A poset $(P,\le)$ can be turned into a category in a standard way. A group $(G,\cdot)$ can also be turned into a category in a standard way. Can we topological space $(X,\tau)$ into a category in a meaningful/useful way? In general, what pairs $(X,\eta)$ consisting of a set $X$ and a collection of subsets $\eta\subseteq\mathcal{P}(X)$ (with some properties) can be turned into a category?

I'm assuming that "meaningful way" means something like 1)you can recover the object from the category and 2)functors between these categories correspond to the appropiate morphisms between the objects


  • 1
    $\begingroup$ For the first one, just define the poset category $(\tau,\subseteq)$. Then, continuous functions seem to correspond exactly to contravariant functors between these categories. $\endgroup$ – WoolierThanThou Feb 16 '20 at 18:46
  • 5
    $\begingroup$ @WoolierThanThou The correspondence isn't "exact". For instance, consider some space $X$ with indiscrete topology $\tau$. Then there are many maps $X\to X$ and they are all continuous, but they induce the same map on the poset $(\tau,\subseteq)$. $\endgroup$ – Wojowu Feb 16 '20 at 19:07
  • $\begingroup$ There's a way to define a topology on $X$ categorically by looking at the category of subobjects of $X$ and restricting to the open ones, continuity of $f:X\to Y$ is then characterized by the fact that the fibered product in Top of every open subobject of $Y$ along $f$ is an open subobject of $X$, but this not as nice as OP was hoping for I fear $\endgroup$ – Alessandro Codenotti Feb 16 '20 at 19:14
  • $\begingroup$ The spezialization preorder gives a concrete faithful functor $\mathsf{Top} \to \mathsf{Pre}$. Composing this with the mentioned functor $\mathsf{Pre} \to \mathsf{Cat}$, we get a concrete faithful functor $\mathsf{Top} \to \mathsf{Cat}$. Notice, however, that it isn't full in general. A more useful functor is the fundamental groupoid $\Pi : \mathsf{Top} \to \mathsf{Cat}$. The general question is too general to have a reasonable answer. $\endgroup$ – Martin Brandenburg Feb 23 '20 at 14:15

Your Answer

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

Browse other questions tagged or ask your own question.