# 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 [18] 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 [18] 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