# Use Whitehead 's theorem to show that a $S^{\infty}$ is contractible.

Use Whitehead 's theorem to show that a $$S^{\infty}= \cup_{n =1}^{\infty} S^n$$ is contractible.

And I know that a space is contractible iff it is homotopy equivalent to a point.

My questions are:

what are the 2 connected CW complexes that I should consider and what is the map between them and how can I show that this map induces isomorphisms from $$\pi_{n}(X)$$ to $$\pi_{n}(Y)$$ for all $$n.$$ Could anyone help me please in this?

• You always have a map from the one-point space to any (nonempty) space, so consider $f: \{*\} \to S^{\infty}$. Then what are the homotopy groups of each of these spaces? Dec 5, 2019 at 15:15
• I do not know I will search the book and answer you @kamills. but could you tell me please which pg of AT give us the information you mentioned in your first statement?
– user591668
Dec 5, 2019 at 15:22
• To compute $\pi_n(S^{\infty})$ take a map $g:S^n\to S^{\infty}$. Note that its image must be contained in some $S^m$, since the image is a compact. Then you can contract $g$, inside $S^{m+1}$, to a constant map. Dec 5, 2019 at 15:23
• Why we should take a map from $S^n$ and how I can contract this $g$ to a constant map ?@conditionalMethod
– user591668
Dec 5, 2019 at 15:26
• You can contract the equator sliding it up to the north pole, for example. Dec 5, 2019 at 15:45

Claim: Any map $$* \to S^\infty$$ is a weak homotopy equivalence, i.e. induces isomorphisms on all homotopy groups.

It suffices to show $$\pi_n S^\infty=0$$ for all $$n$$. To do this we can use the following lemma (whose proof I leave to you as an exercise):

Lemma: Any continuous function $$X\to S^k$$ which is not surjective is null-homotopic.

Proof sketch for Claim: To compute $$\pi_n S^\infty$$, consider a continuous function $$f\colon S^n \to S^\infty$$. Since the image of $$f$$ is compact and $$\cup_k S^k$$ is given the colimit (weak) topology, the image of $$f$$ must be contained in some $$S^k$$ for a finite $$k$$, and hence $$f$$ is not surjective onto $$S^{k+1}$$. Therefore by Lemma $$f$$ is null-homotopic in $$S^{k+1}$$ and hence in $$S^\infty$$.

Now we can conclude $$S^\infty$$ is contractible by Whitehead's theorem.

• Is this claim proved in AT? if so on which pg.?
– user591668
Dec 5, 2019 at 22:47
• Is this Lemma found in AT? if so at which page?
– user591668
Dec 5, 2019 at 22:48
• I feel like these can be found in Hatcher but I'm not sure where, it's possible at least one of them is an exercise. I will give you a hint for the lemma though: supposing $f\colon X\to S^k$ is continuous and $p\in S^k$ is not in the image of $f$, construct a homotopy between $f$ and the function which is constantly $-p$ (write down a straight-line homotopy, but normalize it so it is on the sphere). Dec 6, 2019 at 3:18