In the proposition mentioned in the title, Hartshorne states that for each $i\geq 0$ the higher direct image sheaf $R^if_*\mathcal F$ is exactly the sheafification of the presheaf given by

$V \mapsto H^i(f^{-1}(V),\mathcal F_{|f^{-1}(V)})$

The proof seems to be quite easy: The statement is obvious for $i=0$ and since the functor defined above is an universal $\delta$-functor, we obtain the result by a well-known theorem in homological algebra.

I tried to come up with the details, but trying to show that the functor defined above admits a long exact sequence, i got stuck at one point:

Given an exact sequence $0 \to \mathcal F' \to \mathcal F \to \mathcal F'' \to 0$ on X, for each open $V \subset Y$ we get an exact sequence $0 \to \mathcal F'_{|f^{-1}(V)} \to \mathcal F_{|f^{-1}(V)} \to \mathcal F''_{|f^{-1}(V)} \to 0$ giving rise to a long exact sequence

$0 \to H^0(f^{-1}(V),\mathcal F'_{|f^{-1}(V)}) \to H^0(f^{-1}(V),\mathcal F_{|f^{-1}(V)}) \to H^0(f^{-1}(V),\mathcal F''_{|f^{-1}(V)}) \to H^1(f^{-1}(V),\mathcal F'_{|f^{-1}(V)}) \to \dotsc$

But for this to induce a long exact sequence of presheaves (which would give me the long exact sequence of sheaves since sheafification is exact), i need the boundary-maps $\delta:H^i(f^{-1}(V),\mathcal F''_{|f^{-1}(V)}) \to H^{i+1}(f^{-1}(V),\mathcal F'_{|f^{-1}(V)})$ to commute with the restriction maps $H^i(f^{-1}(V),\mathcal F_{|f^{-1}(V)}) \to H^i(f^{-1}(W),\mathcal F_{|f^{-1}(W)})$ whenever $W \subset V$ is an inclusion of two opens. But i don't see this.

My questions:

1) How exactly do i get (or how do i have to think about) those restriction maps?

2) How can i make sure the restriction maps commute with the $\delta$-maps?

  • 3
    $\begingroup$ Boundary maps are natural transformations of functors; they commute with all induced maps. $\endgroup$ – Alex Degtyarev Nov 20 '14 at 17:50

Let $f: X \to Y$ be a morphism. There is a factorisation $f_*: Sh(X) \to Sh(Y)$: $f_* = (-)^\# \circ f_{p,*}$ with $(-)^\#: PSh(Y) \to Sh(Y)$ exact over $PSh(Y)$ and $\Gamma(f^{-1}(U),-): Sh(X) \to Ab$ factors as $\Gamma(U,-) \circ f_{p,*}$ over $PSh(Y)$ with $\Gamma(U,-): PSh(Y) \to Ab$ exact. It follows that $R^if_* = (R^if_{p,*})^\#$ and $\Gamma(U,R^if_{p,*}) = R^i\Gamma(f^{-1}(U),-)$.


Probably, the easy way to think of your question is the following. Let $f:Y\to X$ be a sufficiently nice morphism. Recall that one can actually work with presheaves on a base of topology $\mathcal{B}$. In particular, let $\mathcal{B}$ consist of affine opens $U\subset X$ and let inclusions be those of principal opens $U_f\subset U$.

Let $U=\mathrm{Spec}\ A$ be an affine open, $Y_U=f^{-1}(U)$. Now, pick an affine covering of $Y_U$ and compute $H^i(Y_U, \mathcal{F})$ by means of the corresponding Cech complex. Each term of the complex is naturally an $A$-module (this is what it means to work over an affine base), cohomologies are $A$-modules, LES's of cohomology are exact sequences of $A$-modules!

Now, figure out yourself what happens when you restrict to $U_f$.


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