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