I am writing my bachelor thesis, mainly about General Topology and Topological Vector Spaces. Moreover I know a little bit about Category Theory: categories, functors, natural transformations, representability and the Yoneda Lemma. A simple consideraation is the following:

Any continuous function between a compact and an Hausdorff space is closed

As an immediate consequence a continuous bijection between two compact Husdorff spaces is automatically a homemorphism. This motivates two facts:

  1. Adding just an open set the topology ceases to be compact and removing one the topology cease to be Hausdorff. Hence the topology of a CHaus space is 'final' with respect to the property of compactness and 'initial' with respect to that of Hausdorffness
  2. A bijective morphism in the category CHaus is automatically an isomorphism

Now my questions are:

  • First of all: are 1 and 2 'categorically' related?
  • Secondly: I think a completely analogous result is interpreting the Banach isomorphism theorem in the category Ban of banach spaces. What's underlying? What do these category share? Can we generalize? Do we have more examples, specially in Topology/Functional Analysis?
  • Third: can someone suggest some nice 'easy' application of category theory to general topology or functional analysis? I mainly saw algebraic and algebraic topological ones.

Thanks in advance

  • 2
    $\begingroup$ I think for the second question, the two things aren't categorically related, at least in a naive analysis. Indeed, "what makes it work" (in a very category theory-biased sense) for CHaus is that this category is monadic over Set, that is, it's equivalent (in fact, concretely isomorphic) to the category of algebras over the ultrafilter monad, which means it has an algebraic definition which implies that "morphisms that are bijections are isomorphisms". (1/2) $\endgroup$ Aug 23 '19 at 11:53
  • 1
    $\begingroup$ On the other hand, Ban_1 (Banach spaces and contractions) is not monadic, see e.g. this MO discussion : mathoverflow.net/questions/8550/… so one might argue that the property comes from different places (2/2) $\endgroup$ Aug 23 '19 at 11:53
  • 1
    $\begingroup$ @WilliamElliot Moreover, consider the identity over a compact Hausdorff space X. If we consider any finer topology on the domain we still have a continuous bijection which clearly is not an homeomorphism: hence it must be that the domain has lost compactness (since it is still Hausdorff, for sure). On the other hand consider any coarser topology on the codomain: the same argument yields that it must have lost Hausdorffness. What is wrong with this? $\endgroup$ Aug 23 '19 at 12:59
  • 3
    $\begingroup$ Number 1 is true : given a compact Hausdorff topology on a space, it is minimal wrt to being Hausdorff, and maximal wrt to being compact. The proof is clear : $id : X\to X$ is a bijection, but not a homeomorphism if you remove/add open sets on the correct side (note : removing open sets preserves compactness, and adding open sets preserves Hausdorffness) $\endgroup$ Aug 23 '19 at 13:02
  • 1
    $\begingroup$ If you know about adjunctions, then all you need to learn is monads so that it should be ok. If you don't, then you first need to learn about them - I don't know if it's "much" or not $\endgroup$ Aug 23 '19 at 13:04

This is more of a long comment than an answer. We call a category $C$ concrete if it's equipped with a forgetful functor $U : C \to \text{Set}$, usually assumed to be faithful; this formalizes the intuitive notion of a category of "sets with extra structure," where $F$ describes the underlying set of an object. The property you want, that a morphism in $C$ which is bijective on underlying sets is an isomorphism, corresponds to $U$ being conservative. A conservative functor is one that reflects isomorphisms, meaning that if $F(f)$ is an isomorphism then $f$ is an isomorphism.

Faithful and conservative functors can be related as follows. First, some nonstandard definitions: say that a morphism is a pseudo-isomorphism if it is both a monomorphism and an epimorphism, and a fake isomorphism if it's a pseudo-isomorphism, but not an isomorphism.

Exercise 1a: Faithful functors reflect epimorphisms and monomorphisms: that is, if $F$ is faithful and $f$ is a morphism, then if $F(f)$ is an epimorphism then $f$ is an epimorphism, and if $F(f)$ is a monomorphism then $f$ is a monomorphism. Hence faithful functors reflect pseudo-isomorphisms.

Exercise 1b: If $F : C \to D$ is a faithful functor and $C$ has no fake isomorphisms (so every pseudo-isomorphism is an isomorphism), then $F$ is conservative.

Hence, if $C$ is a concrete category whose forgetful functor isn't conservative, then $C$ must have fake isomorphisms. $C = \text{Top}$ is a well-known example; in this category fake isomorphisms exist because we can add open sets to a topology and get another topology, which allows us to construct continuous bijections which are not homeomorphisms.

In addition, while it's not true in general that pseudo-isomorphisms are isomorphisms, there are many statements of the form "a morphism which is both a monomorphism and a (some special kind of epimorphism) is an isomorphism." A reasonably useful one in practice is:

Exercise 2a: A morphism which is both a monomorphism and an effective epimorphism is an isomorphism.

Exercise 2b: If $F : C \to D$ is a faithful functor and every epimorphism in $C$ is effective, then $F$ is conservative.

The condition that every epimorphism is effective holds in some categories of algebraic objects, such as $\text{Vect}$ and $\text{Grp}$, but not in others, such as $\text{Ring}$.

It turns out that in $\text{CHaus}$ every epimorphism is effective; what this says somewhat more concretely is that every continuous surjection $X \to Y$ between compact Hausdorff spaces is a quotient map, or in other words that $Y$ has the quotient topology (note that this is emphatically not true in $\text{Top}$!). so this is one way of explaining why $\text{CHaus}$ has a conservative forgetful functor. I don't think this is true in the category of Banach spaces though.

The comments allude to the fact that monadic functors are conservative, and while this covers the case of compact Hausdorff spaces it doesn't cover the case of Banach spaces.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.