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

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:

1. 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:

1. 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 $$m = 2$$.

• Maybe for context one should point out that for stable cohomology operations, the answer is yes. 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 $\Sigma^n H\mathbb Z / m \to \Sigma^{n+k} H\mathbb Z / m$. Aug 24, 2020 at 18:44
• sorry- I deleted my previous answer because I can't seem to make it work without basically just doing the whole computation anyway... maybe one needs to extract something from the Cartan seminar? It would be nice if there was a clean argument like Tim's though... Messing around it seems like $\mathbb{Z}/m[B\mathbb{Z}]\to\mathbb{Z}/m[B\mathbb{Z}/m]$ has an $\mathbb{E}_{\infty}$-$\mathbb{Z}/m$-retract? (which would do it) But I don't really trust that I didn't obscure some error while doing that... if I end up trusting that, I will update the post. Aug 24, 2020 at 21:23
• Along with $m=2$, there is also a positive answer for $n=1$ :) Aug 25, 2020 at 7:28
• @JohnGreenwood nice! but I guess there's a negative answer for n=k=0 ! Aug 26, 2020 at 1:27
• @DylanWilson well played, sir Aug 26, 2020 at 1:39

The answer is (also) yes when $$m=p$$ is an odd prime, by Theoreme 2 in

Cartan, H. Détermination des algèbres $$H_*(\pi, n; Z_p)$$ et $$H^*(\pi, n; Z_p)$$, $$p$$ premier impair.
Séminaire Henri Cartan, Tome 7 (1954-1955) no. 1, Exposé no. 9, 10 p.
http://www.numdam.org/item/SHC_1954-1955__7_1_A9_0/


which gives $$H^*(K(\pi, n); Z/p)$$ as a free graded commutative algebra on a sum of copies of $$Hom(\pi, Z/p)$$ (indexed by certain words of the "first kind") plus a sum of copies of $$Hom({}_p \pi, Z/p)$$ (indexed by certain words of the "second kind"). Here $${}_p \pi$$ denotes the subgroup of elements of exponent $$p$$. Applying this with $$\pi = Z$$ and $$\pi = Z/p$$ shows that $$H^*(K(Z/p,n); Z/p) \to H^*(K(Z,n); Z/p)$$ is surjective.

Cartan first proves the dual statement in homology (Theoreme Fondamental on page 9-03), showing that $$H_*(K(Z,n); Z/p) \to H_*(K(Z/p,n); Z/p)$$ is injective.

• Thanks for your answer. This gives me some hope that it may be true in general, although I am worried about powers of primes. Feb 20, 2021 at 17:12

I Steenrod's 1957 Colloquium Lectures, published as

Steenrod, Norman E.
Cohomology operations, and obstructions to extending continuous functions.
Advances in Math. 8 (1972), 371–416.


he ends Section 17 with:

There are certain elementary cohomology operations which are taken for granted but must be mentioned in order to state the main result. These are: addition, cup products, homomorphisms induced by homomorphisms of coefficient groups, and Bockstein coboundary operators associated with exact coefficient sequences $$0 \to G' \to G \to G'' \to 0$$. Then the main result becomes:

The elementary operations and the operations $$Sq^i$$, $$\beta_2$$, $$P_p^i$$, $$\beta_p$$ generate all reduced power operations by forming compositions.

Thereafter, at the end of Section 21, he writes:

Using the full strength of Cartan's result, Moore  has shown that all cohomology operations, whose initial coefficient groups are finitely generated, are generated by the cohomology operations listed at the end of Section 17.

Here, reference  is: "J. Moore, Seminar notes 1955/1957, Princeton University."

Moore's result implies that the answer to Question 1 is "yes", because any Bockstein operation will vanish on any integral cohomology class. Unfortunately, I do not know if Moore's seminar notes are available somewhere.

• There was a copy of Moore's notes in the Stanford library when I was a graduate student there. Maybe one could request it via interlibrary loan. (Also I think I have a photocopy in my office, though I haven't been there in a long time...) Mar 5, 2021 at 16:20
• Dmitri Pavlov has a nice scan of these notes (better quality than what I could get from my 2002 scan): dmitripavlov.org/scans/moore-algebraic-homotopy-theory.pdf Aug 17, 2021 at 17:33
• Caveat to my previous comment: The above notes might not be what Steenrod was referencing. (They're from the right time period, at least.) Aug 18, 2021 at 16:48