Does $\pi_0$ commute with sequential colimits?

Question

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!

Answer 1

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 $Y_n$.