Action on cohomology of the power map of $K(Z,n)$

Question

Consider the "$m$-th power" map $f:K(\mathbb Z,n)\to K(\mathbb Z,n)$ given by $m\in \mathbb Z\cong H^n(K(\mathbb Z,n),\mathbb Z)\cong [K(\mathbb Z,n), K(\mathbb Z,n)]$. Is it true that in any degree the map $f^*$ on integral cohomology sends any element to a multiple of $m$? It's obviously true in degree $n$ where $f^*$ is just multiplication by $m$ but what about higher degrees? It's enough to consider the case when $m=p$ is prime.

Answer 1

As you say, we can reduce to the case where $m=p$ is prime. By the universal coefficient theorem we know that $H^*(K(\mathbb{Z},n);\mathbb{Z})/p$ injects in the ring $A^*=H^*(K(\mathbb{Z},n);\mathbb{Z}/p)$, so we just need to show that $f^*$ acts as zero on $A^*$ in positive degrees. The kernel of $f^*$ is an ideal and is closed under Steenrod operations (including the Bockstein). There is a tautological class $u\in A^n$ with $f^*(u)=pu=0$. It is known that $A^*$ is the free unstable algebra over the Steenrod algebra generated by $u$ subject only to the relation $\beta(u)=0$. That shows that $f^*=0$ on $A^{>0}$ as required.

Answer 2

When $n=2$, $K(\mathbb Z,2) = \mathbb CP^{\infty}$, and one can see exactly what is going on: if $x \in H^2(K(\mathbb Z,2);\mathbb Z)$ is the fundamental class, then $m^*(x^k) = m^k x^k$.

For higher $n$, things are complicated I think, in part because $H^*(K(\mathbb Z,n);\mathbb Z)$ is too messy to easily describe. (If it was too messy for Serre, it is too messy for the rest of us!) It is easy to calculate $m^*$ on one type of element though: if $y \in H^*(K(\mathbb Z,n);\mathbb Z)$ is in the image of $e^*$, where $e: \Sigma K(\mathbb Z,n) \rightarrow K(\mathbb Z, n+1)$ is the canonical map, then $m^*(y)=my$.