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If $A$, $B$, and $C$ are topological groups, and $f: A \to B$ and $g: B \to C$ are two continuous group homomorphisms, what does it usually mean for

$$1 \to A \stackrel{f}{\to} B \stackrel{g}{\to} C \to 1$$

to be exact?

I would expect that one at least expects exactness as abstract groups (as opposed to, say $g(B)$ just being dense in $C$).

Does one also expect $f$ to be a homeomorphism of $A$ onto $f(A)$? And if so, does one further expect $C$ to be homeomorphic to $B/f(A)$ with the quotient topology>?

Would one's expectations change if $A$, $B$, and $C$ were all abelian topological groups? (And what if they were all topological $G$-modules for some topological group $G$ -- that is, if each was an abelian topological group and additionally we had continuous group action maps $G \times A \to A$ and same for $B$ and $C$, and $f$ and $g$ were $G$-equivariant?)

This seems to just be a question of semantics. I definitely see at least some authors require $A$, $C$ to have the subspace, quotient topology. But then something I read somewhere else seems to suggest that the condition should only be on the topology of $A$ only, or maybe no additional topological condition at all --- I can't quite tell...

So, what do you expect when you read or hear that $1 \to A \to B \to C \to 1$ is an exact sequence of topological groups? Is there a consensus?

UPDATE: The question has been reduced to understanding what it means for a sequence to be exact in a category that is not abelian but has kernels and cokernels (and a zero object). There are some ideas in Matthew Daws's answer and the following comments below, but a reference would be great.

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The writer of this question seems to expect $f$ to be a closed map and $g$ to be open... But I don't quite see yet what that implies for the topologies of $A$ and $C$. – sibilant Nov 29 '11 at 5:22
Ok: for an algebraically exact sequence of topological groups with continuous maps, $g$ being open is equivalent to $C$ having the quotient topology. And $f$ being closed (or open) does imply that $A$ has the subspace topology. In the case where $f(A)$ is closed in $B$, $f$ being closed is equivalent to $A$ having the subspace topology. – sibilant Dec 1 '11 at 21:42
Here is a reference (a paper recently submitted to the arxiv which discusses exactness in general categories. They suggest that the proper answer, one which gets you the useful properties of exact sequences, is more subtle than the obvious generalization. Note that as a way of having zero objects/morphisms, they adopt the language of categories enriched over the category of pointed sets. If you haven't seen enriched categories, you might want to check out the first chapter of Kelly's book. – Aaron Dec 9 '11 at 21:08
@Aaron: This seems quite helpful, though too complicated for me at the moment. This is not my time to learn about enriched categories. I'm quite surprised, though -- the question seems quite simple and I expected it to have a simple answer. I guess I'm still holding out hope, though maybe math.stackexchange isn't the right place for it. – sibilant Dec 10 '11 at 15:17
@Aaron: The categories I'm interested in at the moment all have zero objects as well as kernels and cokernels. They're often even additive. They're just not normal, so that requiring $\mathop{\rm coker} f = g$ (Matthew Daws's suggestion below) is more restrictive than $\ker (\mathop{\rm coker} f) = \ker g$ (my suggestion below)... – sibilant Dec 10 '11 at 23:40

Edit: Added some references; thanks to t.b. for many of these.

Suppose I take as my category topological (Hausdorff) groups and continuous homomorphisms. This has a "zero object"-- just the trivial group $\{1\}$. Then the (group theoretic) kernel of a continuous homomorphism will be a closed subgroup, and hence a topological group in its own right. It's easy to see that this shows that this category has "kernels" in the abstract sense.

Forming cokernels is a little trickier: given $f:G\rightarrow H$ let $Q$ be the closed normal subgroup of $H$ generated by $f(G)$. Then $H/Q$ is a Hausdorff topological group as well (See Hewitt+Ross, Thm. 5.26). The cokernel of $f$ is then the quotient homomorphism $q:H\rightarrow H/Q$. Indeed, clearly $qf=0$; if $q'f=0$ then $f(G) \subseteq \ker q'$ and so as $\ker q'$ is a closed normal subgroup, $Q\subseteq \ker q'$.

Similarly, $f:G\rightarrow H$ is "monic" if and only if $f$ is (set-theoretic) injective. $f$ is "epi" if it has dense range (Edit: the "only if" claim fails for locally compact groups, see an example of Reid; but it's true for locally compact groups in the category of Hausdorff groups, see a paper of Nummela, "On epimorphisms of topological groups" (1978); for merely Hausdorff topological groups, "only if" is false, see paper of Uspenskij).

I then believe that we have an abstract, category-theoretic definition of "exact sequence", namely $$ 0 \rightarrow G \xrightarrow{f} H \xrightarrow{g} K \rightarrow 0 $$ with $f$ monic, $g$ epi, $f$ is the kernel of $g$ and $g$ is the cokernel of $f$. Translating, this means precisely that $f$ is a homeomorphism onto its range, which is the usual kernel of $g$, and $K$ is isomorphic to $H/f(G)$, with $g$ the quotient map. This definition is copied from Mac Lane, "Categories for the working mathematician", Chapter VIII, Section 3. Now, that section is all about abelian categories, but the definition works in more generality.

Is this the "correct" definition? I absolutely wouldn't want to say that. But if you wanted something to start with, this seems reasonable. If you want something slightly different for an application? Then what's wrong with just spelling out, very clearly, what you mean by "exact", and maybe saying something to the effect that this is not an entirely standard defintion?

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The "only if" part of your characterization of epis turns out to be wrong outside the locally compact realm: See Uspenskij, The epimorphism problem for Hausdorff topological groups Re: IIRC. The quotient of an arbitrary topological group by a closed subgroup is Hausdorff, see e.g. Hewitt-Ross, Thm (5.21), page 38. – t.b. Dec 1 '11 at 10:51
@t.b. Thanks! I knew it would be in Hewitt-Ross, but my copy was not with me at the time... – Matthew Daws Dec 1 '11 at 13:27
So are you saying that the right definition of a short exact sequence in any category where zero objects, kernels, cokernerls, monomorphisms, and epimorphsims make sense is what you've proposed: $0 \to G {\stackrel{f}{\to}} H {\stackrel {g}{\to}} K \to 0$ is exact if $f$ is monic, $g$ is epi, $f$ is the kernel of $g$ and $g$ is the cokernel of $f$? In particular this would imply that many monomorphisms and epimorphisms can never be part of any short exact sequence... – sibilant Dec 1 '11 at 21:24
@sibilant: I'm not a category theorist in any sense; and I would always hesitate to say that this was the "right" definition. But I think (IIRC; I got this from some private notes I made 6 months ago) that this definition is that which Mac Lane gives in his book (for an abstract category with kernels and cokernels, and with (non-topological) groups as an example). – Matthew Daws Dec 2 '11 at 8:50
@Matthew: I think for topological groups it's in a paper of Graeme Segal, Cohomology of topological groups, Symposia Mathematica IV (1970), 377-387. See also here page 516. The abstract categorical nonsense might be in Eilenberg-Moore. Unfortunately, I can't check at the moment. – t.b. Dec 4 '11 at 10:27

In an additive (or even preadditive) category, a map $\phi: A \to B$ is proper if the natural map from the kernel of the cokernel of $\phi$ to the cokernel of the kernel of $\phi$ is an isomorphism. Note that part of this definition is the requirement that all these kernels and cokernels exist. In this case, both objects are sometimes called the "image" (though various sources mean one or the other in general). Usually a sequence $A \to B \to C$ is defined to be exact if the maps are proper and the natural map from the image of $A \to B$ to the kernel of $B \to C$ is an isomorphism. See, for example, Yoneda's paper "On Ext and Exact Sequences" or "Quasi-Abelian Categories and Sheaves" Jean-Pierre Schneiders.

A monomorphism is proper iff it is the kernel of some morphism, and dually for epimorphisms. If the category has all kernels and cokernels, then a monomorphism is proper iff it is the kernel of its cokernel, and dually for epimorphisms. So for a sequence $0 \to A \to B \to C \to 0$, if you require that the maps be proper, then the sequence is exact if and only if either (1) $A \to B$ is the kernel of $B \to C$ and $B \to C$ is an epimorphism, or (2) $B \to C$ is the cokernel of $A \to B$ and $A \to B$ is a monomorphism.

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