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/ ]
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/ ]
On Thu, Jun 21, 2012 at 9:34 AM, Michael Barr <barr@math.mcgill.ca> 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.
An easy proof is given in:
A short proof of Eilenberg and Moores theorem Maria Nogin CENTRAL EUROPEAN JOURNAL OF MATHEMATICS Volume 5, Number 1 (2007), 201-204, DOI: 10.2478/s11533-006-0040-7 http://www.springerlink.com/content/ev4756g8n3p81541/ -FL
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/ ]
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). [For admin and other information see: http://www.mta.ca/~cat-dist/ ]
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-... (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. [For admin and other information see: http://www.mta.ca/~cat-dist/ ]
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
On Fri, 22 Jun 2012 01:32:10 +0200, George Janelidze wrote, inter alia: 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.] [For admin and other information see: http://www.mta.ca/~cat-dist/ ]
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
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: 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.] [For admin and other information see: http://www.mta.ca/~cat-dist/ ]
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 [For admin and other information see: http://www.mta.ca/~cat-dist/ ]
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
[For admin and other information see: http://www.mta.ca/~cat-dist/ ]
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 [For admin and other information see: http://www.mta.ca/~cat-dist/ ]
Any group can be embedded into a simple group hence no non-trivial injective groups. (Yes, you'll need the fact that any group can be embedded in a non-simple group.) For an infinite group G embed it first into the group of all permutations of G and then reduce by killing the subgroup of all permutations with support of cardinality less than that of G. The fact that any subgroup is an equalizer is an easy consequence of the pushout theorem for groups: to wit, the pushout of a pair of monics is a pullback [and all monic]. Hence the pushout of a monic with itself yields a pair of maps [indeed monic maps] for which it is an equalizer. The only proof I know for this lemma is via the construction of "free products with amalgamated subgroup" (that is, the construction of the pushouts). (On the top of page 36 of ftp://ftp.sam.math.ethz.ch/hg/EMIS/journals/TAC/reprints/articles/3/tr3.pdf it is mentioned in passing that all epics between groups are onto. Without a proof, alas.) Peter Quoting Michael Barr <barr@math.mcgill.ca>:
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/ ]
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/ ]
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-... Paul Taylor [For admin and other information see: http://www.mta.ca/~cat-dist/ ]
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-...
Paul Taylor
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participants (11)
-
Andrej Bauer -
Florian Lengyel -
Fred E.J. Linton -
George Janelidze -
Joyal, André -
Michael Barr -
Michael Shulman -
Paul Taylor -
Peter LeFanu Lumsdaine -
pjf@seas.upenn.edu -
Ronnie Brown