Recall that a *cohomology operation* is a natural transformation $H^n(-; \pi) \to H^{n+k}(-; G)$ defined on CW complexes.

Does every cohomology operation $H^n(-; \mathbb{Z}) \to H^{n+k}(-; \mathbb{Z}/m)$ factor through $H^n(-; \mathbb{Z}/m)$?

The cohomology operations $H^n(-; \mathbb{Z}) \to H^n(-; \mathbb{Z}/m)$ are all multiples of the map induced on cohomology by the quotient map $\mathbb{Z} \to \mathbb{Z}/m$. In particular, if the above question has a positive answer, then for any such cohomology operation $\theta$, we have $\theta(mx) = 0$.

One can rephrase the above question in terms of Eilenberg-MacLane spaces:

- Does every map $K(\mathbb{Z}, n) \to K(\mathbb{Z}/m, n + k)$ factor through $K(\mathbb{Z}/m, n)$?

I would actually be content with a positive answer to the following broader question:

- Does every map $K(\mathbb{Z}, n) \to K(\mathbb{Z}/m, n + k)$ factor through $K(\mathbb{Z}/r, n)$ for some $r$?

I initially asked question 1 in the Homotopy Theory chatroom. Piotr Pstrągowski's comments, see here, explain why there is a positive answer for $m = 2$.

stablecohomology operations, the answer isyes. That is, any map of spectra $\Sigma^n H\mathbb Z \to \Sigma^{n+k} H\mathbb Z / m$ commutes with multiplication by $m$ (which is null on $\Sigma^{n+k} H\mathbb Z / m$), and so descends, via the cofiber sequence $\Sigma^n H\mathbb Z \xrightarrow m \Sigma^n H\mathbb Z \to \Sigma^n H\mathbb Z / m$, to a map $\Sigma^n H\mathbb Z / m \to \Sigma^{n+k} H\mathbb Z / m$. $\endgroup$