This question is inspired by this Mathoverflow question.

Ignoring size issues, there is a natural way to view a category as a first-order structure in a finite language. In light of this we can ask about the computational complexity of the first-order theory of a given category. Note that a lot of important structure is lost when passing to the first-order level; nevertheless, it still seems interesting to me.

Plenty of categories are easily checked to have very complicated first-order theories - for example, assuming the axiom of choice in the background we can identify the finite sets as those for which every self-injection is a surjection, and since Cartesian products and disjoint unions give the usual arithmetic we have that the first-order theory of the category Sets is not computable.

Incidentally, choice isn't needed here, but the non-choice argument is a bit more tedious.

My question is:

Is the first-order theory of the category Top of topological spaces (with morphisms being continuous maps) computable?

My suspicion is that the answer is no. An obvious way to prove this would be to show that the finite discrete spaces are first-order characterizable here, since then we could run the same argument as for Sets; however, I don't see how to do this. (In particular, the notion of "compact objcet" of a category is not obviously first-order expressible, so that characterization of finite discrete spaces doesn't seem to help.)

There are weaker versions of this question which could be asked:

I am also interested in comments along these lines; however, my main question is specifically about the first-order theory of Top.

EDIT: David Roberts brought to my attention Schlomiuk's paper An elementary theory of the category of topological spaces; among other things, this gives a computable (indeed, finite) theory characterizing Top up to categorial equivalence amongst complete categories, as well as in a more subtle sense (any such category is equivalent to Top$^\mathfrak{S}$ for some model $\mathfrak{S}$ of ETCS).


2 Answers 2


Let me assume choice. It is possible to distinguish discrete spaces from other spaces: those are precisely the spaces $X$ such that for any epimorphism $f:Y\to X$ admits a section $g:X\to Y$ (i.e. $g\circ f=\mathrm{id}_Y$). To see this, recall first that epimorphisms in Top are just the surjective continuous maps.

If $X$ is discrete, then for any epimorphism $Y\to X$ there is a set-theoretical section by AC, which is automatically continuous.

Conversely, suppose every epimorphism into $X$ splits. Let $Y$ have the same underlying space as $X$, but discrete topology. Then the identity on points is a continuous map from $Y\to X$, and since it's an epimorphism, it admits a section, which shows that $X$ is discrete too.

The subcategory of discrete topological spaces is equivalent to the category of sets, and you already know it's uncomputable. It follows the first-order theory of Top is undecidable.

It's not clear to me if this proof can be modified so as to avoid the use of AC.

Edit: a proof without the use of AC: we will again distinguish the subcategory of discrete spaces. Let $1$ be a one-point space (a terminal object) and $2$ a two-point discrete space (a space with exactly two maps from $1$ to it, and with $4$ maps from it to itself) and $i:1\to 2$ any map. Then $X$ is discrete iff for any map $f:1\to X$ there is a map $g:X\to 2$ such that $g\circ f=i$, and for any $f':1\to X,f'\neq f$, we have $g\circ f'\neq i$.

This expresses that for any point $x\in X$, there is a map $X\to 2$ which takes $x$ to one point, and all other points to a different one. Since $2$ was discrete, this implies $\{x\}$ is open, so $X$ is discrete.

Since you say you know Set has uncomputable theory even without AC, we deduce the same for Top.

  • $\begingroup$ Neat, I didn't think of that! I'll hold off on accepting for a bit to see if a choice-free answer can be whipped up. $\endgroup$ Mar 3, 2019 at 21:14
  • $\begingroup$ @NoahSchweber Added a different, AC-free proof :) $\endgroup$
    – Wojowu
    Mar 3, 2019 at 21:24
  • $\begingroup$ Beautiful - accepted. $\endgroup$ Mar 3, 2019 at 21:26
  • $\begingroup$ Per your last line, see Eric's answer, which is what I had in mind for Sets. $\endgroup$ Mar 3, 2019 at 21:50

Here is a method of defining natural numbers in $\mathbf{Set}$ (without Choice) which generalizes immediately to $\mathbf{Top}$. Namely, we can just express the second-order Peano axioms directly in our categorical language. Let us fix a terminal object $1$. Then we can define the natural numbers as a triple $(N,s,0)$ where $N$ is an object, $0:1\to N$ is a morphism, and $s:N\to N$ is a morphism such that they satisfy the second-order Peano axioms, namely:

  • $s$ is injective (i.e., monic).
  • $0$ is not in the image of $s$ (that is, $0$ cannot be factored as a composition $s\circ f$ for any $f:1\to N$).
  • No proper subobject of $N$ contains $0$ and is closed under $s$. That is, if $i:A\to N$ is any monic morphism such that $0$ factors through $i$ and $s\circ i$ factors through $i$, then $i$ is an isomorphism.

For $\mathbf{Set}$, these properties pick out the usual structure of natural numbers (up to isomorphism), and for $\mathbf{Top}$, they pick out the natural numbers with the discrete topology (proof: given any space $N$ with element $0$ and map $s:N\to N$ satisfying the first two conditions, the subset generated by $0$ under $s$ can be given the discrete topology to get a space $A$ and a continuous injection $i:A\to N$, and then the third condition implies $i$ is an isomorphism).

(As an aside, while these axioms work in $\mathbf{Set}$ or $\mathbf{Top}$, there are other axioms for natural numbers besides the Peano axioms which are more natural from a categorical perspective, such as the notion of a natural numbers object.)

We can then fix any such $(N,s,0)$ and define a natural number as a morphism $1\to N$. We can define morphisms $+:N\times N\to N$ and $\cdot:N\times N\to N$ as the unique morphisms satisfying the usual recursive definitions, and these give binary operations on our natural numbers, so we can interpret arithmetic in our category.

  • $\begingroup$ This was the argument I had in mind for the choice-free version w/r/t Sets - stupid though it seems, I didn't notice that it worked for Top as well! $\endgroup$ Mar 3, 2019 at 21:50

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