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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](http://math.stackexchange.com/questions/52164/short-exact-sequences-of-topological-groups-and-lie-groups) 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 $C$.2011-11-29
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    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.2011-12-01
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    [Here](http://arxiv.org/abs/1111.0330) 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](http://tac.mta.ca/tac/reprints/articles/10/tr10abs.html).2011-12-09
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    @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.2011-12-10
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    The good news and bad news with a lot of category theory is that in a lot of familiar categories, many different notions collapse together, e.g. there are many different kinds of [epimorphisms](http://ncatlab.org/nlab/show/epimorphism), several of which are equivalent in $\mathcal{Set}$. This means that in familiar categories, familiar notions can be described/transported simply. Unfortunately, it also means that the obvious generalization won't always be the right one, and won't have all the properties you want it to have (even if it works in the categories you care about).2011-12-10
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    Also, regarding enriched category theory, it's worthwhile to learn (even if only to see closed symmetric monoidal categories), but you can start by thinking of categories whose hom sets have additional structure. You've seen hom sets where morphisms can be added, and likely where they are actually modules over some ring. In the paper, they use pointed sets and have the base point in each hom set correspond to the zero morphism. This lets them have zero morphisms without a zero object. N.B. The product in pointed sets isn't the product in set. The difference models how zero morphisms work.2011-12-10
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    @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)...2011-12-10

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