Why is TopGrp the category of topological groups and continous homomorphisms protomodular? 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!
 A: 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. 
A: 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...
