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!

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 "betareducing" a proof or computation, akin to cut elimination in proof theory. Let's take a look. The proof of theorem 50 in BorceuxClementino (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. 


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... 

