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!

  • $\begingroup$ I would also like to know the answer if the spaces are assumed to be compactly generated weak Hausdorff, or something stronger. $\endgroup$ – Justin Young Aug 8 '13 at 21:41
  • $\begingroup$ 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 $\endgroup$ – Julian Kniephoff Aug 9 '13 at 12:14

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.