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1. • CommentRowNumber1.
   • CommentAuthorUrs
   • CommentTimeOct 5th 2010
   • (edited Oct 5th 2010)
   • PermaLink
   Author: Urs
   Format: MarkdownItexI am being bombarded by questions by somebody who is desiring details on the proofs of the statements listed at [[regular monomorphism]], e.g. that * in Grp all monos are regular; * in Top it's precisely the embeddings etc. I realize that I would need to think about this. Does anyone have a nice quick proof for some of these?

   I am being bombarded by questions by somebody who is desiring details on the proofs of the statements listed at regular monomorphism, e.g. that

   • in Grp all monos are regular;

   • in Top it’s precisely the embeddings

   etc.

   I realize that I would need to think about this. Does anyone have a nice quick proof for some of these?

2. • CommentRowNumber2.
   • CommentAuthorTodd_Trimble
   • CommentTimeOct 5th 2010
   • (edited Oct 5th 2010)
   • PermaLink
   Author: Todd_Trimble
   Format: MarkdownItexIn case your interlocutor didn't know, there's first a lemma: in a category with equalizers and cokernel pairs, a regular mono is the same as a map that is an equalizer of its cokernel pair. In the case of $Top$, this gives a quick proof. If $i: X \to Y$ is a subspace embedding, then we form the cokernel pair $(i_1, i_2)$ by taking the pushout of $i$ against itself (in the category of sets, and using the quotient topology on a disjoint sum). The equalizer of that pair is the set-theoretic equalizer of that pair of functions endowed with the subspace topology. Since monos in $Set$ are regular, we get the function $i$ back with the subspace topology, and we are done. The case of $Grp$ it takes more analysis (and the result is a bit more surprising I think), but I think it can proceed along similar lines, where the cokernel pair is given by an amalgamated coproduct or free product. I don't have any more time to think about this now!

   In case your interlocutor didn’t know, there’s first a lemma: in a category with equalizers and cokernel pairs, a regular mono is the same as a map that is an equalizer of its cokernel pair.

   In the case of $Top$, this gives a quick proof. If $i: X \to Y$ is a subspace embedding, then we form the cokernel pair $(i_1, i_2)$ by taking the pushout of $i$ against itself (in the category of sets, and using the quotient topology on a disjoint sum). The equalizer of that pair is the set-theoretic equalizer of that pair of functions endowed with the subspace topology. Since monos in $Set$ are regular, we get the function $i$ back with the subspace topology, and we are done.

   The case of $Grp$ it takes more analysis (and the result is a bit more surprising I think), but I think it can proceed along similar lines, where the cokernel pair is given by an amalgamated coproduct or free product. I don’t have any more time to think about this now!

3. • CommentRowNumber3.
   • CommentAuthorUrs
   • CommentTimeOct 5th 2010
   • PermaLink
   Author: Urs
   Format: MarkdownItexThanks Todd!! I only have a few minutes online. So I forwarded this information and pasted it, just slightly edited, into the entry [[regular monomorphism]]. I'l try to get back to that later...

   Thanks Todd!!

   I only have a few minutes online. So I forwarded this information and pasted it, just slightly edited, into the entry regular monomorphism.

   I’l try to get back to that later…

4. • CommentRowNumber4.
   • CommentAuthorUrs
   • CommentTimeOct 8th 2010
   • PermaLink
   Author: Urs
   Format: MarkdownItexI filled in the proof that every mono in $Grp$ is regular, following the one in "The Joy of Cats". It's an easy elementary construction, the harder part is making the right ansatz, I suppose. But I don't think I have really understood the deeper meaning of it.

   I filled in the proof that every mono in $Grp$ is regular, following the one in “The Joy of Cats”.

   It’s an easy elementary construction, the harder part is making the right ansatz, I suppose. But I don’t think I have really understood the deeper meaning of it.

5. • CommentRowNumber5.
   • CommentAuthorMike Shulman
   • CommentTimeOct 8th 2010
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexDid you mean to require K to be a *finite* subgroup?

   Did you mean to require K to be a finite subgroup?

6. • CommentRowNumber6.
   • CommentAuthorUrs
   • CommentTimeOct 8th 2010
   • PermaLink
   Author: Urs
   Format: MarkdownItex> Did you mean to require K to be a finite subgroup? Probably not.

   Did you mean to require K to be a finite subgroup?

   Probably not.