Let $f \colon X \to Y$ be a proper morphism of (Noetherian) schemes, $\mathcal{F} \in \mathop{Coh}(X)$. Let $i\_Z \colon Z \hookrightarrow Y$ be a closed subscheme and take the inverse image $W := X \times\_Y Z \overset{i\_W}{\hookrightarrow} X$.


If we assume that $\dim X\_y \leq k - 1$ for all $y \in Y \setminus Z$, then the sheaf $R^k f\_{\*}(\mathcal{F})$ is concentrated on $Z$ by flat base change for the open inclusion $Y \setminus Z \to Y$.



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> Is it true in this case that $$R^k f\_{\*}(\mathcal{F}) \cong i\_{Z\*} R^k (f|\_W)\_{\*}(\mathcal{F}|\_{W})$$ where $\mathcal{F}|\_{W} = i\_W^{\*}(\mathcal{F})$?
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I was not able to derive this from the standard results about base change, nor to find any counterexamples.


**Special case:** What I actually need is a rather special case of the former. Namely in the case I am insterested in:


* all objects are varieties over an algebraically closed field
* $X$, $Z$ and $W$ are smooth
* $\mathcal{F}$ is a line bundle
* $k = 1$
* the map $X \setminus W \to Y \setminus Z$ is an isomorphism.


It interesting to know something about the general case, though.



A form of cohomology and base change
Here are more details for the second example from comments. Set $Y={\mathbb A}^2$, $Z=0$ (the origin), $X=$ blow-up of $Y$ and $Z$, so that $W$ is the exceptional divisor. Take $F\_n=O\_X(nW)$. Since the self-intersection of $W$ is $-1$, we see that $F\_n/F\_{n-1}=F\_n|\_W\simeq O\_W(-n)$
(more properly, the direct image of this sheaf to $X$). Looking at the short exact sequence
$$0\to F\_{n-1}\to F\_n\to F\_n/F\_{n-1}\to 0\qquad(n>0),$$
we see that $R^0f\_\*(F\_n)=R^0f\_\*(F\_{n-1})$, while $R^1f\_\*(F\_n)$ is an extension
$$0\to R^1f\_\*(F\_{n-1})\to R^1f\_\*(F\_n)\to R^1f\_\*(F\_n/F\_{n-1})\to 0.$$
The right term of this exact sequence is $i\_{Z\*}R^1f\_\*(F\_n|\_W)$, so your condition fails as long as $R^1f\_\*(F\_{n-1})\ne0$. But this happens for $n>2$ (again, follows from the same sequence).


