Proposition 20.21.7 (Grothendieck). Let $X$ be a Noetherian topological space. If $\dim (X) \leq d$, then $H^ p(X, \mathcal{F}) = 0$ for all $p > d$ and any abelian sheaf $\mathcal{F}$ on $X$.

[Theorem 3.6.5, Tohoku].

**Proof.**
We prove this lemma by induction on $d$. So fix $d$ and assume the lemma holds for all Noetherian topological spaces of dimension $< d$.

Let $\mathcal{F}$ be an abelian sheaf on $X$. Suppose $U \subset X$ is an open. Let $Z \subset X$ denote the closed complement. Denote $j : U \to X$ and $i : Z \to X$ the inclusion maps. Then there is a short exact sequence

see Modules, Lemma 17.7.1. Note that $j_!j^*\mathcal{F}$ is supported on the topological closure $Z'$ of $U$, i.e., it is of the form $i'_*\mathcal{F}'$ for some abelian sheaf $\mathcal{F}'$ on $Z'$, where $i' : Z' \to X$ is the inclusion.

We can use this to reduce to the case where $X$ is irreducible. Namely, according to Topology, Lemma 5.9.2 $X$ has finitely many irreducible components. If $X$ has more than one irreducible component, then let $Z \subset X$ be an irreducible component of $X$ and set $U = X \setminus Z$. By the above, and the long exact sequence of cohomology, it suffices to prove the vanishing of $H^ p(X, i_*i^*\mathcal{F})$ and $H^ p(X, i'_*\mathcal{F}')$ for $p > d$. By Lemma 20.21.1 it suffices to prove $H^ p(Z, i^*\mathcal{F})$ and $H^ p(Z', \mathcal{F}')$ vanish for $p > d$. Since $Z'$ and $Z$ have fewer irreducible components we indeed reduce to the case of an irreducible $X$.

If $d = 0$ and $X = \{ *\} $, then every sheaf is constant and higher cohomology groups vanish (for example by Lemma 20.21.2).

Suppose $X$ is irreducible of dimension $d$. By Lemma 20.21.4 we reduce to the case where $\mathcal{F} = j_!\underline{\mathbf{Z}}_ U$ for some open $U \subset X$. In this case we look at the short exact sequence

where $Z = X \setminus U$. By Lemma 20.21.2 we have the vanishing of $H^ p(X, \underline{\mathbf{Z}}_ X)$ for all $p \geq 1$. By induction we have $H^ p(X, i_*\underline{\mathbf{Z}}_ Z) = H^ p(Z, \underline{\mathbf{Z}}_ Z) = 0$ for $p \geq d$. Hence we win by the long exact cohomology sequence. $\square$

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