Why are continuous functions the “right” morphisms between topological spaces?

Recently, someone mentioned to me that given a function $f: X \to Y$ there are two natural functions between the powersets $P(X)$ and $P(Y)$. Namely $f: U \subset X \mapsto f(U)$ and $f^{-1}: V \subset Y \mapsto f^{-1}(V)$. Then if we consider maps between $P(P(X)), P(P(Y))$, each of the above maps induce two more, so there are four natural maps.

Thus, it seems on the face of it like there are four natural choices for morphisms of topological spaces (since a topology on $X$ is an element of $P(P(X))$). Why is it that continuous functions are the morphisms we choose and not one of the other four maps?

I understand that the theory we get from taking continuous functions as morphisms is incredibly rich and so this alone provides adequate justification. However, I am looking for a different sort of justification along the lines of "is there some property of continuous functions that immediately suggests they are the 'right' choice of morphisms between topological spaces?".

• Although I quite like "abstract mathematics", I think this kind of question is a little bit misguided. Continuous maps should formalize the intuitive meaning of continuity. It is not their purpose of being morphisms of a category. There are many other classes of morphisms which could do that, for example open maps, closed maps and specialization-order-preserving maps. There are no notions in mathematics which are natural objectively; they can only be natural with regard to what we want to do with them. – Martin Brandenburg Sep 15 '14 at 23:35
• @MartinBrandenburg Sure, I am aware of why continuous maps arise. I just wonder if there are some abstract reasons why they are a good choice. It seems like you are suggesting that there is no such reason, in which case that already is something of an answer. – Alexander Sep 15 '14 at 23:43
• Yes there are many abstract reasons, and probably others will tell you about simplicial sets (morphisms are pointwise and geometric realization should be a functor), topoi and stuff like that (continuous maps should induce pullback functors), but I don't think that anything like that should be considered an answer. Continuous maps are a good choice simply because we are interested in continuous maps. We don't look at them because they fit into a nice formal framework, but because they model geometric transformations. – Martin Brandenburg Sep 15 '14 at 23:46
• Yeah, I totally agree with you. I'm just interested in knowing these other perspectives. Many thanks – Alexander Sep 15 '14 at 23:49
• Note that the inverse image is distinguished from the direct image by the fact that it's a representable functor on sets: it's $\text{Hom}(-, 2)$, whereas the direct image functor is not representable. – Qiaochu Yuan Sep 16 '14 at 15:14

I had the same question as you when I was studying topological spaces: in particular, it annoyed me that the definition didn't look like "preserves some structure" in the sense that I'd become familiar with in abstract algebra, e.g. preserving a group operation in the case of morphisms of groups. Here were my thoughts at the time. Here are two proposals I currently have for answers to this question.

Kuratowski

There is an alternative and equivalent axiomatization of topological spaces called the Kuratowski closure axioms. Here a topology on a space $X$ is described in terms of the operation $\text{cl}$ on the power set that sends a subset of $X$ to its closure in the topology, and continuity becomes "preserves the closure operator" in the sense that $f(\text{cl}(A)) \subseteq \text{cl}(f(A))$.

Vickers

General topology is actually a kind of logic. I don't know who this insight is due to, but see Vickers' Topology via Logic for much more on this theme. In particular, the open subsets of a topological space should be thought of as axiomatizing semidecidable properties: properties that you can confirm but not necessarily disconfirm, given limited tools (e.g. finite time and precision).

For example, you can confirm whether two things are less than $5$ inches apart by measuring the distance between them to finite precision and seeing if it's less than $5$, so an open ball of radius $5$ in a metric space describes a semidecidable property, but you can't confirm whether two things are less than or equal to $5$ inches apart by measuring the distance between them to finite precision because if you get $4.99 \pm 0.2$ inches you don't know whether that's over or under $5$.

Semidecidability can be used to justify all of the topological space axioms, which is a nice exercise. For example, arbitrary unions of open sets are open because given a method of confirming whether you're in each of those open sets, you get a method of confirming whether you're in any of them by running all of the methods simultaneously and waiting for one to finish. But you only get finite intersections when you try to do the same thing for waiting for all of the methods to finish because method $n$ might take $n$ seconds finish.

Continuous functions then axiomatize "computable functions": for $f$ to be continuous means that it should be possible to compute $f(x)$ "to arbitrary precision" by computing $x$ "to arbitrary precision," where going off of the example of metric spaces "to arbitrary precision" means "to within an arbitrary open set," since it's semidecidable whether $f(x)$ is contained within an open set. In other words, to locate $f(x)$ within some open set $U$, it suffices to locate $x$ within some open set $V$. After a moment's thought you'll see that this is precisely the condition that $f^{-1}(U) = V$.

(I particularly like this justification of topological spaces and continuity because, unlike the justification coming from thinking about metric spaces, it continues to apply to spaces that aren't Hausdorff, and in fact it tells you what it means for a space to not be Hausdorff. One equivalent definition of being Hausdorff is that the diagonal $\{ (x, x) \in X \times X \}$ is closed in $X \times X$. This is equivalent to "$x \neq y$" being semidecidable, so a space fails to be Hausdorff precisely when "$x \neq y$" fails to be semidecidable.)

• See also en.wikipedia.org/wiki/Pointless_topology. Here the resolution is to think of the lattice of open sets as the important structure rather than the underlying set of points. – Qiaochu Yuan Sep 17 '14 at 19:16
• In other words, the continuous maps preserve limit points. – user50229 Sep 7 '16 at 23:20

I am a litte late to answer this question which already has an answer, but I want to make more clear some points left uncleared for the following reader.

First, there are not two functions between $P(X)$ and $P(Y)$ but three. That is there are three functors $P_i$ from $\mathbf{Set}$ to $\mathbf{Pos}$, one contravariant and two covariant: \begin{align} P_1:\mathbf{Set} &\to \mathbf{Pos} & P_2:\mathbf{Set}^{op} &\to \mathbf{Pos} & P_3:\mathbf{Set} &\to \mathbf{Pos}\\ f&\to\exists f& f&\to I(f)& f&\to\forall f, \end{align} Where $\exists f(U)=f(U)$ is the image, $I(f)(U)=f^{-1}(U)$ is the inverse image and $\forall f(U)=(f(U^c))^c$ is something different. When you embed $\mathbf{Pos}$ in $\mathbf{Cat}$ then there is a nice adjoint relation between these 3 functors: $$\exists f\dashv I(f)\dashv\forall(f).$$ From this relation one would get the usual: the image commutes with unions and the inverse image commutes with unions and intersections.

Second point in this answer is that you look at morphisms between objects of the form $P(P(X))$, because they contain among other things the topologies on the set $X$. From this reasoning it seems that you would look at all possible group structures $G(X)$ on a set $X$ and for morphisms you would look for functions between $G(X)$ and $G(Y)$, that clearly associate to a group structure another group structure and do not even begin to "preserve" anything inside the group, since it sends structure to structure.

Now I'd like to answer the question in two ways, one is very satisfactory and the other is not.

The first and satisfactory one is that the definition of a topological space via the open sets is not the only feasible one. We can define a topological space in at least 8 different ways. I would like to look at two of them to see that continuous functions preserve some structure.

The first one is obviously that a topological space is characterized by its converging nets. If you do not know what a net is we can change the aforementioned statement to: "first-countable topological spaces are characterized by their converging sequence."

I'll list the axioms because they are not common to find in the literature. In the following we will use the notion of directed set $\Lambda$ that is a partially ordered set such that $$\forall a,b\in\Lambda\quad\exists c\in\Lambda\ s.t. a\leq c,\ b\leq c.$$ A net is but a function $\Lambda\to X$ from a directed set $\lambda$ to a set $X$ ($\Lambda$=$\mathbb{N}$ being the usual definition of a sequence.). The first difference arise in the definition of subnet, because we would need a function respecting the directed structure: Given two directed sets $\Lambda$ and $M$ a morphism of directed sets will be a functions $f:\Lambda\to M$ such that $\forall \mu\in M\ \exists \lambda\in\Lambda$ such that $\mu\leq f(\lambda)$.

1)For all $x\in X$ the constant nets $s:\Lambda\to X$, $s_\lambda=x$ for all $\lambda$ converge to $x$.

2)If $s:\Lambda\to X$ is a net converging to $x$ then all its subnets converge to x.

3)If $s:\Lambda\to X$ is a generic net (not a priori convergent) and all its subnets admint a subnet converging to $x$ then $s$ is converging to $x$.

4)If $s:\Lambda\to X$ is a net converging to $x$ and for all $s_\lambda$ I take a net $m^\lambda:M_\lambda\to X$ converging to $s_\lambda$ then there exists a diagonal net $t:\Lambda\to X$ defined by $t_\lambda=m^\lambda_{\mu_\lambda}$ converging to $x$.

From this point of view, given two sets $X$ and $Y$ and families $S(X)$ and $S(Y)$ of nets satisfying the above axioms we get that the continuous functions are those functions between $X$ and $Y$ sending converging nets to converging nets.

The second point of view is that of neighbourhoods. I won't state the axioms since wikipedia gives a definition: https://en.wikipedia.org/wiki/Neighbourhood_(mathematics)

A "little" informally, but not that much, we can say that a topological space is set with a directed set $N_x$ at each point $x\in X$ (saying that the elements of this directed set lives in a certain $P(X)$ is not really necessary for the same reason we do not use all the neighbourhoods but just basis of them.) So a continuous function between $X$ and $Y$ is a function between the sets and a morphism of directed sets at each point (when we look at the neighbourhoods as elements of a set $P(X)$ then this morphism of directed sets must be image function.).

Even in this case it is "more" evident how a continuous function preserves the defined structure on the space.

The less satisfactory way to address your old question is by looking at pointless topology. Someone pointed it out but didn't explain why. Formally the relation is the same to the question of morphism of schemes: you take the category of commutative rings and take its opposite. On commutative rings you have structure preserving morphisms and on scheme you have continuous functions and natural transformation of sheaves.

In pointless topology you define a frame to model your definition of the set of open sets: a frame is a partially ordered set with supremum for all subsets and infimum only for finite subsets, with the distributive property.

You define a function to be a morphism if it preserves the arbitrary sup and only finitary inf. The point is that now you take the opposite category to get what are called Locales which contain the category of topological spaces and continuous functions as a full subcategory.

To filter only those objects which are topological space you should consider injective morphisms to frames of the type $P(X)$ and then make the same construction.

It is not in any way satisfactory, but it is natural in some way.

Here is one way to answer the question. The data of a continuous map between sober topological spaces is equivalent to the data of a geometric morphism of the associated toposes of sheaves.

Though the restriction to sober spaces may seem unsatisfying, keep in mind that the full category of all topological spaces is full of pathologies.

An answer among others. A topology on $X$ is obtained by the relation $r\subseteq X\times \mathcal P(X)$ where $(a,S)\in r\iff a\in \overline S$. Then a "canonical morphism" $(X,r)\to (Y,s)$ is a function $f:X\to Y$ such that $x\in \overline{f^{-1}(T)}\implies f(x)\in\overline T$, which means continuity.

But of course it is possible to chose other morphisms as long as the axioms for a category is fulfilled. In metrical spaces there are several different types morphisms that interests mathematicians.

There is probably more than one way to answer this question, so I make no claim to mine being the definite one.

A morphism of topological spaces should be one that is somehow compatible with the topological structure, and indeed as you say there are two natural notions of "compatibility" to consider. To decide which one to use, you should ask yourself what you want isomorphisms of topological spaces to do.

I think it is natural to require that an isomorphism from a topological space $(X,\mathscr{T}_X)$ to another topological space $(Y,\mathscr{T}_Y)$ preserve the topologies exactly, in the sense that the induced morphism $\mathscr{T}_X\to\mathscr{T}_Y$ be an isomorphism in some appropriate category (certainly in $\mathbf{Set}$, but possibly in a more restrictive category as well). The usual notion of a homeomorphism will satisfy this property. Therefore continuous functions are natural morphisms in $\mathbf{Top}$ in the sense that the resulting isomorphisms, i.e. homeomorphisms, preserve the interesting structures of objects in $\mathbf{Top}$ (namely the topologies) exactly.

It's not immediately obvious that the other natural notion of morphism (I think those are open maps?) you consider will have this property of preserving the structure of interest (the topology) when you consider the consequent notion of isomorphisms - or indeed if they preserve any interesting structure at all. (Any homeomorphism will be such a morphism however, since homeomorphisms are open maps.)

This line of reasoning also suggests why measurable functions can be considered "natural" morphisms in the category of measurable spaces - the concept of isomorphism for this notion of morphism provides an isomorphism of the underlying relevant structures.

Of course, this is putting the cart before the horse - whatever notion of morphism for topological spaces we have should be intuitive first, and I think whatever category theory follows we should accept as what it is.

• I'm pretty sure that if you take open maps as your morphisms you get exactly the same "isomorphisms." That is, in the category of topological spaces with open maps as morphisms, the isomorphisms are homeomorphisms in the usual sense of Top. – Joshua Ruiter Apr 21 '17 at 13:13

A perspective not mentioned in the other answers is that we define Top to be the category topological spaces and continuous maps because we are interested in continuous maps.

That is, continuous maps are the right morphisms for the category because they are the morphisms we are interested in.

I think it's actually worth turning the question around:

Is "topological space" a good notion for describing objects this category?

And it turns out that the answer is yes. Let $*$ be the one point space and $S$ the Sierpinski space.

With a little modification, the definition of topological space can be seen to be the axiomatization of the properties of the functors $\hom_{\mathbf{Top}}(*, X)$ and $\hom_{\mathbf{Top}}(X, S)$, which are the "set of points" and "family of open sets" respectively.

(if you've not seen the latter correspondence, given a map $f:X \to S$, the corresponding open set is $f^{-1}(o)$, where $o$ is the open point of $S$)

And furthermore, these functors completely determine the objects: a map $f:X \to Y$ of spaces is an isomorphism if and only if $\hom(*, f) : \hom(*,X) \to \hom(*,Y)$ and $\hom(f,S) : \hom(Y,S) \to \hom(X,S)$ are bijections.

For comparison, sometimes you see people consider the category of metric spaces and continuous maps, and it's quickly clear that this category isn't telling us much about metrics.

Topological Structure

Consider some none-empty sets $X, X^*$. We get a topological structure (i.e. a topology) on $X$, by taking some collection of subsets $\tau \subseteq \mathcal{P}(X)$ which fulfills the following statements

$(1)~~~$ $\emptyset, X \in \tau$ $~~~~~~~~~~~~(2)~~~$ $\mathcal{U} \subseteq \tau ~\Longrightarrow~ \bigcup \mathcal{U} \in \tau$ $~~~~~~~~~~~(3)~~~$ $\mathcal{U} \subseteq_{\mathrm{finite}} \tau ~\Longrightarrow~ \bigcap \mathcal{U} \in \tau$

Similarly, we can equip $X^*$ with a topology $\tau^*$.

A morphism is supposed to preserve the structure, and as I will try to show, there are at least two reasonable ways to define them between topologies. I will call them ,,direct'' and ,,induced'' morphisms.

,,Direct Morphism''

Take the two sets $X = \{ a, b \}$ and $X^* = \{1 , 2 , 3 \}$ with topologies $\tau = \{ \emptyset, \{a\}, X \}$ and $\tau^* = \{ \emptyset , \{1 , 2 \}, X^* \}$ respectively. Then the map \begin{align} \varphi :~ \tau \longrightarrow \tau^* ~~~;~~~ \emptyset &\longmapsto \emptyset \\ \{ a \} &\longmapsto \{1,2\} \\ X &\longmapsto X^* \end{align} is a bijection between the topologies and preserves the structure, since in this (finite) case $$\varphi(U \cap V) = \varphi(U) \cap \varphi(V) ~~~\land~~~ \varphi(U \cup V) = \varphi(U) \cup \varphi(V)$$ Since it does its structure-preserving job so well, we pat it on the back and call it a morphism. The inverse of $\varphi$ preserves the structure similarly well, so we promote it with the name isomorphism in this example.

Based on this idea we could in genral define, that $\varphi : \tau \longrightarrow \tau^*$ is a direct morphism iff

$(1')$ $~~\varphi(\emptyset) = \emptyset ~~\land~~ \varphi(X) = X^*$

$(2')$ $~~\forall~\mathcal{U} \subseteq \tau : \varphi \left( \bigcup \mathcal{U} \right) = \bigcup \varphi \left( \mathcal{U} \right)$

$(3')$ $~~\forall~\mathcal{U} \subseteq_{\mathrm{finite}} \tau : \varphi \left( \bigcap \mathcal{U} \right) = \bigcap \varphi \left( \mathcal{U} \right)$

and a direct isomorphism iff $\varphi$ and its inverse are direct morphisms.

Note here that in our example there is no bijection between $X$ and $X^*$. However they are still isomorphic if we define morphisms in the above way.

,,Induced Morphism''

Let's now consider the situation where we might study the relation of $X$ and $X^*$ via a map $f : X \longrightarrow X^*$. Then we also want to know if there is a reasonable/natural way in which the structure (i.e. topology) on $X$ is compatible with the one on $X^*$.

There are two very natural ways, in which we can get possible candidates for direct morphisms using $f$. Namely \begin{align*} \operatorname{im}_f : \mathcal{P}(X) \rightarrow \mathcal{P}(X^*) ~~~&,~~~ A \mapsto \{ f(a) \mid a \in A \} \\ \operatorname{pre}_f : \mathcal{P}(X^*) \rightarrow \mathcal{P}(X) ~~~&,~~~ B \mapsto \{ x \in X \mid f(x) \in B \} \end{align*} when fittingly restricted to $\tau$ or $\tau^*$. We will now want to see, what conditions $f$ might has to fulfill, so that they really are direct morphisms.

One condition on $f$ is: It has to be such that the maps $\operatorname{im}_f$ and $\operatorname{pre}_f$ really map from one structure into the other structure. More precisely: \begin{align*} \operatorname{im}_f(\tau) \subseteq \tau^* ~~~\text{and}~~~ \operatorname{pre}_f(\tau^*) \subseteq \tau \end{align*} Otherwise the structure could not be preserved. For $\operatorname{im}_f$ we already run into some trouble, since in general we only have $$\operatorname{im}_f \left( \bigcap \mathcal{U} \right) \subseteq \bigcap \operatorname{im}_f(\mathcal{U})$$ and demanding equality would lead to more restrictions on $f$.

However, for $\operatorname{pre}_f$ and any $\mathcal{U} \subseteq \tau^*$ we have \begin{align*} \operatorname{pre}_f(\emptyset) = \emptyset ~~&,~~ \operatorname{pre}_f(X^*) = X \\ \operatorname{pre}_f \left( \bigcup \mathcal{U} \right) &= \bigcup \operatorname{pre}_f(\mathcal{U}) \\ \operatorname{pre}_f \left( \bigcap \mathcal{U} \right) &= \bigcap \operatorname{pre}_f(\mathcal{U}) \end{align*} This shows that $\operatorname{pre}_f$ suffices $(1'), ~(2')$ and $(3')$ i.e. is a direct morphism. Since it was based on the map $f$ we could say $f$ iduces a morphism. Most imporantly is - as noted before - that $\operatorname{pre}_f$ only gives us an induced morphism by $f$ if $$\operatorname{pre}_f(\tau^*) \subseteq \tau$$ holds to be true. This is equivalent to $$\forall ~U^* \in \tau^* :~ f^{-1}(U^*) \in \tau$$ which is the familiar definition of $f$ being continuous.

We define $f$ to induce an isomorphism iff $f$ and its inverse both induce a morphism.

Note that in the above example then, you will not be able to find an induced isomorphism, since there can be no bijective map $f: X \rightarrow X^*$.