Re: Two questions

Re: Two questions

[Fred E.J. Linton]

On Fri, 22 Jun 2012 01:32:10 +0200, George Janelidze wrote, inter alia:

> 6. Concerning your second question: What you saw (with a special argument
> for 2) is on the same Page 21 of the above-mentioned Eilenberg--Moore
paper.
> The argument was used to prove that every epimorphism of groups is
> surjective with no mention of regular monomorphisms, but in fact they prove
> that, for every homomorphism
> 
> j : H --> G, there exist two homomorphisms
> 
> k, l : G --> P with k(g) = L(g) only when g is in j(H).
> 
> This also appears as Exercise 5 of Section 5 of Chapter I in Mac Lane's
> "Categories for the Working Mathematician", with very precise hints.

What the origins of the argument Mac Lane outlines here are, I don't know. 
I do seem to recall that I first saw more or less such an argument in a 
graduate course Sammy gave at Columbia during the "golden era" 1958-1963,  
and that Sammy himself, probably at a Bowdoin NSF summer semester on 
categories, revealed his deft trick by which to convert the argument for 
the "index greater than 2" case to a uniform argument ignoring the 
subgroup's index.

Let me record that argument here.

Given are a group G, a subgroup H of G, and an element a in G \ H.

P is to be the group of permutations of the underlying set of the left 
G-set got by forming the coproduct 

G/H + 1

of the principal left G-set G/H of left cosets xH of H in G (x ∈ G) 
with a (trivial) terminal left G-set 1 = {*} (assume * ∉ G/H):

P = perm(|G/H| ∪ {*}) = (|G/H| ∪ {*})! .

One permutation in particular is to be singled out for attention: the 
transposition t ∈ P interchanging * with the coset H itself.

Let me use r: G --> P for the result of composing the regular 
representation of G by left action (g, xH) |-> (gx)H on the cosets of H 
with the obvious injection of (|G/H|)! into (|G/H| ∪ {*})! -- thus:

[r(g)](xH) = (gx)H  ,
[r(g)](*) = * .

And let me write s: G --> P for the result of conjugating by t the various 
assorted values of r -- thus:

[s(g)](xH) = t([r(g)](t(xH))) ,
[s(g)](*) = t([r(g)](t(*))) = t([r(g)](H)) = t(gH) .

The end-game strategy is now this:

(i) for h ∈ H: r(h) = s(h) ; yet
(ii) for g = a, [r(a)](*) = * but [s(a)](*) = aH (whence r ≠  s).

Details: Bear in mind that if, for x in G and h in H, we have (hx)H = H, 
we must have hx in H, whence also x in H so that xH = H. Then for (i):

(a) if xH ≠ H -- [s(h)](xH) = t([r(h)](xH)) = t((hx)H) = (hx)H = [r(h)](xH)
;
(b) if xH = H -- [s(h)](H) = t([r(h)](*)) = t(*) = H = [r(h)](H) ;
(c) and at * -- [s(h)](*) = t([r(h)](H)) = t(H) = * = [r(h)](*) .

And bear in mind also that aH ≠ H because a ∉ H; thus:

for (ii) -- [s(a)](*) = t([r(a)](H)) = t(aH) = aH ≠ * = [r(a)](*) .

By the way, if the subgroup H had index 3 or more in G, one need not 
require recourse to any external element * as above -- one could let any 
coset other than H and aH (when there are such) play the role of *, and, 
with but a few additional wrinkles (unless I am mistaken (which is always  
possible :-) )), I believe that is essentially how the proof George cites  
from Mac Lane's exercise works.

Cheers, -- Fred

[NB: I'm writing ∈ for an element symbol, ∉ for a crossed-out 
element symbol, ∪ for a union symbol, and ≠ for a crossed-out equal 
sign. With luck these HTML glyph constructs will simply display as the 
glyphs they're meant to represent; and if not, they're no more painful to  
decipher than their TeX counterparts, which HTML can't display as glyphs.]




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Re: Two_questions

[Joyal, André]

Thank you Fred for your nice description of Sammy's proof
that every subgroup is an equaliser (is a regular subgroup).
I would like to suggest a small variant which might be simpler.
Sammy's proof can be decomposed in three steps:

(1) the pullback of a regular subgroup is regular;

(2) Every subgroup of a group G is the stabiliser of a point in some G-set;

(3) If E! is the full permutation group of a set E,
then the stabiliser S(p) of a point p in E is a regular subgroup of E!.

The key argument lies in step (3). Let E' be the set obtained from E by
adding copy p' of p. There are two embeddings u,u':E-->E',
the first u is the inclusion of E in E', and the second u' is defined 
by putting u'(p)=p' and u'(x)=x for x different than p.
This leads to a pair of homomorphisms h,h':E!-->E'! the equaliser of which
is the stabiliser S(p) of p in E!.

This last argument seems to differ from the argument you have presented.
 
Am I making an error?

Best, 
André





-------- Message d'origine--------
De: Fred E.J. Linton [mailto:fejlinton@usa.net]
Date: ven. 22/06/2012 23:09
À: categories
Objet : categories: Re: Two questions
 

On Fri, 22 Jun 2012 01:32:10 +0200, George Janelidze wrote, inter alia:

> 6. Concerning your second question: What you saw (with a special argument
> for 2) is on the same Page 21 of the above-mentioned Eilenberg--Moore
paper.
> The argument was used to prove that every epimorphism of groups is
> surjective with no mention of regular monomorphisms, but in fact they prove
> that, for every homomorphism
> 
> j : H --> G, there exist two homomorphisms
> 
> k, l : G --> P with k(g) = L(g) only when g is in j(H).
> 
> This also appears as Exercise 5 of Section 5 of Chapter I in Mac Lane's
> "Categories for the Working Mathematician", with very precise hints.

What the origins of the argument Mac Lane outlines here are, I don't know. 
I do seem to recall that I first saw more or less such an argument in a 
graduate course Sammy gave at Columbia during the "golden era" 1958-1963,  
and that Sammy himself, probably at a Bowdoin NSF summer semester on 
categories, revealed his deft trick by which to convert the argument for 
the "index greater than 2" case to a uniform argument ignoring the 
subgroup's index.

Let me record that argument here.

Given are a group G, a subgroup H of G, and an element a in G \ H.

P is to be the group of permutations of the underlying set of the left 
G-set got by forming the coproduct 

G/H + 1

of the principal left G-set G/H of left cosets xH of H in G (x ? G) 
with a (trivial) terminal left G-set 1 = {*} (assume * ? G/H):

P = perm(|G/H| ? {*}) = (|G/H| ? {*})! .

One permutation in particular is to be singled out for attention: the 
transposition t ? P interchanging * with the coset H itself.

Let me use r: G --> P for the result of composing the regular 
representation of G by left action (g, xH) |-> (gx)H on the cosets of H 
with the obvious injection of (|G/H|)! into (|G/H| ? {*})! -- thus:

[r(g)](xH) = (gx)H  ,
[r(g)](*) = * .

And let me write s: G --> P for the result of conjugating by t the various 
assorted values of r -- thus:

[s(g)](xH) = t([r(g)](t(xH))) ,
[s(g)](*) = t([r(g)](t(*))) = t([r(g)](H)) = t(gH) .

The end-game strategy is now this:

(i) for h ? H: r(h) = s(h) ; yet
(ii) for g = a, [r(a)](*) = * but [s(a)](*) = aH (whence r ?  s).

Details: Bear in mind that if, for x in G and h in H, we have (hx)H = H, 
we must have hx in H, whence also x in H so that xH = H. Then for (i):

(a) if xH ? H -- [s(h)](xH) = t([r(h)](xH)) = t((hx)H) = (hx)H = [r(h)](xH)
;
(b) if xH = H -- [s(h)](H) = t([r(h)](*)) = t(*) = H = [r(h)](H) ;
(c) and at * -- [s(h)](*) = t([r(h)](H)) = t(H) = * = [r(h)](*) .

And bear in mind also that aH ? H because a ? H; thus:

for (ii) -- [s(a)](*) = t([r(a)](H)) = t(aH) = aH ? * = [r(a)](*) .

By the way, if the subgroup H had index 3 or more in G, one need not 
require recourse to any external element * as above -- one could let any 
coset other than H and aH (when there are such) play the role of *, and, 
with but a few additional wrinkles (unless I am mistaken (which is always  
possible :-) )), I believe that is essentially how the proof George cites  
from Mac Lane's exercise works.

Cheers, -- Fred

[NB: I'm writing ? for an element symbol, ? for a crossed-out 
element symbol, ? for a union symbol, and ? for a crossed-out equal 
sign. With luck these HTML glyph constructs will simply display as the 
glyphs they're meant to represent; and if not, they're no more painful to  
decipher than their TeX counterparts, which HTML can't display as glyphs.]




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Re: Two questions

[Joyal, André]

I thank you again Fred for your nice explanations. 

I have a question for you and the categories list.

The category of Lie algebras over a field
is very similar to the category of groups.
I wonder if every monomorphism of Lie algebras is regular?

Best wishes,
andré
 


-------- Message d'origine--------
De: Fred E.J. Linton [mailto:fejlinton@usa.net]
Date: dim. 24/06/2012 21:34
À: Joyal, André
Cc: categories
Objet : Re: categories: Re: Two_questions
 
Correction/clarification to previous note: in what follows, the sloppy phrase

> ... the resulting u' is the same as the result of conjugating all the values
of u by the transposition ...

should be replaced by the more accurate expanded version

> ... the h' (as below) resulting from u' is the same as what Sammy gets by
conjugating all the values of the h that arises from u by the transposition
... 

Sorry for the sloppy writing. 

Cheers, -- Fred


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Re: Two_questions

[Fred E.J. Linton]

Correction/clarification to previous note: in what follows, the sloppy phrase

> ... the resulting u' is the same as the result of conjugating all the values
of u by the transposition ...

should be replaced by the more accurate expanded version

> ... the h' (as below) resulting from u' is the same as what Sammy gets by
conjugating all the values of the h that arises from u by the transposition
... 

Sorry for the sloppy writing. 

Cheers, -- Fred

------ Original Message ------
Received: Sun, 24 Jun 2012 08:09:45 PM EDT
From: "Fred E.J. Linton" <fejlinton@usa.net>
To: André"  <joyal.andre@uqam.ca>Cc: "categories" <categories@mta.ca>
Subject: categories: Re: Two_questions

> Salut, André,
> 
> I'm ashamed how long it took me to come to this realization, but
> you were absolutely correct in your surmise that, when one lets ...
> 
>> ... E' be the set obtained from E by
>> adding copy p' of p. There are two embeddings u,u':E-->E',
>> the first u is the inclusion of E in E', and the second u' is defined  
>> by putting u'(p)=p' and u'(x)=x for x different than p.
> 
> ... the resulting u' is the same as the result of conjugating all the
values
> of u by the transposition t that exchanges p with p'. Thus, the ...
> 
>> ... pair of homomorphisms h,h':E!-->E'! the equaliser of which
>> is the stabiliser S(p) of p in E!.
> 
> ... that arises is exactly the same as the pair Sammy's argument 
> would adduce, and the only respect in which ...
> 
>> This last argument seems to differ from the argument you have presented.
> 
> is that for Sammy it was enough to observe that h and h' differ SOMEWHERE  
> when E is more than just {p}, while what you observe is rather more, namely,

> that h and h' actually differ EVERYWHERE other than on S(p), i.e., that  the

> ONLY place where h and h' do NOT differ is S(p) :-) .
> 
>> Am I making an error?
> 
> Only in thinking that seeming difference makes any real difference :-) .
> 
> Cheers,  -- Fred



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Re: Two_questions

[Fred E.J. Linton]

Salut, André,

I'm ashamed how long it took me to come to this realization, but
you were absolutely correct in your surmise that, when one lets ...

> ... E' be the set obtained from E by
> adding copy p' of p. There are two embeddings u,u':E-->E',
> the first u is the inclusion of E in E', and the second u' is defined 
> by putting u'(p)=p' and u'(x)=x for x different than p.

... the resulting u' is the same as the result of conjugating all the values
of u by the transposition t that exchanges p with p'. Thus, the ...

> ... pair of homomorphisms h,h':E!-->E'! the equaliser of which
> is the stabiliser S(p) of p in E!.

... that arises is exactly the same as the pair Sammy's argument 
would adduce, and the only respect in which ...

> This last argument seems to differ from the argument you have presented.

is that for Sammy it was enough to observe that h and h' differ SOMEWHERE  
when E is more than just {p}, while what you observe is rather more, namely, 
that h and h' actually differ EVERYWHERE other than on S(p), i.e., that the 
ONLY place where h and h' do NOT differ is S(p) :-) .

> Am I making an error?

Only in thinking that seeming difference makes any real difference :-) .

I hope this clarifies matters, and I think it's worth showing categories@mta.


Cheers,  -- Fred




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Re: Two questions

[Peter LeFanu Lumsdaine]

On Thu, Jun 21, 2012 at 2:54 PM, Michael Shulman <mshulman@ucsd.edu> wrote:
> On Thu, Jun 21, 2012 at 6:34 AM, Michael Barr <barr@math.mcgill.ca> wrote:
>> Somewhere I have seen a proof that all monics in the category of groups
>> are regular.  I think it was in a paper by Eilenberg and ??? and it  needed
>> a special argument if there were elements of order 2.  Can someone help me
>> find this?
>
> I don't know the original reference, but this is exercise 7H in The
> Joy of Cats, which contains a substantial hint for a proof (not
> involving a special case for order-2 elements, so maybe it is a
> different proof).

There was some discussion of this a couple of years ago on Andrej
Bauer’s blog and mathoverflow:

http://math.andrej.com/2010/11/10/subgroups-are-equalizers-constructively/
http://mathoverflow.net/questions/41208/are-all-group-monomorphisms-regular-constructively/

(the chronology/causality between the two is a little non-obvious: the
MO question predates the blog post, but the eventual solution at MO
follows it)

The discussion there is on proving this constructively, but as often,
the first step is looking at the classical proof with a magnifying
glass.

–p.


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Re: Two questions

[George Janelidze]

text/plain;format=flowed;charset="iso-8859-1";
Content-Transfer-Encoding: 7bit
Sender: categories@mta.ca
Precedence: bulk
Reply-To: "George Janelidze" <janelg@telkomsa.net>

Dear Michael,

My answer consists of several parts:


1. Indeed, Mac Lane gives no hints, and indeed, many people claim that
Baer's result is unpublished. But look at Theorem 2 in

[R. Baer, Absolute retracts in group theory, Bulletin AMS 52, 1946, 501-506]

- it says: "The identity is the only group which is a retract of every
containing group.",

and its proof consists of one page. This fact, to which I refer below as not
having non-trivial absolute retracts, of course immediately implies that
there are no non-trivial injective objects.


2. Next, let us look at the last complete sentence on Page 21 and its
explanation on the next page in

[S. Eilenberg and J. C. Moore, Foundation of relative homological algebra,
Memoirs AMS 55, 1965].

That sentence says "Not only is the category G [of groups] not injectively
perfect, but we shall show that every injective object in G is trivial."
There no reference to Bear's 19 year old (then) paper, but there is a
reference on Baer's 31 year old paper (hence from 1934), where Baer
constructs large simple groups. And of course knowing that there are
arbitrary large simple groups with infinite cyclic subgroups makes the
result trivial.


3. Next a surprise: look at Page 78 of

[Maria S. Voloshina, On the holomorph of a discrete group, PhD Thesis,
University of Rochester, 2003, arXiv:math/0302v2 [math.GR] 14 Jan 2004].

Surely not knowing about Baer's work, Voloshina says in the Abstract there:
"In chapter 6 we give a short proof of the well-known fact due to S.
Eilenberg and J. C. Moore that the only injective object in the category of
groups is the trivial group." Her argument is something I have never seen
before, and to me it is the best proof - so let me rephrase it here keeping
the same notation but writing x* for the inverse of x:

Let G be an injective group, x an element in G, F[a,b] and F[c,d] free
groups on two-element sets {a,b} and {c,d} respectively, and i, f, g
homomorphisms defined as follows:

i : F[a,b] --> F[c,d] has i(a) = c and i(b) = dcd*

f : F[a,b] --> G has f(a) = 1 and f(b) = x;

g : F[c,d] --> G is any homomorphism with gi = f, which exists since G is
injective.

We have g(c) = gi(a) = f(a) = 1, and so

x = f(b) = gi(b) = g(dcd*) = g(d)g(c)(g(d))* = g(d)(g(d))* = 1.

Is not it wonderful, and is it possible that nobody have noticed it before
2003?!


4. Actually there is an important additional reason why I like Voloshina's
argument. Let us look at her F[c,d] as Z+Z, that is, the coproduct of the
additive group Z of integers with itself.

Let p : Z+Z --> Z be the homomorphism induced by the identity homomorphism
and the trivial homomorphism, and let

k : ZbZ --> Z+Z be the kernel of p ("b" stands for "bemol").

This ZbZ occurs from a monad (Zb(-),e,m) on the category of groups, whose
algebras are exactly Z-groups, and this is a special case of a categorical
story presented in

[D. Bourn and G. J., Protomodularity, descent, and semidirect products, TAC
4, 2, 1998, 37-46],

with more categorical links in my papers with Francis Borceux and Max Kelly,
and further developments by various authors. According to that story, k is
the "the best example of a bad normal monomorphism" in a sense, and so it is
a natural thing to use it "against injectivity".

Back to Voloshina's notation, ZbZ is BETWEEN F[a,b] and F[c,d]. Indeed it is
the subgroup in F[c,d] freely generated by the set

S = {xcx* | x is an integer powers of d},

and I say "between" since S contains i(a) = c = 1c1* and i(b) = dcd*; that
is, there is a homomorphism

j : F[a,b] --> ZbZ with kj = i.

Since ZbZ is free on S, Voloshina's homomorphism f extends to a homomorphism

f' : ZbZ --> G, and so

k : ZbZ --> Z+Z, which belongs to the categorical story, can be used instead
of i in Voloshina's proof.


5. Another interesting thing is that k above is a normal monomorphism, and
so my simple modification of Voloshina's argument proves that there are no
non-trivial groups that are injective with respect to normal monomorphisms.
And it is interesting because here we see the big difference between
injective objects and absolute retracts in the category of groups. Indeed,
going back to Baer's paper of 1946, we see

"THEOREM 1. The group G is complete if, and only if, it meets the following
requirement:
(*) If G is a normal subgroup of the group E, then G is a retract of E."

and then inside Remark on Page 503:

"...This shows in particular that every group is a subgroup of a complete
group..."

Well, I must confess I have not checked Baer's proofs, but I hope they are
correct.


6. Concerning your second question: What you saw (with a special argument
for 2) is on the same Page 21 of the above-mentioned Eilenberg--Moore paper.
The argument was used to prove that every epimorphism of groups is
surjective with no mention of regular monomorphisms, but in fact they prove
that, for every homomorphism

j : H --> G, there exist two homomorphisms

k, l : G --> P with k(g) = L(g) only when g is in j(H).

This also appears as Exercise 5 of Section 5 of Chapter I in Mac Lane's
"Categories for the Working Mathematician", with very precise hints.
However, a group theorist would probably prefer to use known facts pushouts
of group monomorphisms.

Regards, George

P.S. While I was writing this, several other answers came. But instead of
changing my message, let me only mention that it is good to have more
references, and that Maria Nogin should be the same person as Maria
Voloshina (Probably one of the two surnames is her husband's surname).




--------------------------------------------------
From: "Michael Barr" <barr@math.mcgill.ca>
Sent: Thursday, June 21, 2012 3:34 PM
To: "Categories list" <categories@mta.ca>
Cc: "Bob Raphael" <raphael@alcor.concordia.ca>
Subject: categories: Two questions

> Googling around, I have come on several claims that there are no
> non-trivial injectives in the category of groups (e.g., Mac Lane in the
> 1950 Duality for groups paper credits Baer with an elegant proof, but
> gives no hint of what it might be and Baer's earlier paper on injectives
> doesn't mention it).  I have not come on any proof of this, however.
>
> Somewhere I have seen a proof that all monics in the category of groups
> are regular.  I think it was in a paper by Eilenberg and ??? and it needed
> a special argument if there were elements of order 2.  Can someone help me
> find this?
>
> Michael
>
> --
> The United States has the best congress money can buy.



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Re: Two questions

[Ronnie Brown]

With regard to the second question, the problem seems to be for epic.  I
found the solution somewhere and put it as exercise 8 of section 6.1 of
my Topology and Groupoids book (all editions). Here is the outline
argument (I have not checked it lately!):

Prove that in the category $\grp$ of groups, a morphism $f : G \to
H$ is monic if and only if it is injective; less trivially, $f$ is
epic if and only if $f$ is surjective.  [Suppose $f$ is not
surjective and let $K = \Im f$.  If the set of cosets $H/K$ has
two elements, then $K$ is normal in $H$ and it is easy to prove
$f$ is not epic. Otherwise there is a permutation $\gamma$ of $H/K$
whose only fixed point is $K$.  Let $\pi : H \to H/K$ be the
projection and choose a function $\theta : H/K \to H$ such that
$\pi \theta = 1$.  Let $\tau : H \to K$ be such that $x = (\tau
x)(\theta\pi x)$ for all $x$ in $H$ and define $\lambda : H \to H$
by $x \mapsto (\tau x) (\theta\gamma\pi x)$.  The morphisms
$\alpha, \beta$ of $H$ into the group $P$ of all permutations of
$H$, defined by $\alpha(h)(x) = hx$, $\beta(h) =
\lambda^{-1}\alpha(h)\lambda$ satisfy $\alpha h = \beta h$ if and
only if $h \in K$.  Hence $\alpha f = \beta f$].

Ronnie

On 21/06/2012 14:34, Michael Barr wrote:
> Googling around, I have come on several claims that there are no
> non-trivial injectives in the category of groups (e.g., Mac Lane in the
> 1950 Duality for groups paper credits Baer with an elegant proof, but
> gives no hint of what it might be and Baer's earlier paper on injectives
> doesn't mention it).  I have not come on any proof of this, however.
>
> Somewhere I have seen a proof that all monics in the category of groups
> are regular.  I think it was in a paper by Eilenberg and ??? and it
> needed
> a special argument if there were elements of order 2.  Can someone
> help me
> find this?
>
> Michael
>


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Two questions

[Michael Barr]

Googling around, I have come on several claims that there are no
non-trivial injectives in the category of groups (e.g., Mac Lane in the
1950 Duality for groups paper credits Baer with an elegant proof, but
gives no hint of what it might be and Baer's earlier paper on injectives
doesn't mention it).  I have not come on any proof of this, however.

Somewhere I have seen a proof that all monics in the category of groups
are regular.  I think it was in a paper by Eilenberg and ??? and it needed
a special argument if there were elements of order 2.  Can someone help me
find this?

Michael

-- 
The United States has the best congress money can buy.


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