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

Question

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.

Answer 1

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.

Answer 2

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.