I am a bit confused about a notation I am seeing while reading a paper. So basically I have a continuous map between topological spaces $f:X\to Y$, and I know that this induces a map on cohomology $f^*: H^*(Y;R)\to H^*(X;R)$ for some chosen coefficient ring $R$. But the author of the paper I am reading tends to write $f^*:H^*(Y;R)\to H^*(X;f^*R)$ and I am not sure what to make of this (is this the same map $f^*$ that I know??) What does $f^*R$ exactly mean? I know that this is not a typographical error because there are more than one occasions where the author uses this. Could someone please explain this map to me? Is this notation usual? I haven't seen them so far in algebraic topology textbooks that I read.

Edit: As requested, I cite the paper (version 1): https://arxiv.org/abs/2009.06023. See Proposition 7.3 and 7.4.

  • $\begingroup$ That is odd. My best guess is that the author is emphasizing $f^*H^*(Y;R)$ has the same coefficient ring $R$. $\endgroup$
    – pancini
    Sep 28, 2020 at 6:25
  • 3
    $\begingroup$ Could $R$ be a sheaf of rings rather than just a ring? $\endgroup$
    – Zhen Lin
    Sep 28, 2020 at 6:27
  • $\begingroup$ I guess this is odd also for topologists.. I don't know if it helps, but so far I have seen the author uses this notation for diagonal maps. I don't know if this extra information changes anything. So if $f:E\to E\times_B E$ is the diagonal map where the fiber product is from a fibration $\pi:E\to B$. Would it come out as unusual or rude if I ask the author personally what this really means? $\endgroup$
    – quantum
    Sep 28, 2020 at 6:40
  • $\begingroup$ @Zhen Lin: I don't think so. The paper has nothing on sheaves. I understand that you think that in this case $f$ is a map of affine schemes and $f^*$ its corresponding ring homomorphism. $\endgroup$
    – quantum
    Sep 28, 2020 at 6:43
  • $\begingroup$ Could you please add a reference to the paper to your question? $\endgroup$
    – Christoph
    Sep 28, 2020 at 10:01

1 Answer 1


The paper states "where $R$ is an arbitrary system of coefficients on $E\times_B E$." This is local system cohomology.

Hence, $R$ is indeed a (locally constant) sheaf on $E\times_B E$ and $\Delta^* R$ is the pullback along $\Delta$ to a sheaf on $E$.

  • $\begingroup$ Oh. So I misunderstood this. But maybe not completely. Can we regard this as a general statement? It will take a while for me to understand local system of cohomology, but is it safe to assume that if my sheaf is a constant sheaf associated to a ring $R$ then I can assume $R$ is a ring (and not a sheaf) and $\Delta^*R$ is also $R$? I ask this because I could follow the proof and succeeding texts assuming this. $\endgroup$
    – quantum
    Oct 9, 2020 at 7:20
  • $\begingroup$ Yes, that is correct. $\endgroup$
    – Christoph
    Oct 9, 2020 at 7:38

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