at.algebraic topology - How to get product on cohomology using the K(G, n)?

Consider the evaluation map $S^1 times [S^1,K(Z,n)] to K(Z,n)$. Since $S^1$ is a $K(Z,1)$, and $[S^1, K(Z,n)]=Omega^1(K(Z,n))$ is a $K(Z,n-1)$, up to homotopy we get a map $K(Z,1)times K(Z,n-1) to K(Z,n)$. I'm not actually sure if this induces the product on cohomology. If it does, there is a natural generalization:

Consider the space of pointed maps $[K(A,n),K(A,m+n)]$. Then $pi_k([K(A,n),K(A,m+n)])=0$ for $k>m$, and $=A$ for $k=m$. To see this, note
that (all maps are pointed)

$$[S^k, [K(A,n),K(A,m+n)]] = [K(A,n), Omega^k(K(A,m+n))]$$
$$=[K(A,n),K(A,m+n-k)]= {check H}^{m+n-k}(K(A,n),A) = 0$$

if $k>m$, and $=A$ if $k=m$ (by Hurewicz).

Thus, we have a map $i: K(A,m) to [K(A,n), K(A,m+n)]$ by obstruction theory sending $pi_m(K(A,m))to pi_m([K(A,n),K(A,m+n)])$ isomorphically, which of course is equivalent to a map (by evaluation)

$$K(A,m) times K(A,n) to K(A,m+n).$$

Maybe someone could explain to me if this gives the correct cohomology operation?

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