Homotopy groups of $S^\infty$

Question

I have seen that it is possible to see $S^\infty$ is contractible, which gives trivial homotopy groups $\pi_k(S^\infty)=0$ for all $k\geq1$.

Are there different proofs to show the homotopy groups are trivial, besides constructing homotopys with the contraction or over the contraction in general? The space seems so odd i would love to see other ways to work with it.

Answer 1

$S^ \infty$ is the colimit of the inclusions $S^n \rightarrow S^{n+1}$. Since it can be arranged that these are all inclusions of CW complexes, the homotopy groups of the colimit are the colimit of the homotopy groups. We know that $\pi_k (S^n)$ is trivial for $k <n$ we know that $\pi_k(S^\infty)$ is trivial. Another way of saying this is that the n-skeleton is just the n-sphere and since the n-1 homotopy group depends only on the n-skeleton, we can easily deduce these are trivial.

Alternatively, there is a fibration $\mathbb{Z}/2 \rightarrow S^\infty \rightarrow \mathbb{R}P^\infty$ and the long exact sequence in homotopy tells us that $S^\infty$ has trivial homotopy groups if we know that the higher homotopy groups of $\mathbb{R}P^\infty$ are trival. Now normally we would show these groups are trivial by knowing that $S^\infty$ is contractible, but it is possible to do it the other way around. $\mathbb{R}P^\infty$ is the classifying space for line bundles which can readily be seen to be the classifying space for $\mathbb{Z}/2$-bundles which implies that $\Omega \mathbb{R}P^\infty= \mathbb{Z}/2$, so the higher homotopy groups are trivial.

A third, related way, is that $S^\infty$ classifies line bundles over simply connected spaces. Since over any simply connected space a line bundle is orientable, it must also be trivial. Hence, supposing you know the fundamental group is trivial, there are no higher homotopy groups since all the other spheres are simply connected. This answer is nice because it only uses the fundamental group of the sphere which can be shown to be trivial by van Kampen's theorem.

Answer 2

Just to add to the conversation a bit, another way to see that the homotopy groups of $S^ \infty$ are trivial is to note that if $K$ is compact and $X = \text{colim} \ X_i$ where $X_0 \rightarrow X_1 \rightarrow ...$ is a sequence of closed inclsions of $T_1-$spaces we have to have that every map $K \rightarrow X$ factors as $K \rightarrow X_j \rightarrow X$ for a large enough $j$.

Take $K = S^n$, $X_i = S^i$ and $X = S^ \infty$.

The triviality of the homotopy groups of $S ^\infty$ now follow since if $f:S^n \rightarrow S^ \infty$ is any map it factors as $S^n \xrightarrow g S^m \rightarrow S^\infty$. If $m < n$ we just replace $m$ by a number greater than $n$. Then $g$ has to be nullhomotopic since $n < m$ so $f$ is nullhomotopic aswell.

There is also a more general result which says that if the maps $X_i \rightarrow X_{i+1}$ are cofibrations then $\text{colim}_i \ \pi_k(X_i) = \pi_k(X)$. Then we can see that since the sequence $\pi_k(S^1) \rightarrow \pi_k(S^2) \rightarrow ...$ eventually becomes zero it's colimit also has to be zero. Thus $\pi_k(S^ \infty) = 0$.