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I'm looking for an example of a sequence of topological spaces $Y_1 \rightarrow Y_2 \rightarrow \cdots$ such that the induced map $$ \text{colim}\, \pi_0(Y_i) \rightarrow \pi_0 (\text{colim}\, Y_i) $$ is not a bijection. I believe it's clear that the map is always surjective. I don't care if the spaces are Hausdorff, if the maps are embeddings, etc. Many thanks for either such an example or an explanation of why one does not exist!

Stefan Hamcke
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CJD
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  • The body and title question don't quite match; the body seems to be about a particular shape of colimit. – Qiaochu Yuan May 29 '14 at 06:26
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    $\pi_0$ is morally the left adjoint to the inclusion functor from discrete spaces to spaces; as such, morally it should preserve colimits. This doesn't quite work out for spaces that are not locally path connected, unfortunately. That might be a clue as to where to look for a counterexample. – Qiaochu Yuan May 29 '14 at 06:28
  • Hi, thanks for the comments! Feel free to edit the title, I actually tried to decide what this sort of colimit was called and failed. – CJD May 29 '14 at 06:36
  • Are there any restrictions to the maps $Y_n\to Y_{n+1}$, for example that they are all injections or embeddings? – Stefan Hamcke May 29 '14 at 13:11
  • Nope, I don't intend any restrictions on the maps (or on the spaces). I suppose as Qiaochu pointed out I have restrictions on the type of colimit. I believe there is an elementary topology argument saying that if the maps were closed embeddings (and maybe the spaces were Hausdorff?) then $\pi_0$ would commute with the colimit. – CJD May 29 '14 at 13:18
  • Yes, if you have a directed system of pointed spaces $(X_\alpha,x_0)$ (meaning that for any two spaces $X_α,X_β$, there's an $X_γ$ with pointed maps $i_{αγ}:X_α→X_γ, i_{βγ}:X_β→X_γ$), and if all $i_{αβ}$ are embeddings, then the colimit $(X,x_0)$ can be seen as the union of all $X_α$. We then have an isomorphism $\text{colim} π_i(X_α,x_0)\cong π_i(X,x_0)$ for all $i\ge 0$. If $i>0$, the colimit is formed in the category of groups, if $i=0$, it's in the category of pointed sets. – Stefan Hamcke May 29 '14 at 19:34

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Here's a simple counterexample. Let $X=\mathbb{N}\cup\{\infty\}$ and let $Y_n$ be the quotient of $X$ that collapses $\{0,\dots,n\}$ to a single point. These naturally form a direct system and the colimit is a two-point space $Y$ (one point being the image of $\mathbb{N}$ and the other point being the image of $\infty$) which is path-connected (since every nonempty open set must contain the image of $\mathbb{N}$). On the other hand, each $Y_n$ is totally disconnected, so the colimit of $\pi_0(Y_n)$ is just the set $Y$ with two points, even though $Y$ is path-connected.

You can also get a counterexample where all the spaces (including the colimit) are Hausdorff. For instance, let $X$ be the Cantor set and one-by-one glue together the pairs of endpoints of each "hole" (the open intervals you remove from $[0,1]$ to form the Cantor set). This gives a sequence of quotients $Y_n$ that are all totally disconnected (in fact, they are also Cantor sets) but the colimit is actually homeomorphic to $[0,1]$ (the quotient map from $X$ to the colimit is the Cantor function).

It is true if the spaces $Y_n$ are $T_1$ and the maps $Y_n\to Y_{n+1}$ are embeddings. That's because under those assumptions, every compact subspace of the colimit will be contained in some $Y_n$ (see here for instance) and in particular this includes the image of any path, so if two points of the colimit are in the same path-component, there is a path between them in some $Y_n$.

Eric Wofsey
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