TITLE: Is it possible that isomorphic $\pi_n$'s not induced by a map?
QUESTION [6 upvotes]: This might be a stupid question, but I want to make sure as a beginner of AT.
A map between CW-complexes $f: A \rightarrow B$ is defined to be a weak homotopy equivalence if it induces isomorphisms $f_*: \pi_n(A) \rightarrow \pi_n(B)$ for all $n$. But is it true that $\pi_n(A) \cong \pi_n(B)$ for all $n$ implies that such a map $f$ exists? If not, what is the reason? And what is a good counterexample?

REPLY [6 votes]: Another easy/interesting counter example: Consider $X=S^1\vee S^3$ and its double cover $X_2$, i.e attach two copies of $S^3$ one in north pole and one in south pole of $S^1$. Then $\pi_1(X) \cong \mathbb{Z} \cong \pi_1(X_2)$ and the covering map induces isomorphisms in $\pi_n$ for all $n\geq 2$. But $X$ and $X_2$ are not homotopically equivalent since their Euler Characteristics are different. So there cannot be any induced isomorphism by Whitehead theorem.