Is the category $\text{LCA}_c$ of connected locally compact Hausdorff abelian groups an abelian category? My feeling says no, however I can't immediately find a counterexample.

Alternatively, I'd also be happy to know whether an injective continuous map in $\text{LCA}_c$ is automatically closed.

  • $\begingroup$ In general, you won't have cokernels, is that right? $\endgroup$ – Randall May 30 '18 at 14:25
  • $\begingroup$ The category $\text{LCA}$ is quasi-abelian and hence has cokernels. The cokernel of $f:A\rightarrow B$ is $B/\overline{A}$, hence there is a surjection $B\rightarrow B/\overline{A}$. Moreover, the image of a connected space is connected. $\endgroup$ – Mathematician 42 May 30 '18 at 15:13

Given an irrational number $r\in\mathbb R$, the image of the continuous injective function $\mathbb R\to\mathbb R^2/\mathbb Z^2$ given by $t\mapsto(t,rt)$ is dense but not surjective.

A reference for how to modify homological algebra so that it applies to locally compact Hausdorff abelian groups is Norbert Hoffmann and Markus Spitzweck's paper "Homological algebra with locally compact abelian groups". A reference for the more general topic of exact structures on additive categories is Theo Buehler's article Exact Categories.

  • $\begingroup$ I see, that's a pretty easy counterexample. $\endgroup$ – Mathematician 42 Jun 1 '18 at 14:27

Let me elaborate a little on Vladimir Sotirov's example for readers who may not see how it immediately answers the question. Suppose you have a continuous injective homomorphism $f:A\to B$ of locally compact abelian groups whose image is dense but not all of $B$, as in his answer. Then the kernel of $f$ is $0$, since $f$ is injective. The cokernel of $f$ is also $0$: if $g:B\to C$ is a morphism such that $gf=0$, then $g$ vanishes on the image of $f$ and hence on all of $B$ since the image of $f$ is dense and $C$ is Hausdorff. So $f$ has trivial kernel and cokernel. In an abelian category, this would imply $f$ is an isomorphism, but it is not an isomorphism since it is not surjective. Thus our category cannot be abelian.


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