What are some natural examples of "bimorphism" classes?

Question

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 a special case of bimorphism is that of an isomorphism i.e a morphism which is invertible. 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!

Answer 1

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.

Answer 2

(Disqualifier: I don't have too much expertise) I would say that, at least at the beginning of one's categorical quest, the most common use of the concept of "bimorphism" is to find conditions where "bimorphism => isomorphism". This isn't quite the same as finding those categories which are balanced as there may be some subcategory where "bimorphism => isomorphism", or it may be that $\mathcal{D}$ is a subcategory of $\mathcal{C}$ such that if $C \to D$ is a bimorphism in $\mathcal{C}$ with $D \in \mathcal{D}$ then it is an isomorphism. A silly example of that being the subcategory of discrete spaces in $Top$: if $C \to D$ is a continuous bijection and $D$ has the discrete topology then $C$ has the discrete topology. Or you can have two subcategories, say $\mathcal{C}$ and $\mathcal{H}$, with the property that a bimorphism $C \to H$ is an isomorphism: think "Compact" and "Hausdorff".

The simplest case that I can think of where "bimorphism => isomorphism" holds and is regularly used is in the category of finite dimensional vector spaces (over some fixed field). We regularly tell our students that $T \colon V \to U$ is invertible if and only if it is injective and surjective. Indeed, we give them whole lists of bizarre conditions for when $T$ is invertible!

Here's an analogy that you might like (being more of a topologist than an algebraist). We like complete metric spaces because then we don't have to bother showing that the limit actually exists. If it "morally" exists, that's good enough. So it saves us some mental effort to just show once and for all that $M$ is complete. Similarly, we like balanced categories because then we don't have to bother showing that the inverse actually exists. If it "morally" exists, that's good enough. So it saves us some mental effort to just show once and for all that $\mathcal{C}$ is balanced.

As has been pointed out already, bimorphisms don't form an equivalence class so your actual question doesn't quite make sense, but as David says, the concept of "What structures can be put on $C$" is very important. I used it in comparative smootheology (though I don't use the word "bimorphism") because in passing from one concept of "smooth space" to another, one often needs to move to the "nearest smooth structure". This generally involves showing that the category of "smooth structures on $C$" forms a complete lattice.

So, in summary:

1. "Bimorphism classes": doesn't make sense, so not studied
2. "Bimorphism => isomorphism": very important, because "bimorphism" is much easier to check than "isomorphism" (don't have to actually find that darned inverse). Tends to be in more algebraic situations.
3. "Bimorphism" as "structures on a thingy": very important, used lots of the time in topological settings.

Answer 3

If the category in question is concrete with underlying set functor $C \to Set$, and bimorphisms are given by one-to-one and onto functions, then the more interesting question is what happens inside a bimorphism class. Namely, what sort of $C$-objects are there for a given underlying set? In other words, what do the fibres of $C\to Set$ look like? This is definitely an interesting question for including in a project.

As Peter points out in his comment, a space is bimorphic to the indiscrete space on the same set of points. But it is also bimorphic to the discrete space on the same set of points, and these form the top and bottom respectively of the poset of topologies on a set. This poset is very interesting, and has interesting topologies. This sort of behaviour would crop up with probably any topological concrete category over $Set$ (beware that these are often called 'topological categories' in the literature, but are neither categories internal to $Top$ nor $Top$-enriched categories)

Edit: I realised last night that we are not looking at the fibres of the underlying set functor, but the preimage of a whole isomorphism class of sets. This makes things a little bit more complicated, but also raises the interesting question of when are (essentially) endomorphisms of a set continuous with respect to two given topologies, or rather, what is the structure on the category consisting of spaces with the same underlying set and all maps between these covering endomorphisms of the underlying set? The poset of topologies is in there as those maps that cover the identity map.

Answer 4

May I point out that the relation "is bimorphic to" is a preorder, that is, a reflexive and transitive binary relation since the identity is a bimorphism and the composition of two bimorphisms is a bimorphism.

Every preorder induces a natural equivalence relation (namely, by forcing symmetry). In this context, we may say "$A$ is bimorphically equivalent to $B$" if there exist bimorphisms $A \rightleftarrows B$, where $A$ and $B$ are objects.

One particular property of such an equivalence relation is that it is weaker than isomorphism. In category theory, we often encounter situations in which isomorphisms tend to be too strict, say, when considering the equivalence of categories rather than the isomorphism of categories.

In a balanced category, bimorphic equivalence is the same as objects being isomorphic.

As for the other part of your question, I must say that I am not sure in which geometric context such an equivalence relation would be useful. However, there is one context that may solidify biomorphic equivalence as being useful.

Bimorphisms in Top are precisely the bijective continuous maps. This specifies bimorphic equivalence in Top. Let us say that a property $P$ of topological spaces is a bimorphism invariant if for every pair $X, Y$ of topological spaces, the statement $P(X)$ and $X$ is bimorphically equivalent to $Y$ implies the statement $P(Y)$.

Every bimorphism invariant is a topological invariant (an isomorphism invariant in Top) since, as I have already mentioned, bimorphic equivalence is weaker than isomorphism.

The converse, that is that every topological invariant is a bimorphism invariant does not hold (Top is not balanced). However, it turns out that many of the classical topological invariants are in fact also bimorphism invariants.

It is sufficient to consider any continuous invariant (i.e. a property of topological spaces "transferred" by surjective continuous maps) such as

• compactness

• path connectedness

• separability

• etc.

We can generalize this to other categories and look for isomorphism invariants that are bimorphism invariants in that category. For the sake of example, in Ord, the following isomorphism invariants are also bimorphism invariants:

• "has a least/greatest element”

• “is a totally ordered set”.

Thus, bimorphic equivalence is a good relaxed notion of isomorphisms.

I hope this helps!

Answer 5

Here is an elementary example of a ubiquitous category (a monoid) in which every morphism is a bi-morphism, but the only isomorphism is the identity morphism.

Fix a set $X$.

Consider the set of all finite sequences $x_1$,$x_2$,..$x_n$ such that each $x_i \in X$. We also allow the empty sequence.

Under cancatantion, the collection of all such sequences form the morphisms of a category with one object, such that each morphism is a bimorphism, and such that the only isomorphism is the identity morphism.

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Every strict poset is naturally a thin category (in which there is at most one morphism between objects), and in particular all nontrivial morphisms are both monic and epic (and hence bimorphisms), but not isomorphisms.

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The category whose morphisms are based covers between connected n-manifolds is not thin, but every morphism is both monic and epic, and often not an isomorphism.