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The category of Hausdorff topological abelian groups are commonly cited as an example of a category which is preabelian, but not abelian.

I think one reason that is is not abelian comes from the following. If we take the standard embedding $\mathbb{Q}\hookrightarrow\mathbb{R}$ with the usual topologies, then the kernel and cokernel is the zero group. Then I conclude $coker(0\to\mathbb{Q})$ is $\mathbb{Q}$, and $\ker(\mathbb{R}\to 0)$ is $\mathbb{R}$, but $\mathbb{R}\not\cong\mathbb{Q}$. (Please correct me if I am mistaken!)

However, why is $\mathsf{HTAG}$ actually preabelian? I guess more specifically, why do kernels and cokernels always exist? Is there a reference where this is laid out?

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  • $\begingroup$ Which axiom are you having trouble verifying? $\endgroup$ – Qiaochu Yuan Apr 15 '14 at 3:05
  • $\begingroup$ @QiaochuYuan The existence of kernels and cokernels primarily. $\endgroup$ – Buble Apr 15 '14 at 6:52
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    $\begingroup$ The kernel is the usual kernel; the cokernel is precisely the quotient “codomain modulo closure of the image”. There's a simpler class of examples of mono-epi that aren't isomorphisms: every group is a topological group when endowed with the discrete topology; the identity from $G$ (discretized) to $G$ with its original topology is obviously mono and epi. $\endgroup$ – egreg Apr 15 '14 at 6:58
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The category cannot be abelian, because a mono-epi is not necessarily an isomorphism. Take any (Hausdorff abelian) topological group $G$ and consider $G_d$, the same group but with the discrete topology; the identity map $G_d\to G$ is certainly mono and epi, but it's not an isomorphism unless $G$ was discrete to begin with.

For the kernel, the usual one works, because it's a (closed) subgroup and, with the relative topology, a topological group; the inclusion map is, of course, continuous.

The cokernel of $f\colon G\to H$ is $H/\overline{\operatorname{Im}(f)}$, the quotient of $H$ modulo the closure of the image of $f$ (with the quotient topology). If you have a continuous morphism $g\colon H\to K$ such that $gf=0$, then $\ker g\supseteq \operatorname{Im}(f)$, so also $$ \ker g\supseteq \overline{\operatorname{Im}(f)} $$ because $g$ is continuous and $K$ is Hausdorff.

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  • $\begingroup$ Interesting, thanks egreg. $\endgroup$ – Buble Apr 16 '14 at 1:03

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