Why is TopGrp the category of topological groups and continous homomorphisms protomodular? I know it is, and I have several indirect proofs, but am not able to prove this directly by showing that the split short five lemma holds. Please help!!! Thank you!

  • $\begingroup$ When you say you have indirect proofs: is this example mentioned in the book of Borceux and Bourn? $\endgroup$ – Yemon Choi Jun 7 '13 at 22:53
  • $\begingroup$ Unfortunately I don't have time to get into it now, but possibly you would find estudogeral.sib.uc.pt/bitstream/10316/4633/1/… useful, especially sections 13 and 14 (and references therein). $\endgroup$ – Todd Trimble Jun 11 '13 at 15:37

Given the fact that Borceux and Clementino (et al. from references) are reasonably explicit and constructive in their proofs, a "direct proof" should be obtainable in any case just by systematically unpacking all the lemmas they use. This is an instance of a general metamathematical method called "beta-reducing" a proof or computation, akin to cut elimination in proof theory. Let's take a look.

The proof of theorem 50 in Borceux-Clementino (specialized to the theory of groups) explains that the short split five lemma is a statement expressible in the language of finite limits, so that by a Yoneda argument, it should hold in $\mathbf{Grp}(\mathrm{Top})$, given that the short split five lemma holds in $\mathbf{Grp}$. We can unpack this Yoneda lemma argument to give a direct proof.

Thus, suppose given a commutative diagram in the category of topological groups

\begin{array}{ccccc} \ker(q) & \xrightarrow{i'} & F & \xrightarrow{q} & B \\ \wr \downarrow & & \downarrow \pi & & \downarrow 1_B\\ \ker(p) &\xrightarrow{i}&E&\xrightarrow{p}& B & \end{array}

where $p$ and $q$ are assumed to be split epic. We want to show $\pi$ is an isomorphism of topological groups. Let $U(E)$ be the underlying topological space of $E$. Then $\hom(U(E), -)$ sends this diagram of topological groups to a diagram of ordinary groups. Since this representable preserves kernels, split epics, etc., we infer from the split five lemma in $\mathbf{Grp}$ that $\hom(U(E), \pi)$ is a group isomorphism. In particular, there is a continuous map $s: E \to F$ which is sent to $1_E$ by $\hom(U(E), \pi)$; in other words, such that $\pi \circ s = 1_E$. We argue similarly that $\hom(U(F), \pi)$ is a group isomorphism, so that there exists a unique continuous map in $\hom_{\mathrm{Top}}(F, F)$ which maps to $\pi$ under $\hom(U(F), \pi)$. Since both $1_F$ and $s \circ \pi$ are such maps, we also have $s \circ \pi = 1_F$. Finally, since the forgetful functor $\mathbf{Grp}(\mathrm{Top}) \to \mathrm{Top}$ reflects isomorphisms, we have that $\pi$ is an isomorphism in $\mathbf{Grp}(\mathrm{Top})$.

But we might as well go whole hog and make it even more direct, by following the diagram chase implicit in the preceding paragraph and constructing the inverse of $\pi$ explicitly. (I'll use additive notation here, even though we're in the context of groups and not abelian groups.) Thus, let $j$ be a section of the split epi $q$. We have $p \pi (j p) = q j p = 1_B p = p$, so $p(1_E - \pi j p) = 0$. It follows that $1_E - \pi j p$ factors through the kernel $i: \ker(p) \to E$; write $1_E - \pi p j = i g$ for some (unique) $g: E \to \ker(p)$. Let $\phi: \ker(q) \to \ker(p)$ be the isomorphism on display above. I claim that the continuous map

$$s = i'\phi^{-1} g + j p: E \to F$$

is inverse to $\pi: F \to E$. Indeed, in one direction, we have

$$\pi(i'\phi^{-1} g + j p) = \pi i' \phi^{-1} g + \pi j p = i \phi \phi^{-1} g + \pi j p = i g + \pi j p = (1_E - \pi j p) + \pi j p = 1_E.$$

In the other direction, to prove $s \pi = 1_F$, we first note that

$$\pi (1_F - s \pi) = \pi - \pi s \pi = \pi - 1_E \pi = 0$$

so that in particular, $0 = p \pi (1_F - s \pi) = q (1_F - s \pi)$. Therefore $1_F - s\pi$ factors through the kernel of $q$: we have $1_F - s\pi = i'h$ for some unique $h: F \to \ker(q)$. From the equation displayed above, we thus have $0 = \pi i' h = i \phi h$. Since $i \phi$ is monic, this implies $h = 0$. Therefore $1_F - s\pi = i' h = 0$, and this completes the proof.

  • $\begingroup$ +1 for the "pedestrian version" (not that there's anything inherently wrong with the more abstract version, but it is good to see both for Brains of Little Bear like myself) $\endgroup$ – Yemon Choi Jun 13 '13 at 7:57
  • $\begingroup$ Thanks, Yemon. I have to admit that I slipped at first and wrote the above as if in the context of abelian groups, but as long as one is careful, it goes through for groups as well. Of course, Borceux and Clementino do things much more generally; their paper is a nice instructive application of categorical universal algebra. $\endgroup$ – Todd Trimble Jun 13 '13 at 8:22
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    $\begingroup$ Made a note on that here: ncatlab.org/nlab/show/topological+group#Protomodularity $\endgroup$ – Urs Schreiber Jun 13 '13 at 13:32

That's a good example of an indirect proof for it. It would be sooo nice if it could be shown directly that the split short five lemma (http://ncatlab.org/nlab/show/five+lemma#short_split_five_lemma_28) holds in TopGrp. All I need to show is that w (in the diagram in the link) is an open map. Not working for me, not sure why...

  • $\begingroup$ Sigh...I'm quite familiar with the proof given in Semi-abelian topological algebras by Borceux/Clementino. The proof is correct, but not direct. I'm also familiar with every single reference in the paper. It is odd that such a simple true topological assertion is being difficult to prove. $\endgroup$ – Lucas Earl Jun 12 '13 at 7:59
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    $\begingroup$ Feel free to comment here again if you are unsatisfied with the argument presented in my answer (I realize you don't have the points yet to comment under answers other than your own -- I'll upvote your question to help you get there). $\endgroup$ – Todd Trimble Jun 13 '13 at 7:12
  • $\begingroup$ @ToddTrimble It seems that Lucas has two accounts, one unregistered with 41 points and one registered with 1 point. If they (belong to the same user and) are merged, he will be able to comment on your post as the owner of the question. (I realize the post is old, but this just came up in the review queue.) $\endgroup$ – Joonas Ilmavirta Feb 26 '18 at 16:20

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