Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

It seems to me that in Travaux de Griffiths (equation 3.2.1), Deligne casually mentions a theorem along the following lines:

Let $f: X \to B$ be a proper map of topological spaces. Suppose $\mathcal{A}$ denotes one of the constant sheaves $\mathbb{Z}, \mathbb{C}$ (on any space). Let $\mathcal{B}$ be an $\mathcal{A}$-module on $B$. Then there exists a canonical isomorphism $$\mathcal{B} \otimes_{\mathcal{A}_B} R^n f_* (\mathcal{A}_X) \to R^n f_* (f^{-1} \mathcal{B}).$$

Where does this morphism come from in the first place (presumably adjointness of $f_*$, $f^{-1}$?), and why is it an isomorphism? It feels to me like this should be a standard fact (but I haven't come across it so far), any reference would be appreciated.

Note: in the context of this article, $X$, $B$ are complex manifolds and $f$ is smooth and submersive. My gut feeling is that this should not matter here.

Thanks in advance, Tom

share|cite|improve this question

Not an answer, just a long comment: This is probably related to what I know as the projection formula. The following is given in Qing Liu's book Algebraic Geometry and Arithmetic Curves on page 190 as Proposition 5.2.32:

Proposition. Let $f:X\to Y$ be a separated and quasi-compact morphism of schemes. Let $\newcommand{\F}{\mathcal F}\F$ be a quasi-coherent sheaf on $X$. Let $\newcommand{\G}{\mathcal G}\G$ be a quasi-coherent sheaf on $Y$. Then for any $p>0$, we have a canonical homomorphism $$(R^pf_\ast\F)\otimes_{\newcommand{\O}{\mathcal O}\O_Y}\G \longrightarrow R^pf_\ast(\F\otimes_{\O_X}f^\ast\G).$$

Now substitute $\G=\newcommand{\B}{\mathcal B}\B$, $\F=\newcommand{\A}{\mathcal A}\A$, $Y=B$ and $n=p$ and this looks very similar.

Edit In fact, I don't have any other reference, but this post on mathoverflow quotes the formula for ringed spaces.

share|cite|improve this answer
This question references Ravil Vakil's notes:… I also sketch where the map comes from. – Matt Sep 19 '12 at 13:03
This doesn't use properness at all, whereas the article would make me believe that assumption is crucial ... – Tom Bachmann Sep 20 '12 at 8:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.