Note: As pointed out in a comment by Peter and echoed by Andrew in his answer, the question as stated does not make sense because "being bimorphic to" is not an equivalence relation. Nevertheless, I will leave the question as stated in case it continues to receive interesting answers.
Justification
This coming academic year I will be supervising a fourth-year undergraduate project on Category Theory. As should be clear to MO regulars, I'm far from an expert in this topic, but I am keen to improve my fluency in this language: hence the project.
Maybe it's only me, but I find that the unifying beauty of category theory can best be appreciated when one has lots of concrete (not necessarily in the categorical sense!) examples. Most books I've seen take this view as well and present a number of examples taken from algebra and topology, mostly. Given the origins of the subject, this is of course not surprising. From time to time, though, I find myself wanting more "realistic" examples for certain concepts. This question is about one such concept.
Background
Recall that a bimorphism in a category is a morphism which is both monic (i.e., left cancellable) and epic (i.e., right cancellable). A closely related notion to (and, in a concrete category, a special case of) bimorphism is that of isomorphism: a morphism $f:A \to B$ is an isomorphism if there exists a morphism $f^{-1}: B \to A$ such that $f \circ f^{-1} = 1_B$ and $f^{-1} \circ f = 1_A$, with $1_{A,B}$ the corresponding identity morphisms. Categories in which all bimorphisms are isomorphisms are said to be balanced.
We say that two objects $A,B$ in a category are isomorphic if $\operatorname{Mor}(A,B)$ contains an isomorphism. Now, if the category is not balanced, it could very well be that $\operatorname{Mor}(A,B)$ contains a bimorphism even if it does not contain an isomorphism. For lack of a better name, let us call such objects bimorphic and let us speak of bimorphism classes of objects,…
Needless to say, the concept of isomorphism classes is very important and sits at the centre of any classification problem, but what about the concept of bimorphism classes?
Question
Is there a natural context in which one is compelled to relax the notion of isomorphism to that of bimorphism? In particular, any geometric context?
Thanks in advance!
Best Answer
I'm going to be deliberately provocative and say that I don't really know of any use for the concept of bimorphism as such. (I also don't really like the name; it sounds to me like something that's both a morphism and a comorphism.)
One use that's been proposed is "to find situations in which bimorphism ⇒ isomorphism." Such situations may be interesting, but as far as I can tell they are rarely (if ever) used. What seems to happen much more often is that we have some factorization system (E,M) and we use the fact that E+M=iso, which is true for any factorization system. The most common case is probably (extremal epi, mono), followed perhaps by (epi, extremal mono); both of these are factorization systems as soon as the relevant factorizations exist.
It might happen, in some case, that E consists of exactly the epimorphisms and M of exactly the monomorphisms (such as when all epis, or all monos, are extremal). But as far as I can tell this fact -- especially the epi part of it -- is hardly ever relevant, because in practice it's quite hard to characterize the epis in a given category or to check that a given morphism is epi, nor is the answer often especially meaningful. Since monadic functors also create limits, and in particular monomorphisms, a morphism in a category monadic over Set is monic iff it is injective -- but this is not true for epis, and even in quite nice categories the epis can be fairly bizarre. It's usually the extremal epis which coincide with the "surjections" and form a factorization system with the monos.
For instance, Andrew cited vector spaces as an example of a balanced category. But as I pointed out in my comment, do we ever use that fact? What we actually teach our undergraduates is that injective+surjective=iso for vector spaces; we (or, at least, I) don't tell them anything about why surjective=epi, or even what epi means. And when doing linear algebra, I might occasionally use the fact that surjections are in particular epi (which just follows because the forgetful functor to Set is faithful), but never the converse. It's just as true for groups, rings, fields, monoids, etc. that injective+surjective=iso, and we use that fact in doing algebra all the time -- but does the non-surjective ring epimorphism Z → Q, showing that rings (unlike vector spaces) are not balanced, ever actually bother us in practice?
In the topological situation, it's true that the epimorphisms in Top are precisely the surjective continuous maps. But does that fact really help you when looking for conditions ensuring that a continuous bijection is an isomorphism, or using that fact in practice? Odds are the property of a continuous bijection you're going to use is that it's continuous and a bijection, not that it's monic and epic in the category Top.
The categorical version of "continuous bijection in Top" is "inverted by the forgetful functor to Set," and I think that in general the property of "being inverted by a forgetful functor" is quite interesting and important. For instance, a forgetful functor with the property that any morphism inverted by it is already an isomorphism is called conservative, and these include all monadic functors. The question about all the different topologies one can put on a given set also seems to me to really be about morphisms inverted by the forgetful functor; is it really important here that continuous surjections are the epis in Top? I expect that if you modify the definition of Top a little, then it may no longer be true that epis coincide with continuous surjections, and in that case I bet that it is the continuous surjections which are of more interest.
At this point, perhaps the most interesting thing I know about bimorphisms is that they often form the middle class of a ternary factorization system. I'll be happy to be proven wrong, however.