3
$\begingroup$

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.

$\endgroup$
11
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ 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. $\endgroup$ – Denis Nardin Oct 13 '18 at 11:46
3
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ 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. $\endgroup$ – Kevin Carlson Oct 13 '18 at 20:03

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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