# Weak homotopy equivalence between $\Omega \underset{\rightarrow}{\lim}Z_n$ and $\underset{\rightarrow}{\lim}\Omega Z_n$

Let $$Z_1 \rightarrow Z_2 \rightarrow\cdots$$ be an arbitrary sequence of CW-complexes and let $$\Omega X$$ denote the loop space over $$X$$. In Allen Hatcher's "Algebraic Topology" (http://pi.math.cornell.edu/~hatcher/AT/AT.pdf, section 4.F, last lines) it's stated that the natural map $$\underset{\rightarrow}{\lim} \ \Omega Z_n \rightarrow \Omega \underset{\rightarrow}{\lim} \ Z_n$$ is a weak homotopy equivalence (the map is given by the universal property of the direct limit); I also recall that the direct limit of a sequence of CW-complexes is the mapping telescope. I'm trying to prove this fact but I don't know how to proceed; in particular, I don't know how to relate the homotopy groups of the mapping telescope to the homotopy groups of the spaces $$Z_n$$. There is a relation for the homology groups, namely $$H_i(\underset{\rightarrow}{\lim}\ Z_n)\simeq\underset{\rightarrow}{\lim}H_i(Z_n)$$ but I don't know how this can help.

First note that Hatcher's exercise says "where direct limits mean mapping telescopes", so he is defining $$\underset{\rightarrow}{\lim}$$ to mean the telescope. I disapprove of that quite strongly. The telescope is the same as the homotopy colimit, and the standard notation for that is $$\underset{\rightarrow}{\text{holim}}$$. Anyway, if we write $$T$$ for the telescope, then it is the union of closed subspaces $$T_n$$ with $$T_n$$ homotopy equivalent to $$Z_n$$, and $$T$$ is topologised as the colimit of the subspaces $$T_n$$. There is a standard lemma about this situation: if $$K$$ is a compact subset of $$T$$ then we can choose a point $$x_n\in K\cap (T_n\setminus T_{n-1})$$ for each $$n$$ such that $$K\cap (T_n\setminus T_{n-1})\neq\emptyset$$, and we find that the set of all $$x_n$$'s is discrete and compact and therefore finite, so $$K$$ is contained in $$T_n$$ for some $$n$$. By applying this to the image of an arbitrary based map $$S^k\to T$$, we see that $$\Omega^k T$$ is the union of the subspaces $$\Omega^k T_n$$. A similar argument with $$[0,1]\times S^k$$ shows that the homotopies match up, so $$\pi_k(T)$$ is the colimit of the groups $$\pi_k(T_n)\simeq\pi_k(Z_n)$$. Together with the isomorphism $$\pi_k\Omega = \pi_{k+1}$$, this will give what you need.
• I think there is a solid case to be made to stop inserting the $\mathrm{ho}$ for homotopy limits and colimits, but I agree that Hatcher's book is probably not the right place for it. – Denis Nardin Oct 13 '18 at 11:46
It's because $$S^1$$ is a compact object in spaces (in this case, literally compact) and therefore maps out of it commute with filtered colimits. I don't know anything about CW complexes, but for simplicial sets, that is actually an isomorphism.
• If by "literally compact" you mean open-cover compact, then this isn't actually the property you want for commutation with filtered colimits. In fact there aren't any compact objects in spaces, although open-cover compact spaces look like compact objects with respect to filtered colimits of sufficiently nice inclusions. Furthermore, open-cover compact spaces are not generally compact objects in the $\infty$-category of spaces (e.g. the so-called Hawaiian earring.) Of course, this all looks pretty pathological from a certain perspective. – Kevin Carlson Oct 13 '18 at 20:03