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I'm looking for concrete topological intuition for the derived pushforward.

Let $f:X\to Y$ be a continuous map. The derived pushforward $\mathbf Rf_\ast$ takes a sheaf $F$ to the sheafification of the cohomology presheaf $V\mapsto \mathrm H^\bullet(f^{-1}V,F)$. When $f$ is the identity, the sheafification is zero for $n\geq 1$.

The sheafification of a presheaf $P$ can be constructed by taking mapping $PU$ to the equivalence class of families of sections of $P$ defined on an open cover of $U$, where we identify families which coincide on sufficiently small open covers. A section $s\in PU$ is mapped to the equivalence class it represents.

Hence the fact the above sheafification is zero when $f=1$ expresses the fact sheaf cohomology is global in nature (every cocycle admits a cover on which it's zero).

Using the above construction of sheafification, a section in $\mathbf Rf_\ast F(V)$ is an equivalence class of a family of cocycles $(\Gamma_i\in \mathrm H^\bullet(f^{-1}V_i,F))$ where $(V_i)\twoheadrightarrow V$ is an open cover, and we identify families if they coincide on a sufficiently small preimage of an open cover.

The sheafification map is non-zero for $n\geq 1$ if there's a cocycle of $F$ on some preimage $f^{-1}V$ which is not killed by any preimage of an open cover of $V$.

When $f:X\subset Y$ is a subspace inclusion the above means there's a cocycle of $F$ on $f^{-1}V=X\cap V$ such which does not restrict to zero on any open nighborhood in $f^{-1}V_0=X\cap V_0$ in $X$ of some problematic point $x_0\in X$.

This can't happen for closed embeddings because their pushforward functor is exact.

Question 1. What's an instructive example of a subspace inclusion whose higher direct images are nonzero?

Question 2. For "what kind of maps" $f$ does one expect non-zero higher direct images non-zero? (Examples welcome.)

Lastly, I'd appreciate references with explicit topological examples the derived pushforward.

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  • $\begingroup$ Are you sure about locally closed embeddings? Because if $X$ is the spectrum of a local ring, then $\Gamma \colon \mathbf{Ab}(X) \to \mathbf{Ab}$ is exact since it agrees with taking the stalk at the closed point. But if $j_* \colon \mathbf{Ab}(U) \to \mathbf{Ab}(X)$ is exact as well, then global sections on $U$ is exact, which we know is not always true. $\endgroup$ Apr 21, 2020 at 19:04
  • $\begingroup$ @R.vanDobbendeBruyn, I corrected the question. Thanks! $\endgroup$
    – Arrow
    Apr 21, 2020 at 19:07

1 Answer 1

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You expect the higher direct images of $f: X \to Y$ to be nonzero if for arbitrarily small neighborhoods $y_0 \in U$ the space $f^{-1}(U)$ has non-vanishing higher cohomology. I.e. $R^i f_* \mathbb Z$ vanishes if and only if all of its stalks do.

For instance, think about the inclusion of $\mathbb R^2 - 0$ into $\mathbb R^2$. Taking $y_0 = 0$, you see that if $U$ is a small ball around $y_0$, then $f^{-1}(U) = U - 0$ is homotopy equivalent to a circle, which has $H^1 = \mathbb Z$

At every other point, the preimages of contractible neighborhoods are contractible. So, from looking at the stalks, we see that $R^1f_* \mathbb Z$ is $i_* \mathbb Z$ where $i$ is the inclusion of the origin.

The takeaway is, to understand $R^i f_*$ you need to understand the cohomology of preimages $f^{-1}(U)$ for $U$ small.

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