Unit and counit of adjunction yield a canonical map This is exercise 4.2 of the "book sheaves in geometry and logic"
If $F: C\rightarrow D$ is left adjoint to $G: D\rightarrow C$, while both $C$ and $D$ are cartesian closed, show that the unit and counit of the adjunction yield a canonical map $F(C × GD)\rightarrow F(C) × D$, while the fact that $G$ preserves products yields by evaluation a canonical map $G(D^E) × G(E)\rightarrow G(D)$. Use the Yoneda lemma to prove that the first canonical map is iso iff the second is.
I know that $C$ and $D$ are cartesian closed is equavalent to say that bogh of them have terminal objects, exponentials and finite products....
The unit of adjunction is  $\eta : id_C \rightarrow G\circ F$
The co-unit of adjunction is  $\epsilon : F\circ G \rightarrow id_D$.
I think I should use the isomorphism in page 162 :
$Hom_{\epsilon}$ $(B×A,\Omega)$$\cong$ $Hom_{\epsilon}$ $(A,PB)$
But I don't know how...
Please help....
 A: Let's start with the two canonical maps. These are both constructed by some standard manipulations with adjoints and homsets.

We start with what we want:
$$F(C \times GD) \to FC \times D$$
by the universal property of the product, this is equivalent to a pair of maps
$$
\begin{align}
(1) & &F(C \times GD) &\to FC \\
(2) & &F(C \times GD) &\to D
\end{align}
$$
It's easy to construct a canonical map for $(1)$: $F \pi_L$ works, where $\pi_L$ is left projection.
To construct a map for $(2)$, we use adjointness to flip $F$ to the other side,
and it suffices to build a map
$$C \times GD \to GD$$
Of course, $\pi_R$ works for this.

The second case is similar. We want
$$ G(D^E) \times GE \to GD$$
But we know that $G$ preserves products. So this is the same thing as a map
$$G(D^E \times E) \to GD$$
But we have a map $\mathsf{eval} : D^E \times E \to D$ (the counit of the cartesian closed adjunction), and so $G \ \mathsf{eval}$ works.

If you like, you can run these arguments backwards and get actual terms pinning down the desired maps, but this tends to be unenlightening so I haven't done it.
Now to the second part of the exercise: why must one isomorphism beget another?
Say we know that
$$F(C \times GE) \cong FC \times E.$$
For reasons that will be apparent later, I've renamed $D$ to $E$.
By yondea this happens if and only if
$$\mathcal{D}(F(C \times GE), D) \cong \mathcal{D}(FC \times E, D)$$
but by applying the $F \dashv G$ and cartesian closed adjunctions to both sides, we get
$$\mathcal{C}(C, GD^{GE}) \cong \mathcal{C}(C \times GE, GD) \cong \mathcal{D}(FC, D^E) \cong \mathcal{C}(C, G(D^E))$$
but by yoneda again, this happens exactly when
$$GD^{GE} \cong G(D^E)$$
Now if you chase through the definitions of all the maps we've been using, you should get the isomorphism above comes from an isomorphism $GD \cong G(D^E) \times GE$ by applying the cartesian closed adjunction one more time. I should warn you, though: I haven't actually checked this.

I hope this helps ^_^
