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* Two questions
@ 2012-06-21 13:34 Michael Barr
  2012-06-21 17:07 ` Ronnie Brown
                   ` (3 more replies)
  0 siblings, 4 replies; 6+ messages in thread
From: Michael Barr @ 2012-06-21 13:34 UTC (permalink / raw)
  To: Categories list; +Cc: Bob Raphael

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.


[For admin and other information see: http://www.mta.ca/~cat-dist/ ]


^ permalink raw reply	[flat|nested] 6+ messages in thread

* Re:  Two questions
  2012-06-21 13:34 Two questions Michael Barr
@ 2012-06-21 17:07 ` Ronnie Brown
  2012-06-21 23:32 ` George Janelidze
                   ` (2 subsequent siblings)
  3 siblings, 0 replies; 6+ messages in thread
From: Ronnie Brown @ 2012-06-21 17:07 UTC (permalink / raw)
  To: Michael Barr; +Cc: Categories list, Bob Raphael

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
>


[For admin and other information see: http://www.mta.ca/~cat-dist/ ]


^ permalink raw reply	[flat|nested] 6+ messages in thread

* Re: Two questions
  2012-06-21 13:34 Two questions Michael Barr
  2012-06-21 17:07 ` Ronnie Brown
@ 2012-06-21 23:32 ` George Janelidze
       [not found] ` <E1ShqME-0001Sf-JM@mlist.mta.ca>
  2012-06-22  7:19 ` regular monos of groups Paul Taylor
  3 siblings, 0 replies; 6+ messages in thread
From: George Janelidze @ 2012-06-21 23:32 UTC (permalink / raw)
  To: Michael Barr, Categories list

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.



[For admin and other information see: http://www.mta.ca/~cat-dist/ ]


^ permalink raw reply	[flat|nested] 6+ messages in thread

* Re: Two questions
       [not found] ` <E1ShqME-0001Sf-JM@mlist.mta.ca>
@ 2012-06-22  0:25   ` Peter LeFanu Lumsdaine
  0 siblings, 0 replies; 6+ messages in thread
From: Peter LeFanu Lumsdaine @ 2012-06-22  0:25 UTC (permalink / raw)
  To: Categories list; +Cc: Andrej Bauer

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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^ permalink raw reply	[flat|nested] 6+ messages in thread

* regular monos of groups
  2012-06-21 13:34 Two questions Michael Barr
                   ` (2 preceding siblings ...)
       [not found] ` <E1ShqME-0001Sf-JM@mlist.mta.ca>
@ 2012-06-22  7:19 ` Paul Taylor
  2012-07-15 21:27   ` Andrej Bauer
  3 siblings, 1 reply; 6+ messages in thread
From: Paul Taylor @ 2012-06-22  7:19 UTC (permalink / raw)
  To: categories

Mike Barr said

> Somewhere I have seen a proof that all monics in the category of groups
> are regular.

Somebody asked this question a while back on MathOverflow
and I gave the following proof, which is both constructive and
more explicit than the replies that have been given here.

http://mathoverflow.net/questions/41208/are-all-group-monomorphisms-regular-constructively/46054#46054

Paul Taylor



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^ permalink raw reply	[flat|nested] 6+ messages in thread

* Re: regular monos of groups
  2012-06-22  7:19 ` regular monos of groups Paul Taylor
@ 2012-07-15 21:27   ` Andrej Bauer
  0 siblings, 0 replies; 6+ messages in thread
From: Andrej Bauer @ 2012-07-15 21:27 UTC (permalink / raw)
  To: categories

My blog post has a hole. I am not convinced by Paul's answer on
Mathoverflow (perhaps I just have to work out the details which Paul
left out). So this leaves the question: is every mono in the category
of groups regular, constructively speaking?

With kind regards,

Andrej

On Fri, Jun 22, 2012 at 9:19 AM, Paul Taylor <pt12@paultaylor.eu> wrote:
> Mike Barr said
>
>> Somewhere I have seen a proof that all monics in the category of groups
>> are regular.
>
> Somebody asked this question a while back on MathOverflow
> and I gave the following proof, which is both constructive and
> more explicit than the replies that have been given here.
>
> http://mathoverflow.net/questions/41208/are-all-group-monomorphisms-regular-constructively/46054#46054
>
> Paul Taylor
>

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^ permalink raw reply	[flat|nested] 6+ messages in thread

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-- links below jump to the message on this page --
2012-06-21 13:34 Two questions Michael Barr
2012-06-21 17:07 ` Ronnie Brown
2012-06-21 23:32 ` George Janelidze
     [not found] ` <E1ShqME-0001Sf-JM@mlist.mta.ca>
2012-06-22  0:25   ` Peter LeFanu Lumsdaine
2012-06-22  7:19 ` regular monos of groups Paul Taylor
2012-07-15 21:27   ` Andrej Bauer

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