Show that $Sh(C,J)^{D^{op}}$ is a Grothendieck topos Let $C$ be a small category and $J$ a Grothendieck topology on $C$. Let $Sh(C,J)^{D^{op}}$ be the category of functors $D^{op}\rightarrow Sh(C,J)$ and natural transformations between them, for some given small category $D$. Show that $Sh(C,J)^{D^{op}}$ is a Grothendieck topos by exhibiting an equivalence of categories $Sh(C\times D,J')\cong Sh(C,J)^{D^{op}}$ for some suitable topology $J'$.
So sorry for asking a quistion like this
with out any ideas...but I really dont know how should I prove this....
 A: First, you should know (or prove for yourself) the general isomorphism of functor categories (which I denote by $\mathrm{Hom}$)
$(1) \qquad \mathrm{Hom}(C \times D,E) \cong \mathrm{Hom}(D,\mathrm{Hom}(C,E)).$
There is only one reasonable way to write down a functor in both directions, and the definitions of all the objects here imply that they are well-defined and inverse to each other.
By replacing $C,D$ by their dual categories and taking $E$ to be the category of sets, we get an isomorphism
$(2) \qquad \mathrm{PSh}(C \times D) \cong \mathrm{Hom}(D^{op},\mathrm{PSh}(C)),$
where $\mathrm{PSh}(C) := \mathrm{Hom}(C^{op},\mathbf{Set})$
Now let us assume that $C$ has a Grothendieck topology $J$. Let's work with bases (Definition III.2 in Maclane-Moerdijk), since I find it much more intuitive than working with sieves. So for every object $X \in C$ we have covering families $\{f_i : X_i \to X\}$ satisfying three axioms (concerning isomorphisms, pullbacks, transitivity). Each such family produces a family $\{(f_i,\mathrm{id}_Y) : (X_i,Y) \to (X,Y)\}$ of morphisms in $C \times D$. We need to make this stable under isomorphisms, so consider more generally $\{(f_i,g) : (X_i,Y) \to (X,Y')\}$ for any isomorphism $g : Y \to Y'$ (but the notion of a sheaf is clearly not changed). It is easy to check that these satisfy the three axioms as well. For instance, the second axiom uses that pullbacks in product categories are calculated in each entry, and that the pullback of an isomorphism is an isomorphism. We get a Grothendieck topology $J'$ on $C \times D$.
A presheaf $F : (C \times D)^{op} \to \mathbf{Set}$ is a sheaf for $J'$ iff for all covering families $\{(f_i,\mathrm{id}_Y) : (X_i,Y) \to (X,Y)\}$ as above the sequence
$(3) \qquad F(X,Y) \to \prod_i F(X_i,Y) \rightrightarrows \prod_{i,j} F(X_i \times_X X_j,Y)$
is exact. In other words, for every $Y \in D$ the presheaf $ F(-,Y) : C^{op} \to \mathbf{Set}$ is a sheaf for $J$.
But this exactly means that (2) restricts to an isomorphism (not just an equivalence, by the way)
$(4) \qquad \mathrm{Sh}(C \times D,J') \cong \mathrm{Hom}(D^{op},\mathrm{Sh}(C,J)).$
