Sober Space and Morphism of Their Topoi? This is the exercise of Martin Olsson's "Algebraic Spaces and Stacks":
Recall that a topological space $X$ is called SOBER if every irreducible closed subset has a unique generic point.
Exercise 2.C. Let $Op(X)$ be the natural site of open sets of $X$. Let $X_{cl}$ be the associated topos. (Recall that a point of a topos $T$ is a morphism of topoi $x:pt\to T$ where $pt$ is the topos of the one-point space (So is actually the category of sets)).
(c) Let $X$ is sober and $j:pt\to X_{cl}$ be a point of $X_{cl}$. Let $U$ be the union of those open sets $V\subset X$ such that $j^*h_V$ is the empty set. Show that $U$ is open and $Z=X-U$ is irreducible. Show that $j=j_{\eta}$ where $\eta$ is the generic point of $Z$ and $j_{\eta}$ is the morphism of topoi induced by point $\eta$;
(d) Let $X$ be ANY topological space and $Y$ is sober. Show that any morphism of topoi
$$f:X_{cl}\to Y_{cl}$$
is induced by a unique continuous map $X\to Y$.

For (c), I may have two ideas:
(c-1) When we describe the condition that $j^*h_V$ is the empty set, we find that the empty set is the initial object of the category of sets. So we can have for any set $S$, we have
$$1=\#(Hom(j^*h_V,S))=\#(Hom(h_V,j_*S))=\#(j_*S(V))$$
by adjointness and Yoneda's lemma. After this I don't know how to do?
(c-2) We may construct the morphism of topological spaces directly, as: let the underlined topological space of $pt$ is $\{a\}$, then we let $f:\{a\}\to X$ that
$a\mapsto \eta$ iff $\forall Y\ni\eta,a\in f^*h_V$. Then this $\eta$ should be the unique generic point of $Z=X-U$. But I don't know how to make this precisely?
Thank you for your help!!!
 A: Note that a subset $F$ is irreducible iff its complement $U$ is "prime": $U = U_1 \cap U_2$ implies $U = U_1$ or $U = U_2$. From this description it should be clear that your $X - U$ is irreducible, because $h$ preserves meets and in the lattice of sets, $a \wedge b = 0$ iff $a = 0$ or $b = 0$.
Now, your calculation is useful for the second part of (c) indeed. First of all, recall that a geometric morphism $Y \to X$ restricts to an adjunction $\text{Op}(X)\substack{\rightarrow \\[-0.1ex] \leftarrow} \text{Op}(Y),$ for these are subobjects of the terminal object and both $j^*$ and $j_*$ are left-exact. [Moreover, by the density theorem, this adjunction determines the geometric morphism.] Thus, $j^*h_U$ can be either $0$ or $1$. But now one can just check that $j_*$ acts on $S \in \operatorname{Set}$ as the pushforward of the skyscraper sheaf: $j_*(S)(U) = Sh(X)(h_U, j_*S) = \operatorname{Set}(j^*h_U, S) = \operatorname{Set}(0,S) = 1$ when $\eta \not\in U,$ and similarly it is $\operatorname{Set}(1,S) = S$ when $\eta \in U$.
