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

Whitehead thm.

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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?

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    $\begingroup$ 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? $\endgroup$
    – kamills
    Dec 5, 2019 at 15:15
  • $\begingroup$ 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? $\endgroup$
    – user591668
    Dec 5, 2019 at 15:22
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    $\begingroup$ 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. $\endgroup$ Dec 5, 2019 at 15:23
  • $\begingroup$ Why we should take a map from $S^n$ and how I can contract this $g$ to a constant map ?@conditionalMethod $\endgroup$
    – user591668
    Dec 5, 2019 at 15:26
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    $\begingroup$ You can contract the equator sliding it up to the north pole, for example. $\endgroup$ Dec 5, 2019 at 15:45

1 Answer 1

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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.

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  • $\begingroup$ Is this claim proved in AT? if so on which pg.? $\endgroup$
    – user591668
    Dec 5, 2019 at 22:47
  • $\begingroup$ Is this Lemma found in AT? if so at which page? $\endgroup$
    – user591668
    Dec 5, 2019 at 22:48
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    $\begingroup$ 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). $\endgroup$
    – William
    Dec 6, 2019 at 3:18

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