# Which functors preserve the number of connected components?

The categories $$\mathbf{Top}$$ of topological spaces, $$\mathbf{sSet}$$ of simplicial sets and $$\mathbf{Cat}$$ of small categories are all equipped with a functor $$\pi_0$$ into the category $$\mathbf{Set}$$ of sets, which is a left adjoint and measures the number of connected components. There are also plenty of functors between the upper categories, which gives rise to the question, if they preserve the number of connected components.

I have already proven this for $$|-|\colon\mathbf{sSet}\leftrightarrows\mathbf{Top}\colon \operatorname{Sing}$$, for $$\tau\colon\mathbf{sSet}\leftrightarrows\mathbf{Cat}\colon N$$ and for $$\operatorname{Sd}\colon\mathbf{sSet}\leftrightarrows\mathbf{sSet}\colon\operatorname{Ex}$$. The functor $$\operatorname{Ex}^\infty=\varinjlim_{n\in\mathbb{N}}\operatorname{Ex}^n\colon\mathbf{sSet}\rightarrow\mathbf{sSet}$$ still works fine since the colimit commutes with the left adjoint $$\pi_0$$. The problem is, that $$\operatorname{Sd}^\infty=\varprojlim_{n\in\mathbb{N}}\operatorname{Sd}^n\colon\mathbf{sSet}\rightarrow\mathbf{sSet}$$ does not work fine because of the limit and since it is not the left adjoint of $$\operatorname{Ex}^\infty$$ and also writing either $$X$$ or $$\operatorname{Sd}^n(X)$$ as a colimit doesn't seem to work out. Does $$\operatorname{Sd}^\infty$$ also preserve the number of connected components? If yes, how is it proven and if no, is there a counterexample?

The fact, that there are many other categories with a forgetful functor to $$\mathbf{Set}$$ and left adjoints (free object functor) or right adjoints (cofree object functor) like we have for the three categories above, gives also rise to some more questions: Are there more categories with a functor like $$\pi_0$$ (like for example the category $$\mathbf{Graph}$$ of graphs) giving the number of connected components and functors to other categories with one, which I have not yet considered? If yes, is the number of connected components preserved? Are there other categories with functors similar to $$\pi_0$$ in the sense that they measure something similar, so we can look at more possible preservations?

• Can the one downvoting and voting to close please elaborate? Are there any rules I have broken or what is the problem? Should I delete the added part with the proofs as it is not necessary for the question? Is there something else wrong with the question? I just don't understand it, I have been preparing this question for two whole weeks now and paid attention that every detail fits. Oct 29, 2022 at 16:55
• I thought it is expected here to provide enough context for the question and for what you know and understand so far? My first question, where I also showed a lot of what I did so far, has attracted two downvotes yet, but the second question, where I just took three minutes to only ask a question without almost any own thoughts did not? Is that how you are supposed to do it here? Oct 29, 2022 at 17:13
• I did not downvote your question, but I see that there is a close vote (five close votes needed to close a question) with the tag "Not research-level mathematics". I think you should interpret the downvote as another user expressing their opinion that this question is a better fit for Math.SE. (My sense, as a non-category theorist, is that this statement is probably in the literature somewhere and that it's also possible to prove by computation using the definitions.) I imagine this is frustrating, especially with unexplained downvotes, but it's best not to interpret as a value judgement.
– mme
Oct 29, 2022 at 17:24
• I also think there are some ways you could have formatted your question to make it more immediately accessible. (a) I think it would be easier to understand exactly what you're asking if your question was on one line instead of in a big diagram: "Do functors F, G, H, ... preserve connected components?" (b) I do not have the sense that the included proofs will help experts understand your question, as they are likely able to reproduce these proofs themselves; the longer a question is the less likely someone will make it to the end, so it may be better to remove them.
– mme
Oct 29, 2022 at 17:24
• An interesting and straightforward question that’s not really research level might get attention, but a sprawling one that’s not very interesting might be treated more strictly. It seems the question here is really, does $\mathrm{Sd}^\infty$ preserve connected components? I think that’s basically a fine question, but there’s no motivation here that makes me think it would be particularly interesting. Also, your bio may make people suspicious of whether you’re going to be easy to have conversations with. I’d at least consider removing sexual comments. Oct 29, 2022 at 17:48

Ignoring issues with what TOP really should be, let me focus on the question whether $$\pi_0(Sd^\infty(X))=\pi_0(X)$$ for a simplicial set $$X$$. If we look at $$X=\Delta^1$$, we should get a counterexample. Here $$\pi_0$$ should denote the equivalence class of vertices, where two vertices are equivalent, if there is a finite path (a list of one simplices ignoring the direction) joining them. In the inverse limit, we have uncountably many zero simplices, but only countably many (non-degenerate) one simplices. Thus $$\pi_0(Sd^\infty(\Delta^1))$$ must be uncountable.

A long time ago I thought a bit about these situations. Morally, I think there category of sets is somehow the wrong category for such questions. We can view any set as a discrete topological space. But then that functor $$SET\to TOP$$ is not compatible with limits.

I believe I could show a statement of the form that if $$(X_n)_{n\in \mathbb{B}}$$ is a inverse system of m-dimensional, simplicial (totally disconnected compact Hausdorff-spaces), then the canonical map $$|\lim_n X_n|\to \lim_n |X_n|$$ is a homeomorphism, where $$m$$-dimensional means that $$Sk^m(X_n)\to X_n$$ is a homeomorphism for all $$n$$ and $$|-|$$ is the non-fat geometric realization (the one that people would call the wrong one).

My motivation was just to write the Hawaiian earrings or compatifications of a tree as the geometric realization of something. That seemes to work, as long as we also allow a topology on the $$n$$-simplices. Funnily then one can also write $$\Delta^1$$ as the geometric realization with the Cantor-Set as its Zero-Skeleton.

I haven't published that but I might look it up if there is interest.

• As someone who's studied category theory for 3+ years, your second to last sentence (beginning with 'funnily enough') is the first time I've ever felt tempted to look into the simplical approach to higher categories. Very cool! Oct 30, 2022 at 4:16
• Another way to see that $Sd^\infty$ is not path connected is to assume that there is a path of some length $m$ connecting the endpoints. Applying the simplicial map $Sd^\infty(\Delta^1)\to Sd^n(\Delta^1)$ yields a path of length $m$ connecting the endpoints in $Sd^n(Delta^1)$ for all $n$. But by construction the shortest path connecting the endpoints has length $2^n$. So we get a contradiction for large $n$. Oct 31, 2022 at 6:57

You don't give your definition of $$\pi_0$$ on $$\text{Top}$$, but since you mention left adjoints I assume it is the left adjoint of the inclusion $$\text{Set} \to \text{Top}$$ of discrete spaces into $$\text{Top}$$. If so, this functor does not give the connected components of a topological space; it gives the quasicomponents. does not exist, because the inclusion of discrete spaces does not preserve infinite products; thanks to Emily in the comments for the correction. The functor $$\pi_0 \text{Sing}(X)$$ also does not give the connected components of a topological space; it gives the path components.

There are topological spaces whose quasicomponents are larger than their connected components, and/or whose connected components are larger than their path components; either of these give a counterexample to your claim that $$\text{Sing}(X) : \text{Top} \to \text{sSet}$$ preserves $$\pi_0$$ as you define it.

• @SamuelAdrianAntz In the last step, it looks like you are implicitly assuming that $X$ equals the colimit over simplices of $|\Delta^n|$. In other words, you're assuming that $X = |Sing(X)|$-- which is not true. Oct 29, 2022 at 18:46
• Right. $X$ isn't even homotopy equivalent to $|\text{Sing}(X)|$ in general (only weakly homotopy equivalent), and the difference can already be detected at the level of $\pi_0$. Oct 29, 2022 at 18:59
• Ah, of course, I oversaw that. I tend to confuse it with the adjunction $\tau\dashv N$, where $N$ is fully faithful, and $\tau N\mathcal{C}\cong\mathcal{C}$. Thank you both! Oct 29, 2022 at 19:00
• @QiaochuYuan Hi Qiaochu! I'm a bit confused about the adjointness property: while you mention that the set of quasicomponents is left adjoint to the discrete space functor, the nLab mentions here (in the "the left adjoint of the discrete space functor" section) that the discrete space functor does not admit a left adjoint since it doesn't preserve infinite products. Do you know what's going on? Feb 4 at 16:33
• @Emily: oops. Yeah, that seems true. I'm not sure what I had in mind there. Feb 4 at 19:20