The following result is due to Grothendieck:
If $X$ is a noetherian topological space of dimension $n$, then for all $i>n$ and all sheaves of abelian groups $\mathscr{F}$ on $X$, we have $H^i(X,\mathscr{F})=0$.
Exercise III.2.1 in Hartshorne is designed to show that the bound is strict. I will restate the exercise here.
(a) Let $X=\mathbb{A}_k^1$ be the affine line over an infinite field $k$. Let $P,Q$ be distinct closed points of $X$, and let $U=X-\{P,Q\}$. Show that $H^1(X,\mathbb{Z}_U)\ne 0$.
(b) More generally, let $Y\subseteq X=\mathbb{A}_k^n$ be the union of $n+1$ hyperplanes in general position, and let $U=X- Y$. Show that $H^n(X,\mathbb{Z}_U)\ne 0$.
Here, $\mathbb{Z}_U:=j_!(\mathbb{Z}|_{U})$ is the sheaf obtained by extending $\mathbb{Z}|_U$ by zero outside of $U$, where $j:U\to X$ is inclusion. Similarly, for the inclusion $i:Y\to X$ where $Y := \{P, Q\}$, define $\mathbb{Z}_Y:=i_*(\mathbb{Z}|_Y)$. I've been able to solve $(a)$ by making use of the exact sequence of sheaves on $X$
$$0\to\mathbb{Z}_U\to\mathbb{Z}\to\mathbb{Z}_Y\to 0$$
This gives rise to a long exact sequence in cohomology, from which we find $H^1(X,\mathbb{Z}_U)\ne 0$.
I'm having trouble with part $(b)$ and would appreciate some help. I'm trying to proceed by induction, with the base case given by part $(a)$. It seems to me that the inductive step should make use of the fact that if $Y=H_1\cup\ldots\cup H_{n+1}$ is the union of $n+1$ hyperplanes in general position in $\mathbb{A}_k^n$, then $Y-(H_2\cup\ldots\cup H_{n+1})$ is the complement of $n$ hyperplanes in general position in $H_1=\mathbb{A}_k^{n-1}$, but I can't seem to work it out. Thanks in advance for any help.
Best Answer
This isn't the inductive proof you were looking for, but hopefully it's illuminating.
First recall the following theorem about Čech resolutions for locally finite closed covers:
Theorem [Godement, II, Thm. 5.2.1]. Let $X$ be a topological space, $\mathscr{F}$ a sheaf of abelian groups on $X$, and $\mathfrak{M} = (M_i)_{ \in I}$ a locally finite closed cover of $X$. Then, the complex $\mathscr{C}^\bullet(\mathfrak{M},\mathscr{F})$ as defined in [Hartshorne, p. 220] is a resolution for $\mathscr{F}$.
In our case, write $Y = \bigcup_{i=0}^n H_i$ where the $H_i$ are hyperplanes in general position; this is a finite closed cover of $Y$. Letting $\mathscr{F} = \mathbf{Z}\rvert_Y$ in the theorem above, and then pushing forward along the inclusion $i\colon Y \to X$, we have a resolution $$ 0 \longrightarrow \mathbf{Z}_Y \longrightarrow \bigoplus_i \mathbf{Z}_{H_i} \longrightarrow \bigoplus_{i<j} \mathbf{Z}_{H_i \cap H_j} \longrightarrow \cdots \longrightarrow \bigoplus_{i_1<i_2<\cdots<i_n} \mathbf{Z}_{H_{i_1} \cap H_{i_2} \cap \cdots \cap H_{i_n}} \longrightarrow 0 $$ of $\mathbf{Z}_Y$ on $X$, since $\bigcap_{i=0}^n H_i = \emptyset$ by the hypothesis that the $H_i$ are in general position. This is also a flasque resolution since each intersection $H_{i_1} \cap H_{i_2} \cap \cdots \cap H_{i_\ell}$ is irreducible, and since direct images of flasque sheaves are flasque.
Now note that after taking global sections in the resolution above, we obtain the chain complex for the simplicial homology of $S^{n-1}$ (here we use that the hyperplanes intersect in the "expected" way, and so the hyperplanes must be chosen generally). This implies $H^{n-1}(X,\mathbf{Z}_Y) = H_0(S^{n-1},\mathbf{Z}) = \mathbf{Z}$ if $n > 1$ and $\mathbf{Z}^2$ if $n = 1$. Now using the long exact sequence on cohomology from your short exact sequence $$0 \longrightarrow \mathbf{Z}_U \longrightarrow \mathbf{Z} \longrightarrow \mathbf{Z}_Y \longrightarrow 0,$$ we have an exact sequence $$\cdots \longrightarrow H^{n-1}(X,\mathbf{Z}) \longrightarrow H^{n-1}(Y,\mathbf{Z}) \longrightarrow H^n(X,\mathbf{Z}_U) \longrightarrow H^n(X,\mathbf{Z}) \longrightarrow \cdots.$$ Finally, if $n > 1$, then $H^{n-1}(X,\mathbf{Z}) = H^n(X,\mathbf{Z}) = 0$ since $\mathbf{Z}$ is flasque on $X$, and so we have that $H^n(X,\mathbf{Z}_U) = \mathbf{Z}$. If $n = 1$, then the exact sequence is $$ 0 \longrightarrow \mathbf{Z} \longrightarrow \mathbf{Z}^2 \longrightarrow H^1(X,\mathbf{Z}_U) \longrightarrow 0 $$ and so and $H^1(X,\mathbf{Z}_U) = \mathbf{Z}$ if $n = 1$ as well.
Since I wrote up a proof of the Theorem I cited above, I thought I might as well put it here for completeness, especially since the only English reference I could find assumes $X$ is $T_3$. First, we show the following analogue of the glueing axiom for locally finite closed covers.
Lemma [Godement, II, Thm. 1.3.1]. Let $\mathscr{F}$ be a sheaf on $X$ and suppose $\mathfrak{M}$ is a locally finite closed cover, and consider $\mathscr{F}$ as its espace étalé $\operatorname{Sp\acute{e}}(\mathscr{F}) \to X$. Suppose we are given continuous maps $s_i \colon M_i \to \operatorname{Sp\acute{e}}(\mathscr{F})$ such that $s_i\rvert_{M_i \cap M_j} = s_j\rvert_{M_i \cap M_j}$ for each $i,j$. Then, there exists a section $s\colon X \to \operatorname{Sp\acute{e}}(\mathscr{F})$ such that $s\rvert_{M_i} = s_i$ for all $i \in I$.
Proof. It is clear that the $s_i$ glue to give a (unique) section $s\colon X \to \operatorname{Sp\acute{e}}(\mathscr{F})$ such that $s\rvert_{M_i} = s_i$; it suffices to show this section is continuous. For any point $x \in X$, since the cover $\mathfrak{M}$ is locally finite, there is an open set $U(x)$ that intersects only finitely many $M_i$; call these $M_{i_1},\ldots,M_{i_n}$. Now on $U(x)$, the $s_{i_1},\ldots,s_{i_n}$ glue to form a section $t \colon U(x) \to \operatorname{Sp\acute{e}}(\mathscr{F})$ by, e.g., [Munkres, Thm. 18.3]; since $s = t$ on $U(x)$, we see that $s$ is continuous since continuity is local. $\blacksquare$
Now we can prove the Theorem.
Proof of Theorem. We define $\epsilon \colon \mathscr{F} \to \mathscr{C}^0$ by taking the product of the natural maps $\mathscr{F} \to f_*(\mathscr{F}\rvert_{M_i})$ for $i \in I$. The map $\epsilon$ is injective: every point $x \in X$ lies in some $M_i$, there the map on stalks $\mathscr{F}_x \to (f_*(\mathscr{F}\rvert_{M_i}))_x$ is an isomorphism, and so $\epsilon$ is injective. Exactness at $\mathscr{C}^0$ follows from the Lemma. It therefore suffices to show exactness at $\mathscr{C}^n$ for each $n \ge 1$; it suffices to show exactness on stalks.
Let $x \in X$. For each $n \ge 1$, we define a map $$ k \colon \mathscr{C}^n(\mathfrak{M},\mathscr{F})_x \longrightarrow \mathscr{C}^{n-1}(\mathfrak{M},\mathscr{F})_x $$ as follows. Given $\alpha_x \in \mathscr{C}^{n-1}(\mathfrak{M},\mathscr{F})_x$, it is represented by a section $\alpha \in \Gamma(U,\mathscr{C}^n(\mathfrak{M},\mathscr{F}))$ over a neighborhood $U$ of $x$; since $\mathfrak{M}$ is a locally finite cover, we can assume that $U$ intersects only finitely many $M_{j_1},\ldots,M_{j_n}$, and since they are closed, we can assume they all contain $x$ after possibly subtracting some $M_{j_\ell}$ off from $U(x)$. By replacing $\mathfrak{M}$ with $\mathfrak{M} \cap U$, we can assume $\mathfrak{M}$ is in fact finite. Now let $j \in \{j_1,\ldots,j_n\}$ be arbitrary; for any $p$-tuple $i_0 < \cdots < i_{p-1}$, we set $$ (\beta_{i_0,\ldots,i_{p-1}})_x = (\alpha_{j,i_0,\ldots,i_{p-1}})_x $$ using the notational convention of Rem. III.4.0.1. Since there are only finitely many indexes $j$, we see by the Lemma that these $\beta$ glue to give a section $\beta_{i_0,\ldots,i_{p-1}} \in \mathscr{F}\rvert_{M_{i_0,\ldots,i_{p-1}}}(M_{i_0,\ldots,i_{p-1}})$ and hence we get an element $\beta \in \Gamma(U,\mathscr{C}^n(\mathfrak{M} \cap U,\mathscr{F}))$. Now we define $k$ above as $\alpha_x \mapsto \beta_x$. By the same argument as in Lem. III.4.2, $k$ is a homotopy operator for the complex $\mathscr{C}^\bullet_x$, that is, we have $(dk + kd)(\alpha_x) = \alpha_x$ for any germ $\alpha_x \in \mathscr{C}^p_x$, and so the identity map is homotopic to the zero map. Finally, it follows that the cohomology groups $h^p(\mathscr{C}^\bullet_x)$ of this complex are $0$ for $p \ge 1$. $\blacksquare$