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 $m = 2$.

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    $\begingroup$ 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 $\Sigma^n H\mathbb Z / m \to \Sigma^{n+k} H\mathbb Z / m$. $\endgroup$
    – Tim Campion
    Aug 24, 2020 at 18:44
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    $\begingroup$ 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. $\endgroup$ Aug 24, 2020 at 21:23
  • $\begingroup$ Along with $m=2$, there is also a positive answer for $n=1$ :) $\endgroup$ Aug 25, 2020 at 7:28
  • $\begingroup$ @JohnGreenwood nice! but I guess there's a negative answer for n=k=0 ! $\endgroup$ Aug 26, 2020 at 1:27
  • $\begingroup$ @DylanWilson well played, sir $\endgroup$ Aug 26, 2020 at 1:39

2 Answers 2


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.

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.

  • $\begingroup$ Thanks for your answer. This gives me some hope that it may be true in general, although I am worried about powers of primes. $\endgroup$ 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.

  • $\begingroup$ 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...) $\endgroup$
    – Dan Ramras
    Mar 5, 2021 at 16:20
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    $\begingroup$ 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 $\endgroup$
    – Dan Ramras
    Aug 17, 2021 at 17:33
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    $\begingroup$ Caveat to my previous comment: The above notes might not be what Steenrod was referencing. (They're from the right time period, at least.) $\endgroup$
    – Dan Ramras
    Aug 18, 2021 at 16:48

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