Local cohomology with respect to a point. (Hartshorne III Ex 2.5) I'm trying to do Hartshorne's exercises on local cohomology at the moment and seem to be stuck in Exercise III 2.5. The problem goes as follows: 

$X$ is supposed to be a Zariski space (i.e a Noetherian and sober topological space) and $P\in X$ a closed point. Let $X_P$ denote the subset of $X$ consisting of all points specializing to $P$, and give it the induced topology. Let $j:X_P\rightarrow X$ be the inclusion map and write $\mathcal F_P=j^{-1}\mathcal F$ for any sheaf (of abelian groups) $\mathcal F$ on $X$. Then for all $i\in\mathbb Z$ and sheaves $\mathcal F$ we have
  \begin{equation}
H_P^i(X,\mathcal F) = H^i_P(X_P,\mathcal F_P)
\end{equation}

I succeeded in showing that the identity holds for $i=0$, so to prove the claim it would suffice to know that the $H^i_P(X_P,(~\cdot~)_P)$ define a universal $\delta$-functor. I tried to do this with the standard procedure by showing effaceability of these functors for $i>0$. So first I tried to show that pulling back along $j$ preserves flsaqueness, but meanwhile I'm quite convinced that this isn't true. I do not see why injectives would go to injectives, either. Then I thought maybe for special sheaves like sheaves of discontinuous sections flasqueness is preserved, but, as before, it didn't lead anywhere. Does anyone have some good advice?
Thanks!
 A: By the excision isomorphism (Exercise III.2.3(f)), we have already
$$
    H^i_P(X,\mathscr{F}) \cong H^i_P(V,\mathscr{F}\rvert_V) \cong \varinjlim_{V \ni P}
    H^i_P(V,\mathscr{F}\rvert_V),
$$
where the direct limit is over the restriction maps induced by the inclusions $P \in W \subset V$. So it remains to show $\varinjlim_{V \ni P} H^i_P(V,\mathscr{F}\rvert_V) \cong H^i_P(X_P,\mathscr{F}_P)$. Note that for any inclusion $P \in W \subset V$, we have the morphism of long exact sequences from Exercise III.2.3(e)
$$\require{AMScd}\begin{CD}
  \cdots @>>> H^{i-1}(V,\mathscr{F}\rvert_V)
  @>>> H^{i-1}(V^P,\mathscr{F}\rvert_{V^P})
  @>>> H^{i}_P(V,\mathscr{F}\rvert_V)
  @>>> H^{i}(V,\mathscr{F}\rvert_V)
  @>>> H^{i}(V^P,\mathscr{F}\rvert_{V^P})
  @>>> \cdots\\
  @. @VVV @VVV @VVV @VVV @VVV\\
  \cdots @>>> H^{i-1}(W,\mathscr{F}\rvert_W)
  @>>> H^{i-1}(W^P,\mathscr{F}\rvert_{W^P})
  @>>> H^{i}_P(W,\mathscr{F}\rvert_W)
  @>>> H^{i}(W,\mathscr{F}\rvert_W)
  @>>> H^{i}(W^P,\mathscr{F}\rvert_{W^P})
  @>>> \cdots
\end{CD}
$$
where $V^P$ denotes the punctured open set $V \setminus \{P\}$, and similarly for $W$. Note that every open set $V$ containing $P$ contains $X_P$ as well by the fact that it is closed under generization by Exercise II.3.17(e), and so the direct limit taken over all $V \ni P$ is the same as the direct limit taken over all $V \supset X_P$. Using the universal property of the direct limit, and the fact that direct limits are exact in the category of abelian groups [Atiyah–Macdonald, Exc. 1.19], we therefore get the corresponding morphism of long exact sequences
$$\begin{CD}
  \cdots @>>> \varinjlim_{V \ni P}H^{i-1}(V,\mathscr{F}\rvert_V)
  @>>> \varinjlim_{V \ni P}H^{i-1}(V^P,\mathscr{F}\rvert_{V^P})
  @>>> \varinjlim_{V \ni P}H^{i}_P(V,\mathscr{F}\rvert_V)
  @>>> \varinjlim_{V \ni P}H^{i}(V,\mathscr{F}\rvert_V)
  @>>> \varinjlim_{V \ni P}H^{i}(V^P,\mathscr{F}\rvert_{V^P})
  @>>> \cdots\\
  @. @VVV @VVV @VVV @VVV @VVV\\
  \cdots @>>> H^{i-1}(X_P,\mathscr{F}_P)
  @>>> H^{i-1}(X_P^P,\mathscr{F}_P\rvert_{X_P^P})
  @>>> H^{i}_P(X_P,\mathscr{F}_P)
  @>>> H^{i}(X_P,\mathscr{F}_P)
  @>>> H^{i}(X_P^P,\mathscr{F}_P\rvert_{X_P^P})
  @>>> \cdots
\end{CD}
$$
Now, we see that if the vertical arrows above are isomorphisms except for the middle arrow, then by the five lemma [Weibel, Exc. 1.33], the middle one would be as well. Thus, it suffices to show that
$$\tag{*}\label{eq:III.2.5}
    \varinjlim_{V \ni P} H^i(V,\mathscr{F}\rvert_V) \cong H^i(X_P,\mathscr{F}_P), \quad
    \varinjlim_{V \ni P} H^i(V^P,\mathscr{F}\rvert_{V^P}) \cong
    H^i(X_P^P,\mathscr{F}_P\rvert_{X_P^P}).
$$
We first show the isomorphisms for $i = 0$ (using the Lemma below):
$$
    \varinjlim_{V \ni P} \Gamma(V,\mathscr{F}\rvert_V) =
    \varinjlim_{V \ni P} \Gamma(V,\mathscr{F}) =
    \varinjlim_{V \supset X_P} \Gamma(V,\mathscr{F}) =
    \Gamma(X_P,\mathscr{F}_P)\\
    \varinjlim_{V \ni P} \Gamma(V^P,\mathscr{F}\rvert_{V^P}) =
    \varinjlim_{V \ni P} \Gamma(V^P,\mathscr{F}) =
    \varinjlim_{V \supset X_P} \Gamma(V^P,\mathscr{F}) =
    \varinjlim_{V^P \supset X_P^P} \Gamma(V^P,\mathscr{F}) =
    \Gamma(X_P^P,\mathscr{F}_P\rvert_{X_P^P})
$$
and so it suffices to show them for $i \ge 1$.
As in the proof of Prop. III.2.9, since the isomorphisms hold for $i = 0$, to show the functors agree for $i \ge 1$, it suffices to show they are both effaceable as functors $\mathfrak{Ab}(X) \to \mathfrak{Ab}$, since in that case they are both universal, hence must be isomorphic by Thm. III.1.3A. So let $\mathscr{G}$ be the sheaf of discontinuous functions of $\mathscr{F}$ from Exercise II.1.16(e). Then, $\mathscr{G}$ is flasque and there is a natural inclusion $\mathscr{F} \hookrightarrow \mathscr{G}$. Now the sheaves $\mathscr{G}\rvert_V$, $\mathscr{G}\rvert_{V^P}$, $\mathscr{G}\rvert_{X_P}$, and $\mathscr{G}\rvert_{X_P^P}$ are flasque by the Lemmas below. Thus, every functor in \eqref{eq:III.2.5} is zero when applied to $\mathscr{G}$ by Prop. III.2.5, hence the functors are effaceable so we are done. $\blacksquare$
EDIT. Andreas pointed out a gap in the proof, and provided the following Lemma to fix it:
Lemma. If $\mathscr{F}$ is a sheaf on a Zariski space $X$, and $Y \subset X$ is a subset closed under generization, then $$\varinjlim_{\tilde{U} \supset U} \Gamma(\tilde{U},\mathscr{F}) \cong \Gamma(U,\mathscr{F}\rvert_Y)$$ where $\tilde{U}$ are open in $X$, and so there is no need to sheafify in the definition of restriction.
Proof. Because open subsets $U$ of $Y$ are closed under generization in $X$, it suffices to show the claim for $U = Y$. Consider the natural map $$\varinjlim_{Y\subseteq U}\Gamma(U,\mathscr{F}) \to \Gamma(Y,\mathscr{F}\rvert_Y)$$ obtained from the definition of sheafification.
Injectivity. This follows from the fact that a separated presheaf injects into its sheafification [Stacks, Tag 0082].
Surjectivity. We want to show every global section of $\mathscr{F}\rvert_Y$ extends to some open neighborhood of $Y$.  So let $s \in \Gamma(Y,\mathscr{F}\rvert_Y)$. Since $Y$ is noetherian, there is a maximal open subset $U \subset Y$ such that $s\rvert_U$ lifts to a section $t\in\Gamma(\tilde{U},\mathscr{F})$ with $\tilde{U} \cap Y=U$. Note it is nonempty since choosing an arbitrary point $z \in Y$, the germ of $s_z$ lives in $\mathscr{F}_z = (\mathscr{F}\rvert_V)_z$, and $s_z$ extends to some open subset of $X$.
Suppose $U \ne Y$, and let $y \in Y \setminus U$. Then, by the argument above using stalks, there exists an open neighorhood $V$ of $y$ such that $s\rvert_V$ lifts to a section $t'\in\Gamma(\tilde{V},\mathscr{F})$ with $\tilde{V} \cap Y=V$.
Now because $t,t'$ are both local lifts of $s$, there exists an open subset $W$ in $X$ such that $U \cap V \subset W \subset \tilde{U} \cap \tilde{V}$ and $t\rvert_W = t'\rvert_W$ (since the support of $t-t'$ is closed in $\tilde{U} \cap \tilde{V}$). Now let $A = \tilde{U} \cap \tilde{V} \setminus W$, and let $A=\bigcup_{i=1}^k A_i$ be an irreducible decomposition. We want to show $\overline{A} \cap Y = \emptyset$, where $\overline{A}$ denotes the closure of $A$ in $X$. For suppose $\overline{A} \cap Y$ is non-empty; then, since $Y$ is closed under generization, we can assume the generic point $x_i$ of some $A_i$ also lies in $\overline{A} \cap Y$. But this is a contradiction, since then, $x_i \in \tilde{U} \cap \tilde{V} \cap Y = U \cap V \subset W$ implies that $x_i\notin A_i$.
Now the two sections $t\rvert_{\tilde{U} \setminus \overline{A}}$ and $t'\rvert_{\tilde{V}\setminus\overline{A}}$ match on the overlap of their domains, which contains $U \cap V$, and hence patch together to give a lift of $s\rvert_{U\cup V}$, contradicting the maximality of $U$. $\blacksquare$
This is used in the following way:
Lemma. If $\mathscr{F}$ is a flasque sheaf on a Zariski space $X$, and $Y \subset X$ is a subset closed under generization, then $\mathscr{F}\rvert_Y$ is flasque.
Proof. This follows since direct limits are exact, and using the previous Lemma. $\blacksquare$
A: Here is how I would prove this:
Let $X$ be a Zariski space. Let $ p \in X$ be a closed point and let $X_{p} \subset X$ be the set of all points $ q \in X$ such that $p \in (q)^{-}$. We observe that $X_{p}$ has the induced topology. Now let $j: X_{p}\hookrightarrow X$ be the inclusion map. For any sheaf $\mathcal F$, let $\mathcal F_{p}=j^{*}\mathcal F$. We then state our claim as: $\Gamma_{p}(X, \mathcal F)=\Gamma_{p}(X, \mathcal F_{p})$. Any open set $U$ containing $p$ also contains $X_{p}$ so gluing sheaves will not affect $\Gamma(X_{p}, \mathcal F_{p})= \lim\underset {\rightarrow p \in U} \Gamma (U, \mathcal F)$. Taking $\pi \in \Gamma_{p} (X, \mathcal F)$, we get a section $\pi^{\prime} \in \Gamma_{p}(X_{p}, \mathcal F_{p})$. We may represent $\pi^{\prime}$ by $\pi \in \Gamma(U, \mathcal F)$. Shrink $U$ such that we may assume $\mathrm{Supp}(\pi)=p$. Glue $\pi$ and $0\in \Gamma (X/p, \mathcal F)$ to obtain a global section. Then there exists a bijection $\Gamma_{p}(X, \mathcal F_{p})\leftrightarrow \Gamma_{p}(X_{p}, \mathcal F_{p})$. 
Finally, if $0 \rightarrow \mathcal F \rightarrow I_{0} \rightarrow I_{1} \rightarrow \cdots$ is a flasque resolution of $\mathcal F$, then $0 \rightarrow \mathcal F_{p} \rightarrow I_{0,p} \rightarrow I_{1,p} \rightarrow \cdots$ is an injective resolution of $\mathcal F_{p}$. By repeating this argument we obtain $\Gamma_{p}(X, I_{i}) \cong \Gamma_{p}(X_{p}, I_{i,p})$ and we have that $H^i_p(X, \mathcal F)=H^{i}_{p}(X_{p}, \mathcal F_{p})$. 
I hope this helps!
