What do continuous maps between $X$ and $Y$ tell us about $X$ and $Y$?

Question

Category theory, roughly, is about the relation between structures of a certain kind, rather than their internal structure.

In the category of groups, I can see how this is the case: the existence of a particular group homomorphism between $G$ and $H$ really tells you alot about $G$ and $H$ as structures.

But I’ve never thought of continuous maps between topological spaces $X$ and $Y$ as “telling us something about $X$ and $Y$”. Admittedly, I’m not very advanced at topology.

I would like to understand this perspective better. Could you give me a broad overview of how a continuous map $f:X\to Y$ “tells us something about $X$ and $Y$”?

Answer 1

In my opinion the easiest way to understand it is to look at the isomorphisms, instead of the continous maps, because they tell much more about the required structure on $X$ and $Y$. If you continue on your example with groups, two isomorphic groups $X$ and $Y$ are such that they will have exactly the same structure for everything regarding groups.

Now it turns out that for topological spaces, the thing that we need to say "$X$ and $Y$ are exactly the same for everything regarding the topology" is the notion of homeomorphism, that is a bijective continuous map whose inverse is continuous. It is not at first obvious why this is the right notion, so let me explain. A topology on $X$ is a class $\tau_X\subset\mathcal{P}X$ of open sets in $X$, satisfying some properties. Now take two topological spaces $X$ and $Y$ with a homeomorphism $f : X \to Y$. $f$ is bijective means that $X$ and $Y$ are the same as sets, $f$ is continuous means that the preimage of $\mathcal{O}_Y\in\tau_Y$ is in $\tau_X$, and $f^{-1}$ is injective means that the preimage of $\mathcal{O}_X\in\tau_X$ is in $\tau_Y$. If you put all this together, it says that $f$ is a bijection between $X$ and $Y$, and the bijection induced by $f$ between $\mathcal{P}X$ and $\mathcal{P}Y$ can restrict to bijection between $\tau_X$ and $\tau_Y$. And this formulation relates much more to the definition of topological space.

Now if $f$ is a continuous map between $X$ and $Y$, by definition, the preimage of every open of $Y$ is open in $X$, so this is already "telling us something relating the topological structure on $X$ with the one on $Y$", but I admit it is pretty vague.

To my knowledge, if you want more general properties to hold, you will want some extra properties for your topological (like separation...), because the category of topological spaces is very weird, but I might be wrong on that point.

Now let's assume that you have a category of "nice" topological spaces (whatever nice means). Studying topology is hard, but we can compute its fundamental group. Now a continuous map $f : X \to Y$ induces a map $\pi_1(f) : \pi_1(X)\to\pi_1(Y)$, which happens to be a group homomorphism, so if you already know how a group homomorphism tells us something about the similarity between the two groups, and if you know (or accept), that the fundamental group tells us a lot about the topology of the spaces, then you see indirectly how a continuous maps tells indeed a lot about the topology of the two spaces.

Note that this explanation I gave is very weak, in the sense that a continuous map tells us a lot more than a group homomorphism on the fundamental groups, it induces also a morphism between the higher homotopy groups (and also teh groupoid version of this statements, for non-connected spaces). So in the end it really says quite a lot, but not much that I can give explicitly

Answer 2

I think it is kind of a philosophy of understanding what is a space. Let's start with the simplest case.

We know that what is a vector space $V$ over some field $k$. And in linear algebra, we concentrate on the linear properties of a linear space. However, the linear properties are those preversed by linear maps, so the set of all linear functions $\mathrm{Hom}(V,k)$ is crucial. But it isn't hard to show that $V\cong\mathrm{Hom}(V,k)$, which means if we know all the linear functions, we already know the linear structure of our original vector space $V$.

Move to a senario a little bit complicated. Suppose $\mathbb{A}_k^n$ be the affine space associated with the vector space $k^n$. Over this space $\mathbb{A}_k^n$, we want to know the affine properties preserved by affine maps. Again, if we know all the affine functions (from $\mathbb{A}_k^n$ to $k$), we can construct the original space $\mathbb{A}_k^n$.

If we keep moving on, to somewhere called differential geometry, things become more clear. We have some coordinate patches, then all functions on the manifold can be (and should be) dealt with by taking the pull-back (locally). Just an example, the cotangent space of a manifold at some point is defined by something on the set of all smooth functions, and its dimension coincide with the dimension of the manifold. You see this is a geometrical property that we can extract if we only know all the functions from a manifold to $\mathbb{R}$.

All the cases are saying, that as long as we know all the functions form some object in our category, to some really simple object in our category, we can always know many properties preserved by the morphisms in this category, even constructing the object again.

Grothendick said something like in the classical algebraic geometries, the way we studied varieties was wrong. This was because varieties should not be defined by the zero locus of some polynomials. Almost all the properties of a space come from studying the set of all possible good functions, so the definition of a space should also be there if we know all the functions. This is basically the scheme. We first know all functions, which form a ring $R$, then we know the space should be $|\mathrm{Spec}~R|$, and the topology on this space is Zariski topology. And $R$ is really the set of functions, while locally we can have more functions.

This pattern also shows in algebraic topology. Suppose we have a compact surface $S$, and we know all the functions from $S^1$ and $S^2$ to $S$, then we already know the fundamental group of $S$ and hence $H_1$. Therefore, we essentially know what $S$ is.

All these do not purely come from algebraic topology, but they are all saying the same thing. In a category, if we know all the morphisms from an object to other object, then this object should be determined uniquely up to isomorphism. I hope this could be helpful.