Is it possible that isomorphic $\pi_n$'s not induced by a map? 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? 
 A: No, there are examples of spaces for which all homotopy groups are isomorphic, but not map induces these isomorphisms for all $n$ simultaneously.
For example, let $A = \mathbb{R}P^2\times S^3$ and let $B = S^2\times \mathbb{R}P^3$.
The universal cover of $A$ and $B$ are both $S^2\times S^3$, which implies all the higher homotopy groups are isomorphic.  Further, both fundamental groups are isomorphic to $\mathbb{Z}/2\mathbb{Z}$.
So, why isn't there a map which induces an isomorphisms on all $n$?  By Whitehead's theorem, if there was such map, then $A$ and $B$ would be homotopy equivalent.  They are not since, for example, the top homology group $H_5$ is trivial for $A$ but non-trivial for $B$.
A: Nah. Take, for instance, $S^3 \times \Bbb{CP}^\infty$ and $S^2$. They have the same homotopy groups (check the long exact sequence of the fibration $S^1 \to S^3 \to S^2$) but are not homotopy equivalent (see the homology).
A: 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.
