Map $\langle X, Y\rangle \to \text{Hom}(\pi_n(X), \pi_n(Y))$ bijection?

Question

I have two questions.

1. How do I see that the map$$\langle X, Y\rangle \to \text{Hom}(\pi_n(X), \pi_n(Y)), \quad [f] \mapsto f_*,$$is a bijection if $X$ is an $(n - 1)$-connected CW complex and $Y$ is a path-connected space with $\pi_i(Y) = 0$ for $i > n$?
2. How does this imply that CW complex $K(G, n)$'s are uniquely determined, up to homotopy type, by $G$ and $n$?

Answer 1

2. Let $n>0$ and $G$ be a group (abelian if $n>1$). Let $K$ and $K'$ be two CW complexes such that $\pi_n(K)=\pi_n(K')=G$ and $\pi_i(K)=\pi_i(K')=0$ for any positive integer $i\ne n$. The functoriality of $\pi_n$ gives the commutative diagram $\require{AMScd}$ \begin{CD} [K,K']\times[K',K] @>{i_{KK'}\times i_{K'K}}>> \mathrm{Hom}(G,G)\times\mathrm{Hom}(G,G)\\ @V{\circ}VV @V{\circ}VV\\ [K,K] @>{i_{KK}}>> \mathrm{Hom}(G,G), \end{CD} where $i_{KK'}$ is the isomorphism from part 1. Commutativity tells us $$i_{K'K}^{-1}(\mathrm{id}_G)\circ i_{KK'}^{-1}(\mathrm{id}_G)=i_{KK}^{-1}(\mathrm{id}_G)=\mathrm{id}_K\in[K,K].$$ Similarly, the same diagram with $K$ and $K'$ swapped tells us $$i_{KK'}^{-1}(\mathrm{id}_G)\circ i_{K'K}^{-1}(\mathrm{id}_G)=\mathrm{id}_{K'}\in[K',K'].$$ Thus, $i_{K'K}^{-1}(\mathrm{id}_G)$ and $i_{KK'}^{-1}(\mathrm{id}_G)$ are homotopy inverses, and so $K$ and $K'$ are homotopy equivalent.