8
$\begingroup$

I know Hilton's result about a finite wedge of spheres, and I know that certain homotopy groups (such as the third homotopy group) can be directly calculated for an infinite wedge too.

My question is -- Is there some general result that gives the homotopy groups of an (uncountable) infinite wedge of 2-spheres in terms of the homotopy groups of 2-spheres (or other simpler spaces)?

Thanks.

$\endgroup$
1
  • 3
    $\begingroup$ Here are a couple of elementary observations which may or may not be helpful to you. Homotopy groups commute with filtered colimits of pointed topological spaces under fairly mild conditions. Moreover, an infinite wedge sum coincides with the filtered colimit of the finite wedge sub-sums. $\endgroup$ Sep 23, 2013 at 20:21

1 Answer 1

14
$\begingroup$

For any infinite wedge $X=\bigvee_{i\in I} X_i$, the homotopy groups are just the colimit $$\pi_n(X)=\operatorname{colim}_F \pi_n\left(\bigvee_{i\in F} X_i\right),$$ where $F$ ranges over all finite subsets of $I$. The reason for this is simple: $S^n$ is compact, so the image of any map $S^n\to X$ must be contained in $\bigvee_F X_i$ for some finite $F$; this means the natural map from the colimit to $\pi_n(X)$ is surjective. Similarly, since $S^n\times I$ is compact, any homotopy also lands in a finite wedge, so the natural map is injective. The same reasoning would apply if you replaced $S^n$ with any compact space, and the infinite wedge with any colimit for which any compact subset factors through some stage (eg, any filtered system of sub-CW complexes).

$\endgroup$
3
  • $\begingroup$ That's of course an important observation, but my impression was that it is not the real difficulty of this problem. Namely, from the statement of the Hilton-Milnor theorem it seems hard to conclude what the maps $\pi_n(\bigvee_{i\in F} X_i)\to\pi_n(\bigvee_{i\in F'} X_i)$ are precisely for two finite sets $F \subseteq F'$. Could anyone shed any light on this? $\endgroup$ Sep 24, 2013 at 5:19
  • $\begingroup$ I may be overlooking some subtlety, but I don't see anything hard about that. The Hilton-Milnor theorem gives a direct sum decomposition of $\pi_*(\bigvee_F X_i)$ (if each $X_i$ is a suspension), and if $F\subset F'$ the set of summands for $\bigvee_F X_i$ are naturally a subset of the set of summands for $\bigvee_{F'} X_i$. The induced map then just comes from the inclusion of these summands. $\endgroup$ Sep 24, 2013 at 6:40
  • 1
    $\begingroup$ After a closer look, I agree that the naturality is exactly as you described. Still, at least to me understanding the naturality of the Hilton-Milnor theorem was at least as hard as understanding that homotopy groups commute with filtered colimits. $\endgroup$ Sep 24, 2013 at 7:26

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.