quite comprehend (although maybe the answer is in the theorem Marco
> mentioned.
>
> Since my curious sequence was an exercise in CWM, it is surprising that
> Saunders never raised the question in the form I did.  The conclusion
> certainly looks like something out of the snake lemma, but I was unable to
> formulate it as a cosequence.
>
> Incidentally, the theorem on acyclic models, as it appears in my book,
> can be described as a map induced by a composite of relations that, in
> homology, becomes functional.
>
> Michael
>

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Re: Question on exact sequence

[Michael Barr]

Actually, on further thought, I agree with you.  I didn't originally want
a slick proof but to understand and then I forgot why I really raised the
question.  After all, I had already proved it.  So what I really wanted
and still want to know is what conditions on a map between two three term
sequences gives the 6 term exact sequence (with or without the end 0s).
The situation of the snake lemma is so different from the situation I
(and, obviously others) discovered that one wonders still what general
conditions could possibly encompass the two cases.  That really was my
initial question and that question now comes back to me.

Michael

On Fri, 13 Nov 2009, George Janelidze wrote:

> All right, then I shall better stop, unless there will be new unexpected
> comments (because what Bill and others say, will take us too far...)
>
> George
>
> ----- Original Message -----
> From: "Michael Barr" <barr@math.mcgill.ca>
> To: "George Janelidze" <janelg@telkomsa.net>
> Sent: Friday, November 13, 2009 4:33 AM
> Subject: Re: categories: Re: Question on exact sequence
>
>
>> Actually that diagram with the sums does really answer the question as I
>> had understood it.  There may be a deeper question, but I am not sure how
>> to formulate it.
>>
>> Michael
>>
>> On Fri, 13 Nov 2009, George Janelidze wrote:
>>
>>> I have further comments to Marco and Steve (maybe tomorrow...), but now
> I am
>>> only answering
>>>
>>>> Since my curious sequence was an exercise in CWM, it is surprising that
>>>> Saunders never raised the question in the form I did.  The conclusion
>>>> certainly looks like something out of the snake lemma, but I was unable
> to
>>>> formulate it as a cosequence.
>>>
>>> from Michael's message:
>>>
>>> Dear Michael,
>>>
>>> Does "formulate" mean "obtain/deduce"? Obtaining the curious sequence as
> a
>>> consequence of the snake lemma is actually easy, and Saunders surely
> knew
>>> it - which probably explains why did not he raise your question. Given
> your
>>> f : A ---> B, h : B ---> C and g = hf, just apply the snake lemma to
>>>
>>>    <1,f>         [f,-1]
>>> A ---> A + B ---> B
>>> |              |              |
>>> | f            | g+1       | h
>>> v             v             v
>>> B ---> C + B ---> C
>>>    <h,1>        [1,-h]
>>>
>>> where + denotes the direct sum, <...> "uses" it as product, and [...]
> "uses"
>>> it as coproduct (and use the fact that Ker(g) = Ker(g+1)).
>>>
>>> However, this does not answer your original question of course.
>>>
>>> George

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Re: Question on exact sequence

[George Janelidze]

Yes, this is how I understood your original question. In order to make some
comments, let me mention/recall:

You say "a map between two three term sequences" and such a diagram in a
category C is of course the same as a composable pair of morphisms in the
arrow category Arr(C). Therefore let me write it as

f ---> g ---> h   (1)

- and so your question is: when - for an abelian C - does (1) induce an
exact sequence

(0 -->) ker f --> ker g --> ker h --> cok f --> cok g --> cok h (--> 0)
(2)

My comments are:

1. I think Steve made a very good comment. Let me repeat it in my words:
When C is abelian, the category Cat(C) of internal categories in C is
equivalent to Arr(C). Writing cat(f) for the internal category corresponding
to f, we have:

(a) ker f = the internal group of automorphisms of 0 in cat(f) = the
fundamental group of cat(f). Therefore, let me write ker f = pi_1(f).

(b) cok f = the object of connected components of cat(f). Therefore, let me
write cok f = pi_0(f).

After that the sequence (2) becomes

(0 -->) pi_1(f) --> pi_1(g) --> pi_1(h) --> pi_0(f) --> pi_0(g) --> pi_0(h)
(--> 0),

and your question becomes a classical question of abstract homotopy theory!
And Steve gave (a partial?) answer that comes out of Enrico's work.

2. As we both agree, Marco also made a very good comment. Namely, Marco has
a general theorem that applies to (1) in the two cases of main interest: to
the standard snake lemma and to the case you called curious.
.
3. My own comment (although I am slightly changing it here) was/is that the
morphism ker h --> cok f (in (2)) should probably be considered as an extra
structure on (1) rather than something "induced". In Marco's approach it is
an induced morphism on subquotients, but it becomes "induced" in the two
cases of interest for different reasons. And those two reasons ONLY BECOME
the same when we present the two cases of interest as special cases of
Marco's theorem in Marco's way. But using Marco's way we also automatically
equip our diagram with an extra structure. Note also, that an extra
structure appears in Steve's second message as z : A --> C' (which surely
helps to create ker h --> cok f, although I have not checked the details).

4. It is amazing how many non-abelian versions of your question might be
asked! Since Dominique has a non-abelian snake lemma it can be about
semi-abelian/homological/protomodular categories - I would call it
"Bourn-nonabelian" although there are older versions with "old" axioms. Or
it can be Grandis-nonabelian - since Marco has own homological categories,
where he develops abstract homological algebra. Or it can be replacing
(all?) morphisms with pairs of parallel morphisms in a category enriched in
commutative monoids - as I understand this is what Bill had in mind asking
his question. Or it can be replacing f, g, h with internal categories (or
groupoids) in a rather large class of categories - as follows from Steve's
suggestion.

And there are related questions that came out of other interesting
messages...

George

----- Original Message -----
From: "Michael Barr" <barr@math.mcgill.ca>
To: "George Janelidze" <janelg@telkomsa.net>
Cc: "Categories list" <categories@mta.ca>
Sent: Friday, November 13, 2009 6:06 PM
Subject: Re: categories: Re: Question on exact sequence


> Actually, on further thought, I agree with you.  I didn't originally want
> a slick proof but to understand and then I forgot why I really raised the
> question.  After all, I had already proved it.  So what I really wanted
> and still want to know is what conditions on a map between two three term
> sequences gives the 6 term exact sequence (with or without the end 0s).
> The situation of the snake lemma is so different from the situation I
> (and, obviously others) discovered that one wonders still what general
> conditions could possibly encompass the two cases.  That really was my
> initial question and that question now comes back to me.
>
> Michael
>
> On Fri, 13 Nov 2009, George Janelidze wrote:
>
> > All right, then I shall better stop, unless there will be new unexpected
> > comments (because what Bill and others say, will take us too far...)
> >
> > George
> >

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Re: Question on exact sequence

[Michael Barr]

I think I have now come to understand this, at least in part.  A couple
things you said triggered this realization.  First the point that this
was taking place in the arrow category.  Second that maybe sometimes you
needed a map to get the connecting homomorphism and sometimes not.

The first point made me think maybe rather than the arrow category I
should perhaps be thinking about the category of chain complexes.  The
second somehow made me think about the difference between homology and
homotopy equivalences.

Perhaps this explanation amounts to shooting flies with elephant guns,
but it satsifies me that it has fully explained things.  Let me start by
discribing an important difference between the two situations.  Both
diagrams:

0 --> A --> B --> C --> 0.......A --> A --> B
......|.....|.....|.............|.....|.....|
......|.....|.....|.............|.....|.....|
......|.....|.....|.............|.....|.....|
......v.....v.....v.............v.....v.....v
0 --> A'--> B'--> C'--> 0.......B --> C --> C

with maps as in earlier posts, give 6 term exact sequences, but the
second continues continues to do so when you apply a homfunctor Hom(D,-)
or Hom(-,D), which is not generally true of the first.  Now this, had I
noticed it earlier, would have immediately put me in mind of the
difference between homotopy and homology.  It is a fact that a homology
equivalence between two chain complexes is a homotopy equivalence iff it
remains an equivalence if you apply any covariant (or any contravariant)
homfunctor.  This fact, along with a couple other things I will mention
below, is proved somewhere in "Acyclic Models".  Next, when I drew the
diagram

0 ---> A ---> A + B ---> B ---> 0
.......|........|........|.......
.......|........|........|.......
.......|........|........|.......
.......|........|........|.......
.......v........v........v.......
0 ---> B ---> B + C ---> C ---> 0

which was marked by a peculiar appearance of - signs, I recognized that
it looked like a mapping cone of something and, had I only worked out of
what, I might have realized sooner what was going on.  It is actually
the mapping cone of the map

0 ---> A
|......|
|......|
|......|
v......v
B ---> B
|......|
|......|
|......|
v......v
C ---> 0

Finally, the observation that the chain complex 0 ---> A ---> C ---> 0
is homotopic to 0 ---> A + B ---> B + C ---> 0 is a complete triviality.

So what is the general situation.  Let me raise the question in this
form.  Suppose f: C' ---> C and g: C ---> C'' is a map of chain
complexes.  When can we expect an exact triangle

  H(C') ------> H(C)
...^............./
....\.........../
.....\........./
......\......./
.......\...../
........\.../
.........\.v
.........H(C'')

Clearly we need something to induce the map H(C'') ---> H(C').  The
obvious thing would be a map S(C'') ---> C' (S is the suspension
functor).  The example of two-length sequences shows that this is too
much to hope for.  It looks like what you need is a relation from S(C'')
--> C' that induces, somehow a map on homology.  This assumption holds
for the case 0 --> C' --> C --> C' --> 0 and also holds for the original
curious sequence (the suspension is what saves the day here, as seen
above).  Is there a better way to express this?  I don't see one.
Perhaps the last word on this hasn't been said yet.

Another question I haven't answered but should be doable is whether the
existence of a relation S(C'') - - - -> C' that you assume induces a
morphism on homology is sufficient to make the triangle exact.  Beyond
this, I have some inchoate ideas that I will explore.  Clearly there has
to be some compatibility between R and f and g.

As I said, this explanation is perhaps a little heavy but I know no
simpler one.

Michael




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Re: Question on exact sequence

[George Janelidze]

1. Let me reformulate (to make it easier for non-experts and to ask for a
reference, because I have a strong feeling that I knew this many years ago)
your new approach to

0 --> ker f --> ker g --> ker h --> cok f --> cok g --> cok h --> 0 in the
case g = hf:

Every morphism u : X ---> Y of complexes in an abelian category induces a
long exact homology sequence involving the mapping cone of u. If

X = (...0 ---> B -h-> C...), Y = (...A -f-> B ---> 0...), and u =
(...,0,1,0,...),

then this long exact sequence is the sequence of homology objects of the
third column (or any other column - but the third column is displayed
better) of the diagram

............................................................................
.....
                                                       h
...0   --->   0   --->   0   --->   B   --->  C   --->   0...
...|              |              |              |              |
|...
...|              |              |              |              |
|...
...v             v             v    f        v             v             v
...0   --->   0   --->  A  --->    B   --->   0  --->    0...
...|              |              |              |              |
|...
...|              |              |              |              |
|...
...v             v             v     d      v             v             v
...0   --->   0  ---> A+B ---> B+C ---> 0  --->   0...
...|              |              |              |              |
|...
...|              |              |              |              |
|...
...v             v             v     -h     v             v             v
...0   --->   0  --->   B    --->  C   --->  0   --->   0...
...|              |              |              |              |
|...
...|              |              |              |              |
|...
...v             v    -f      v             v             v             v
...0   --->  A   --->   B   --->   0   --->  0   --->   0...
...|              |              |              |              |
|...
...|              |              |              |              |
|...
...v             v     -d    v             v             v             v
...0  ---> A+B ---> B+C ---> C  --->   0   --->   0...
...|              |              |              |              |
|...
...|              |              |              |              |
|...
...v             v      h     v             v             v             v
...0  --->   B    --->  C   --->  0   --->   0   --->   0...
...|              |              |              |              |
|...
...|              |              |              |              |
|...
...v     f       v             v             v             v             v
...A   --->   B   --->   0   --->  0   --->   0   --->   0...
............................................................................
......

where the vertical arrows are either the coproduct injections, or the
product projections, and d, written in the language of elements, is defined
by d(a,b) = (b+f(a),-h(b)). And this way of establishing our exact sequence
is 'more straightforward' than to use snake lemma because the choice X =
(...0 ---> B -h-> C...), Y = (...A -f-> B ---> 0...), and u =
(...,0,1,0,...) is 'more straightforward' then the choice of the "peculiar
diagram" (I mean what you call "the diagram ... marked by a peculiar
appearance of - signs").

Now my comment: Well, considering long exact homology sequences produced by
mapping cones as a part of "basic education", nobody will disagree. But
considering snake lemma as a starting point, one would try to find a hidden
instance of the snake lemma in the big diagram above. And it is there of
course - to be applied to the diagram

0   --->   A   --->  A+B   --->   B   --->  0
|              |                 |                |              |
|              | f               | d             |-h           |
v             v                v               v             v
0   --->   B   --->  B+C  --->    C   --->   0

formed by f, d, and -h in the second, third, and forth row respectively of
the big diagram. Is this (small) new diagram less peculiar than the peculiar
one? Yes and No for obvious reasons, and moreover, the two diagrams are
isomorphic of course. This, however, does not mean that I am trying to
criticize your new approach and/or I am proposing anything better.

2. 'By the way' you mentioned: "...the difference between homotopy and
homology.  It is a fact that a homology equivalence between two chain
complexes is a homotopy equivalence iff it remains an equivalence if you
apply any covariant (or any contravariant) homfunctor..." As you surely
know, this fact also is an abelian version of something important in
internal category theory: say, for internal 1-categories it is about the
functors that are "internally fully faithful  and essentially surjective
on objects" versus the internal category equivalences.

George

----- Original Message -----
From: "Michael Barr" <barr@math.mcgill.ca>
To: "George Janelidze" <janelg@telkomsa.net>; <categories@mta.ca>
Sent: Saturday, November 14, 2009 6:24 PM
Subject: categories: Re: Question on exact sequence


> I think I have now come to understand this, at least in part.  A couple
> things you said triggered this realization.  First the point that this
> was taking place in the arrow category.  Second that maybe sometimes you
> needed a map to get the connecting homomorphism and sometimes not.
>
> The first point made me think maybe rather than the arrow category I
> should perhaps be thinking about the category of chain complexes.  The
> second somehow made me think about the difference between homology and
> homotopy equivalences.
>
> Perhaps this explanation amounts to shooting flies with elephant guns,
> but it satsifies me that it has fully explained things.  Let me start by
> discribing an important difference between the two situations.  Both
> diagrams:
>
> 0 --> A --> B --> C --> 0.......A --> A --> B
> ......|.....|.....|.............|.....|.....|
> ......|.....|.....|.............|.....|.....|
> ......|.....|.....|.............|.....|.....|
> ......v.....v.....v.............v.....v.....v
> 0 --> A'--> B'--> C'--> 0.......B --> C --> C
>
> with maps as in earlier posts, give 6 term exact sequences, but the
> second continues continues to do so when you apply a homfunctor Hom(D,-)
> or Hom(-,D), which is not generally true of the first.  Now this, had I
> noticed it earlier, would have immediately put me in mind of the
> difference between homotopy and homology.  It is a fact that a homology
> equivalence between two chain complexes is a homotopy equivalence iff it
> remains an equivalence if you apply any covariant (or any contravariant)
> homfunctor.  This fact, along with a couple other things I will mention
> below, is proved somewhere in "Acyclic Models".  Next, when I drew the
> diagram
>
> 0 ---> A ---> A + B ---> B ---> 0
> .......|........|........|.......
> .......|........|........|.......
> .......|........|........|.......
> .......|........|........|.......
> .......v........v........v.......
> 0 ---> B ---> B + C ---> C ---> 0
>
> which was marked by a peculiar appearance of - signs, I recognized that
> it looked like a mapping cone of something and, had I only worked out of
> what, I might have realized sooner what was going on.  It is actually
> the mapping cone of the map
>
> 0 ---> A
> |......|
> |......|
> |......|
> v......v
> B ---> B
> |......|
> |......|
> |......|
> v......v
> C ---> 0
>
> Finally, the observation that the chain complex 0 ---> A ---> C ---> 0
> is homotopic to 0 ---> A + B ---> B + C ---> 0 is a complete triviality.
>
> So what is the general situation.  Let me raise the question in this
> form.  Suppose f: C' ---> C and g: C ---> C'' is a map of chain
> complexes.  When can we expect an exact triangle
>
>   H(C') ------> H(C)
> ...^............./
> ....\.........../
> .....\........./
> ......\......./
> .......\...../
> ........\.../
> .........\.v
> .........H(C'')
>
> Clearly we need something to induce the map H(C'') ---> H(C').  The
> obvious thing would be a map S(C'') ---> C' (S is the suspension
> functor).  The example of two-length sequences shows that this is too
> much to hope for.  It looks like what you need is a relation from S(C'')
> --> C' that induces, somehow a map on homology.  This assumption holds
> for the case 0 --> C' --> C --> C' --> 0 and also holds for the original
> curious sequence (the suspension is what saves the day here, as seen
> above).  Is there a better way to express this?  I don't see one.
> Perhaps the last word on this hasn't been said yet.
>
> Another question I haven't answered but should be doable is whether the
> existence of a relation S(C'') - - - -> C' that you assume induces a
> morphism on homology is sufficient to make the triangle exact.  Beyond
> this, I have some inchoate ideas that I will explore.  Clearly there has
> to be some compatibility between R and f and g.
>
> As I said, this explanation is perhaps a little heavy but I know no
> simpler one.
>
> Michael
>
>
>
>
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Re: Question on exact sequence

[Marco Grandis]

Dear George (and others),

Continuing what you and others have already said, there are (at
least) two ways of extending those two lemmas.

1. Within homological algebra, there is the 'general form' of my
first msg on this point.
As I was saying, I do not like very much this extension; unless one
can find some 'meaning' for the 'generalised homologies'  H'(f,g,h)
and  H"(f,g,h)  that appear there.

2. Within homotopical algebra, one can find a deeper solution, if
more involved.
         The question is now about a triple  (f, g, alpha)  where  f,
g  are consecutive arrows and  alpha  is a nullhomotopy of their
composite  gf.  The hypothesis that we want to express is that this
'h-differential sequence' is 'h-exact' (h for homotopically, or
homotopical).
         This is studied in my paper [*] (Como 2000), in a general
setting and more particularly for a category of morphisms (Section
4); the latter is the case we are interested in for our extension. I
will only sketch what is of interest here; the interested reader can
look at [*].
         Let  D  be a category with pullbacks and pushouts, and  D'
its category of morphisms. Think of an object as a truncated chain
complex   A = (d_A: A1  -->  A0).  (In [*],  D  is also assumed to be
additive, but this is not necessary here.)
         There are obvious nullhomotopies of morphisms: inserting a
diagonal  A0  -->  B1  in a square  f: A = B. Nullhomotopies can be
whiskered with morphisms.
         Every morphism has an h-kernel (= homotopy kernel) and an h-
cokernel, defined by universal properties and constructed with the
pullback and pushout 'inside' the square.
         An object is *contractible* if its identity is
nullhomotopic; ie if its differential is iso.
         Given two consecutive morphisms  f: X  -->  A,  g: A  -->
Y,  and a nullhomotopy  alpha  of  gf,  there is a *homotopical
homology* object  H(f, g, alpha) =(w: B --> Z),  constructed via h-
kernels and h-cokernels ([*], thm 4.5).  Think of  B  as h-boundaries
and of  Z  as h-cycles.
         Say that the h-differential sequence  (f, g, alpha)  is *h-
exact* if   H(f, g, alpha)  is contractible,  ie  w  is iso.  This
yields a nullhomotopy from the h-kernel of  g  to the h-cokernel of  f.

Now (this is not written in [*]), assuming that  D  is pointed, there
is a composed nullhomotopy (by whiskering)

         Omega(Y)  -->  hKer(g)  ==>  hCok(f)  --> Sigma(X),

where:

         Omega(Y)  =  hKer(0 --> Y (sic!) )  =  (0 --> Ker(d_Y)),
         Sigma(X)  =  hCok(X --> 0)  =  (Cok(d_X) --> 0).

Thus our nullhomotopy gives the connecting morphism

         H_1(Y)) = Ker(d_Y))    -->    H_0(X) = Cok(d_X).

Exactness (of the three Omegas followed by the three Sigmas) has to
be studied.

Best
Marco

(PS. Notice that, if D is not additive, we have nullhomotopies
without homotopies!)
[*] M. Grandis, A note on exactness and stability in homotopical
algebra, Theory Appl. Categ. 9 (2001), No. 2, 17-42


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Re: Question on exact sequence

[F William Lawvere]

The remarks of Clemens together with Murray Adelman's
construction as alluded to by Ross suggest the following.

The 1960 triumph of abelian categories was followed by a
decade  which Barr and Grothendieck showed that exactness 
has little to to with additivity. But homology itself would seem also to
have little to do with additivity, if we take seriously the following definition
(that is actually mentioned in passing in many books without the
specific mention of that other 50-year old triumph of category theory).

Given a full inclusion that has both left and right adjoints, there is 
a resulting map from the right adjoint to the left; the image H of that 
map is a further invariant of objects in the bigger category, recorded in  
the smaller. 

For example  A^C will have a full subcategory determined by
a given surjective functor C->D so if A is complete the two adjoints and 
the image exist (in the traditional example, let C be a generic sequence 
and let D be the sequence of zeroes; restricting to the part of A^C where 
d^2=0 may make computing H easier but will not change the definition).

Of course if the left adjoint preserves products, then so will H and
hence H will preserve any kind of algebraic structure.  But the simplest example
(reflexive graphs) also satisfies the "Nullstellensatz " of my 2007 TAC paper
on cohesion, which is just a way of saying that H reduces to the right adjoint
 itself.

For (nonreflexive) iterated graphs , I.e., for  C a sequence of parallel pairs,
(for example sums of front vs back faces of cubes), an interesting subcategory
of the functor category is the part where the two are equal. This may be useful
for homology of additive objects where rigs of coefficients are not necessarily
rings.

How can exactness and "long exact sequences" be meaningfully  treated for such
functors H in non-abelian contexts ?

Bill

On Thu 11/12/09  7:41 AM , Michael Barr barr@math.mcgill.ca sent:
> I do appreciate the example since I wondered if the "connecting
> homomorphism" could be induced by a composite of relations as in the
> snakelemma.  I thought not and George has provided an example.  Since
> Tuesday,we have had house guests so I really have not had time to absorb all
> thereplies, but when I have time, I plan to collect them all and try to
> seeif there is a satisfactory general answer of which the two instances I
> described are special cases.  There is something going on here that I
> don't quite comprehend (although maybe the answer is in the theorem
> Marcomentioned.
> 
> Since my curious sequence was an exercise in CWM, it is surprising that
> Saunders never raised the question in the form I did.  The conclusion
> certainly looks like something out of the snake lemma, but I was unable
> toformulate it as a cosequence.
> 
> Incidentally, the theorem on acyclic models, as it appears in my book,
> can be described as a map induced by a composite of relations that, in
> homology, becomes functional.
> 
> Michael
> 
> On Wed, 11 Nov 2009, George Janelidze wrote:
> 
> > The "curious discovery" is Exercise 6
> at the end of Chapter VIII ("Abelian> Categories") of Mac Lane's "Categories
> for the Working Mathematician"...>
> > However, I think it is an interesting question,
> and:>
> > When for the standard snake lemma Michael says
> "...there is an exact> sequence
> > 0 --> ker f --> ker g --> ker h -->
> cok f --> cok g --> cok h --> 0", what> does "there is" mean?
> >
> > There are two well known answers:
> >
> > ANSWER 1.  ker f --> ker g --> ker h   and
>   cok f --> cok g --> cok h are> the obvious induced morphisms and there exists a
> "connecting morphism" d :> ker h ---> cok f making the sequence above
> exact. Such a d is not unique:> for instance if d is such, then so is -d.
> However, since the snake lemma> holds in functor categories, the unnaturality of
> d does not make big> problems in concrete situations.
> >
> > ANSWER 2. ker f --> ker g --> ker h   and 
>  cok f --> cok g --> cok h are> the obvious induced morphisms as before, while
> THE "connecting morphism" d :> ker h ---> ker f is the composite of the
> zigzag>
> > ker h ---> C <--- B ---> B' <---A'
> ---> cok f>
> > (where the arrows are considered as internal
> relations). This "canonical> connecting morphism" d works even in the
> non-abelian case of Dominique Bourn> as I learned from my daughter Tamar who
> developed the "relative version".> Note also, that the desire to have such a
> canonical d (in the abelian case)> was a big original reason for developing what we
> call today "calculus of> relations" (at the beginning with great
> participation of Saunders himself).>
> > And... in the "curious case = Exercise
> 6" the "canonical d" does not work!> For, consider the simplest case of the composite
> 0 ---> B ---> 0: the exact> ker-cok sequence will become
> >
> > 0 --> 0 --> 0 --> B --> B --> 0
> --> 0 --> 0,>
> > where B --> B must be an isomorphism, while
> it is easy to check that the> "canonical d" will become the relation
> opposite to the zero morphism B -->> B.
> >
> > A possible conclusion is that the "master
> theorem" should involve some kind> of "d" as an extra
> structure.>
> > To Steve's message: does Enrico really
> generalize the standard snake lemma> and the "curious case"
> simultaneously?>
> > George
> 
> [For admin and other information see: http://www.mta.ca/~cat-dist/ ]
> 
> 
> 
> 



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Question on exact sequence

[Michael Barr]

I have finally worked a way of reducing my observation to an instance of
the snake lemma.  Begin with f: A ---> B and g: C ---> D and form the
diagram (I have made the background dots so that no demented mailer will
collapse the spaces
...................................................
...............(1).................................
...............(f)............(-f 1)...............
0 -----> A ----------> A + B ----------> B -----> 0
.........|...............|...............|.........
.........|...............|...............|.........
.........|...............|(0  1).........|.........
.........|f..............|(gf 0).........|-g.......
.........|...............|...............|.........
.........|...............|...............|.........
.........|.....(1).......|...............|.........
.........v.....(g).......v....(-g 1).....v.........
0 -----> B ----------> B + C ----------> C -----> 0

Exactness of the rows is easily checked, as is the fact that the kernel
of the middle column is ker(gf) + 0. Dually, its cokernel is essentially
cok(gf).  The snake lemma associated to this is the exact sequence I
wrote originally.

Thanks to all who responded.

Michael



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Re: Question on exact sequence

[Marco Grandis]

Dear Clemens,

Thank you for your comments.

 > Are there generalizations to n composable arrows ?
The lemma I was proposing works for any sequence of composable arrows.

It can be rewritten in a notation similar to yours, but based on
three consecutive arrows  f, g, h.
One would use, alternatively, two kind of 'generalised homologies'

    H'(f,g,h) = Ker(hg) / Im(f),

    H"(f,g,h) = Ker(h) / Im(gf),

where, again,  H/K  is meant as in my previous msg.

 > What about generalizations to non-abelian categories ?

The proof I was mentioning works for Puppe-exact categories, and is
obvious (AFTER one has
constructed the universal model of a sequence of consecutive arrows).
This is the natural setting of distributive homological algebra, that
cannot be developed under the assumption of products.
It would be too long to explain this here; please see my three papers
on this subject, in Cahiers 1984-85
(if interested, of course.)

Best wishes

Marco


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Re: Question on exact sequence

[Marco Grandis]

Dear George,

I have not yet received the message you are referring to.

But I can reply to your question:
 >And do you, Marco, have a canonical connecting morphism?

The morphisms I was mentioning in 'my general lemma' are canonically
induced.
The connecting morphism of the Snake Lemma, or of the other Lemma,
are particular instances.

Best wishes

Marco

On 11 Nov 2009, at 17:36, George Janelidze wrote:

> Dear All,
>
> As I understand, my previous message was written after Marco's and
> Steve's
> messages, although I saw them only afterwards. Unfortunately I
> don't have
> time now, but putting these things together would be very interesting.
> Specifically:
>
> 1. Marco vs Steve: Marco's result contains what Michael is mentioning
> (=Exercise VIII.4.6 in Mac Lane's book) plus Snake Lemma, but does it
> contain what Steve describes as Enrico's result?
>
> 2. Marco vs George: And do you, Marco, have a canonical connecting
> morphism?
>
> 3. Steve vs George: Do you, Steve - hence Enrico - have a canonical
> connecting morphism depending on what you call z : A ---> C' in
> your (Steve)
> message?
>
> George
>
>



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Re: Question on exact sequence

[George Janelidze]

Dear All,

As I understand, my previous message was written after Marco's and Steve's
messages, although I saw them only afterwards. Unfortunately I don't have
time now, but putting these things together would be very interesting.
Specifically:

1. Marco vs Steve: Marco's result contains what Michael is mentioning
(=Exercise VIII.4.6 in Mac Lane's book) plus Snake Lemma, but does it
contain what Steve describes as Enrico's result?

2. Marco vs George: And do you, Marco, have a canonical connecting morphism?

3. Steve vs George: Do you, Steve - hence Enrico - have a canonical
connecting morphism depending on what you call z : A ---> C' in your (Steve)
message?

George



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Re: Question on exact sequence

[Clemens.BERGER]

Marco Grandis wrote:
> Dear Michael,
>
> The following lemma extends both results.
>
> We have a sequence of consecutive morphisms indexed on the integers
>
>     ...  ---->   An  ---->  An+1  ---->  An+2  ---->  ...
>
> (if your sequence is finite, you extend by zero's). Call  f(n,m)  the
> composite
> from  An  to  Am  (n < m).
>
> Then, writing  H/K  a subquotient  H/(H intersection K),
> there is an unbounded exact sequence of induced morphisms:
>
>    ...  ---->   Ker f(n,n+2) / Im f(n-1,n)
>        ---->   Ker f(n+1,n+2) / Im f(n-1,n+1)
>        ---->   Ker f(n+1,n+3) / Im f(n,n+1) ---->  ...
>
> where morphisms are alternatively induced by an 'elementary' morphism
> (say An  -->  An+1) or by an identity.
>
> At each step, one increases of one unit the first index in the numerator
> and the second index in the denominator, or the opposite
> (alternatively); after
> two steps, all indices are increased of one unit, and we go along in
> the same way.
>
> -  Your lemma comes out of a sequence  A ----> B ----> C  (extended
> with zeros).
>
> -  Snake's lemma, with your letters, comes out of a sequence of three
> morphisms
> whose total composite is 0
>
>                           A ----> B ----> B' ----> C
> taking into account that  A' = Ker(B' ----> C')  and  C = Cok(A ---->
> B).
>
> I like your lemma (and the Snake's). The form above does not look
> really nice.
> Perhaps someone else will find a nicer solution?
>
> However, if one looks at the universal model of a sequence of
> consecutive morphisms,
> in my third paper on Distributive Homological Algebra, Cahiers 26,
> 1985, p.186,
> the exact sequence above is obvious. (Much in the same way as for the
> sequence of
> the Snake Lemma, p. 188, diagrams (10) and (11).) This is how I found
> it.
>
> Best wishes
>
> Marco
>
Dear Michael and Marco,

   as addition to Marco's answer, I would propose the following
notation: for any pair of composable arrows f:A-->B and g:B-->C denote
by H(g,f) the cokernel of ker(gf)-->ker(g), or (what amounts to the same
in an abelian category),  the kernel of coker(f)-->coker(gf). Thus the
object H(g,f) is precisely the one which allows one to glue together the
short exact sequences 0-->ker(f)-->ker(gf)-->ker(g) and
coker(f)-->coker(gf)-->coker(g)-->0.

   If gf=0 then H(g,f)=ker(g)/im(f) is precisely the homology object at
B, but as Ross already mentioned, this object is well defined for any
composable pair of arrows. Now, if we have three composable arrows
f:A-->B, g:B-->C, h:C-->D, then there is a  4-term exact sequence
0-->H(g,f)-->H(hg,f)-->H(h,gf)-->H(h,g)-->0. The snake lemma can be
derived from the special case hgf=0 of this 4-term exact sequence, where
g corresponds precisely to the middle vertical arrow of Michael's diagram.

It is interesting to observe that a proof (without elements !) of this
4-term exact sequence uses some nice composition properties of
Hilton-exact squares (a common generalization of pullback and pushout
squares). Some more details can be found at pg. 24 of
http://math.unice.fr/~cberger/structure1.

Several questions arise naturally: this 4-term exact sequence looks like
a "cocycle". Are there generalizations to n composable arrows ? What
about generalizations to non-abelian categories ?

All the best,
                         Clemens.


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Re: Question on exact sequence

[Steve Lack]

Dear Michael,

I worked out what exactly Vitale's exactness condition says in this case.
A commutative diagram

   i'    p'
 A'-->B' --> C'
f|   g|     h|
 v    v      v
 A --> B --> C
   i     p

induces a long exact sequence in the way you suggested, when there is
a morphism z:A-->C' for which the induced

0--> A' --> A+B' --> B+C'-->C-->0

is exact.

Steve.


On 10/11/09 2:22 PM, "Steve Lack" <s.lack@uws.edu.au> wrote:

> Dear Michael,
>
> This is the sort of thing that Enrico Vitale has been working on with
> various people for a number of years. I'm sure he'll provide more  precise
> references, but the idea is that you think of the vertical morphisms in your
> diagrams as internal categories:
>
> A               A+A'
> |               | |
> | f    <--->    | |
> v               v v
> A'               A'
>
> (an internal category in Ab amounts to just a morphism - I'll abbreviate
> this to just (A,A').)
>
> and then an exact sequence of internal categories, in a suitably defined
> sense of exactness, induces a long exact sequence involving the pi_0's and
> pi_1's. (pi_0 of an internal category is the cokernel of the corresponding
> morphism, while pi_1 is the kernel.)
>
> In your diagram (the "curious" one), the morphism 1:C-->C is saying that
> the corresponding internal functor (A,C)-->(B,C) is (not just essentially
> surjective but) the identity on objects. This is the relevant notion of
> "epi". The morphism 1:A-->A says that the corresponding internal functor
> (A,B)-->(A,C) is (among other things) faithful. This is the relevant notion
> of "mono". There is also an exactness condition at (A,C).
>
> Vitale, with various coauthors, has studied such exactness conditions at
> varying levels of generality, but the simplest of these is just internal
> categories in Ab.
>
> Steve.
>
> On 10/11/09 9:57 AM, "Michael Barr" <barr@math.mcgill.ca> wrote:
>
>> I have recently discovered a curious fact about abelian categories.
>> First, let me briefly describe the well-known snake lemma.  If we have a
>> commutative diagram with exact rows (there are variations without the 0
>> at the left end of the top and without the 0 at the right end of the
>> bottom, but here is the strongest form)
>>
>>       0 ---> A ----> B ----> C ----> 0
>>              |       |       |
>>              |       |       |
>>              |f      |g      |h
>>              |       |       |
>>              v       v       v
>>       0 ---> A' ---> B' ---> C' ---> 0
>>
>> then there is an exact sequence
>>   0 --> ker f --> ker g --> ker h --> cok f --> cok g --> cok h --> 0
>>
>> The curious discovery is that you have any pair of composable maps f: A
>> --> B and h: B --> C and you form the diagram (with g = hf)
>>                  1       f
>>              A ----> A ----> B
>>              |       |       |
>>              |       |       |
>>              |f      |g      |h
>>              |       |       |
>>              v       v       v
>>              B ----> C ----> C
>>                  h       1
>> you get the same exact sequence.  So I would imagine that there must be
>> a "master theorem" of which these are two cases.  Does anyone know what
>> it says?  The connecting map here is just the inclusion of ker h into B
>> followed by the projection on cok f.
>>
>> Michael
>>


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Re: Question on exact sequence

[Ross Street]

Dear Michael

Murray Adelman used to use a generalized homology where the complex
need not have consecutive composites zero.

For any pair of maps u : A-->B and v:B-->C in an abelian category,
there is a homology at B: take the image of the canonical map ker(v)--
 >coker(u).

Notice that the rows of the diagram in the snake lemma are exact and
so have zero homology.
The rows of your diagram (below) also have zero homology in the middle.

Ross

On 10/11/2009, at 9:57 AM, Michael Barr wrote:
>
>              1       f
>        A ----> A ----> B
>            |       |       |
>            |       |       |
>            |f      |g      |h
>            |       |       |
>            v       v       v
>         B ----> C ----> C
>                h       1


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Re: Question on exact sequence

[Michael Barr]

That is really fascinating.  I assume that when you wrote A --> B --> B'
--> C, that last should have been C'.  Here is how it arose.  Call a map A
--> B cotorsion if its cokernel is torsion (for a generalized notion of
torsion).  Assuming that torsion objects are closed under extension (and
that a quotient of torsion is torsion) this allows a proof that a
composite of cotorsion maps is cotorsion.

Anyway, thanks.

Michael

On Tue, 10 Nov 2009, Marco Grandis wrote:

> Dear Michael,
>
> The following lemma extends both results.
>
> We have a sequence of consecutive morphisms indexed on the integers
>
> 	...  ---->   An  ---->  An+1  ---->  An+2  ---->  ...
>
> (if your sequence is finite, you extend by zero's). Call  f(n,m)  the
> composite
> from  An  to  Am  (n < m).
>
> Then, writing  H/K  a subquotient  H/(H intersection K),
> there is an unbounded exact sequence of induced morphisms:
>
>   ...  ---->   Ker f(n,n+2) / Im f(n-1,n)
>  	 ---->   Ker f(n+1,n+2) / Im f(n-1,n+1)
>  	 ---->   Ker f(n+1,n+3) / Im f(n,n+1) ---->  ...
>
> where morphisms are alternatively induced by an 'elementary' morphism
> (say An  -->  An+1) or by an identity.
>
> At each step, one increases of one unit the first index in the numerator
> and the second index in the denominator, or the opposite (alternatively);
> after
> two steps, all indices are increased of one unit, and we go along in the same
> way.
>
> -  Your lemma comes out of a sequence  A ----> B ----> C  (extended with
> zeros).
>
> -  Snake's lemma, with your letters, comes out of a sequence of three
> morphisms
> whose total composite is 0
>
>    					 A ----> B ----> B' ----> C
> taking into account that  A' = Ker(B' ----> C')  and  C = Cok(A ----> B).
>
> I like your lemma (and the Snake's). The form above does not look really
> nice.
> Perhaps someone else will find a nicer solution?
>
> However, if one looks at the universal model of a sequence of consecutive
> morphisms,
> in my third paper on Distributive Homological Algebra, Cahiers 26, 1985,
> p.186,
> the exact sequence above is obvious. (Much in the same way as for the
> sequence of
> the Snake Lemma, p. 188, diagrams (10) and (11).) This is how I found it.
>
> Best wishes
>
> Marco
>
>


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Re: Question on exact sequence

[Marco Grandis]

Dear Michael,

The following lemma extends both results.

We have a sequence of consecutive morphisms indexed on the integers

	...  ---->   An  ---->  An+1  ---->  An+2  ---->  ...

(if your sequence is finite, you extend by zero's). Call  f(n,m)  the
composite
from  An  to  Am  (n < m).

Then, writing  H/K  a subquotient  H/(H intersection K),
there is an unbounded exact sequence of induced morphisms:

    ...  ---->   Ker f(n,n+2) / Im f(n-1,n)
    	---->   Ker f(n+1,n+2) / Im f(n-1,n+1)
    	---->   Ker f(n+1,n+3) / Im f(n,n+1) ---->  ...

where morphisms are alternatively induced by an 'elementary' morphism
(say An  -->  An+1) or by an identity.

At each step, one increases of one unit the first index in the numerator
and the second index in the denominator, or the opposite
(alternatively); after
two steps, all indices are increased of one unit, and we go along in
the same way.

-  Your lemma comes out of a sequence  A ----> B ----> C  (extended
with zeros).

-  Snake's lemma, with your letters, comes out of a sequence of three
morphisms
whose total composite is 0

      					 A ----> B ----> B' ----> C
taking into account that  A' = Ker(B' ----> C')  and  C = Cok(A ---->
B).

I like your lemma (and the Snake's). The form above does not look
really nice.
Perhaps someone else will find a nicer solution?

However, if one looks at the universal model of a sequence of
consecutive morphisms,
in my third paper on Distributive Homological Algebra, Cahiers 26,
1985, p.186,
the exact sequence above is obvious. (Much in the same way as for the
sequence of
the Snake Lemma, p. 188, diagrams (10) and (11).) This is how I found
it.

Best wishes

Marco




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Re: Question on exact sequence

[Steve Lack]

Dear Michael,

This is the sort of thing that Enrico Vitale has been working on with
various people for a number of years. I'm sure he'll provide more  precise
references, but the idea is that you think of the vertical morphisms in your
diagrams as internal categories:

A               A+A'
|               | |
| f    <--->    | |
v               v v
A'               A'

(an internal category in Ab amounts to just a morphism - I'll abbreviate
this to just (A,A').)

and then an exact sequence of internal categories, in a suitably defined
sense of exactness, induces a long exact sequence involving the pi_0's and
pi_1's. (pi_0 of an internal category is the cokernel of the corresponding
morphism, while pi_1 is the kernel.)

In your diagram (the "curious" one), the morphism 1:C-->C is saying that
the corresponding internal functor (A,C)-->(B,C) is (not just essentially
surjective but) the identity on objects. This is the relevant notion of
"epi". The morphism 1:A-->A says that the corresponding internal functor
(A,B)-->(A,C) is (among other things) faithful. This is the relevant notion
of "mono". There is also an exactness condition at (A,C).

Vitale, with various coauthors, has studied such exactness conditions at
varying levels of generality, but the simplest of these is just internal
categories in Ab.

Steve.

On 10/11/09 9:57 AM, "Michael Barr" <barr@math.mcgill.ca> wrote:

> I have recently discovered a curious fact about abelian categories.
> First, let me briefly describe the well-known snake lemma.  If we have a
> commutative diagram with exact rows (there are variations without the 0
> at the left end of the top and without the 0 at the right end of the
> bottom, but here is the strongest form)
>
>       0 ---> A ----> B ----> C ----> 0
>              |       |       |
>              |       |       |
>              |f      |g      |h
>              |       |       |
>              v       v       v
>       0 ---> A' ---> B' ---> C' ---> 0
>
> then there is an exact sequence
>   0 --> ker f --> ker g --> ker h --> cok f --> cok g --> cok h --> 0
>
> The curious discovery is that you have any pair of composable maps f: A
> --> B and h: B --> C and you form the diagram (with g = hf)
>                  1       f
>              A ----> A ----> B
>              |       |       |
>              |       |       |
>              |f      |g      |h
>              |       |       |
>              v       v       v
>              B ----> C ----> C
>                  h       1
> you get the same exact sequence.  So I would imagine that there must be
> a "master theorem" of which these are two cases.  Does anyone know what
> it says?  The connecting map here is just the inclusion of ker h into B
> followed by the projection on cok f.
>
> Michael
>
>
>
> [For admin and other information see: http://www.mta.ca/~cat-dist/ ]



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Question on exact sequence

[Michael Barr]

I have recently discovered a curious fact about abelian categories.
First, let me briefly describe the well-known snake lemma.  If we have a
commutative diagram with exact rows (there are variations without the 0
at the left end of the top and without the 0 at the right end of the
bottom, but here is the strongest form)

      0 ---> A ----> B ----> C ----> 0
             |       |       |
             |       |       |
             |f      |g      |h
             |       |       |
             v       v       v
      0 ---> A' ---> B' ---> C' ---> 0

then there is an exact sequence
  0 --> ker f --> ker g --> ker h --> cok f --> cok g --> cok h --> 0

The curious discovery is that you have any pair of composable maps f: A
--> B and h: B --> C and you form the diagram (with g = hf)
                 1       f
             A ----> A ----> B
             |       |       |
             |       |       |
             |f      |g      |h
             |       |       |
             v       v       v
             B ----> C ----> C
                 h       1
you get the same exact sequence.  So I would imagine that there must be
a "master theorem" of which these are two cases.  Does anyone know what
it says?  The connecting map here is just the inclusion of ker h into B
followed by the projection on cok f.

Michael



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