# Loop spaces and filtered colimits

I have read many things now that lead me to believe that the loop space functor preserves filtered (and/or directed) colimits. Is this true? And can somebody give a (sketch of a) proof or point me in the direction of one?

I see why this is true for a directed system of (closed?) inclusions, but I explicitly want to know the answer for arbitrary maps.

I also know that this is true in a "homotopical sense" (meaning that $\pi_1$ commutes with filtered colimits and some other stuff about homotopy colimits which I don't quite understand, yet), but before going down that road, I want to be certain about the space level statement.

I think I got it; thanks to this discussion: http://golem.ph.utexas.edu/category/2009/05/journal_club_geometric_infinit_3.html#c023790

The question boils down to whether or not the natural map

$$\operatorname{colim}_i\Omega X_i \to \Omega \operatorname{colim}_i X_i$$

is an isomorphism, i.e. a homeomorphism for any filtered/directed system $\{X_i\}$. In this strict sense and in this generality however, this is false!

Remember that -- as a set -- $\Omega X = \hom_{\mathrm{Top_*}}(S^1, X)$ for any space $X$, so we can generalize the situation a bit and proof the following: If for every directed system $\{Y_i\}$ the natural map

$$h\colon \operatorname{colim}_i \hom_{\mathrm{Top}_*}(X,Y_i) \to \hom_{\mathrm{Top}_*}(X,\operatorname{colim}_i Y_i)$$

is a bijection, then $X$ is discrete!

To show this, let $U \subseteq X$ be any subset. We want to show that this is open. For that we are going to construct an appropriate directed system: Let $Y_i := X \coprod \mathbb{N}$ equipped with the topology $\{U \coprod \{n\in\mathbb{N} | n \ge k\} | k \ge i\} \cup \{Y_i, \emptyset\}$. The maps $Y_i \to Y_{i+1}$ are given by the (setweise) identity, which is obviously continuous.

Then $\operatorname{colim}_i Y_i = X \coprod \mathbb{N} =: Y$ with the indiscrete topology, since if $V \subset X \coprod \mathbb{N}$ is open, then it is already open in all the $Y_i$, so it has to be empty actually.

Let's now look at the inclusion $j\colon X \to Y$. Since $h$ is a bijection, we get a continuous map $[j]:= h^{-1}(i) \in \operatorname{colim}_i \hom_{\mathrm{Top}_*}(X, Y_i)$ with representative $j\colon X \to Y_i$ for some $i$. such that the composite $X\stackrel{j}{\to} Y_i \to Y$ is $i$.

But then $U = j^{-1}(U \coprod \{n \in \mathbb{N} | n \ge i\})$ and this is open!

• I would also like to know the answer if the spaces are assumed to be compactly generated weak Hausdorff, or something stronger. – Justin Young Aug 8 '13 at 21:41
• Well at least the filtered case seems to be "hopeless"; thinking and searching a bit more about this brought me to the following example: A compact metric space $K$ is the filtered colimit of its countable compact subspaces $K_i$. Now $S^1 \to K$ cannot factor through $K_i$ since $S^1$ is uncountable. Also there seems to be a way to "replace" a filtered diagram with a directed one? I haven't read this yet, but this might also "kill" the directed case: matwbn.icm.edu.pl/ksiazki/bcp/bcp9/bcp919.pdf – Julian Kniephoff Aug 9 '13 at 12:14