How is it obvious that $\times : C \times C \to C$ is right adjoint to the diagonal functor? This is from "Sheaves in Geometry & Logic".
$\times : C \times C \to C$ is the cartesian product of two objects.   So assume that finite products exist in $C$ the above is a functor. 
To say that it is right adjoint to a functor $L$, is to say that for all $a \times b \in C \times C$ and all $d \in C$ we have a bijection of sets:
$$
\text{Hom}_{C}(d, a \times b) \simeq \text{Hom}_{C\times C}((d, d), (a,b))
$$
where we've used $(x,y)$ to distinguish elements in $C \times C$ from those in $C$. 
This is supposed to be obvious, but it isn't to me.
Further, we have to show that the above is natural in $d, (a,b)$.  Is there some sort of composition of natural maps and functors we can find here?

Is it:
$\text{Hom}_{C\times C}((d,d), (a,b)) = \text{Hom}_C(d, a) \times \text{Hom}_C(d, b) \simeq \text{Hom}_C(d, a\times b)$.
?
But how do I show naturality in $(a,b)$?
 A: You are on the right track. The diagonal functor is defined by $\ \Delta: C\to  C\times C$. On objects: $c\mapsto c\times c$. On arrows: $(f,f):(c, c)\to (c', c')$.
From this, we can construct a right adjoint simply by enforcing the rules: 
A right adjoint is a functor $G:C\times C\to C$ such that
$\text{Hom}_C(c,G(a,b))\cong \text{Hom}_{C\times C}(\Delta c,(a,b)).$
The left-hand side of this is clear. The right-hand side is a set of arrows of the form $(c,c)\to (a,b)$ in the category $C\times C.$ But these arrows, by definition, are $pairs\ (f,g)$ where $f:c\to a$ and $g:c\to b$, so there is the obvious bijection $\text{Hom}_{C\times C}(\Delta c,(a,b))\cong \text{Hom}_{C}( c,a)\times \text{Hom}_{C}(c,b)$. This is natural as well, but that's not important here. It just gives us an idea of what $G should $ be.
That is, $G=\times$, the product functor. On objects: $(a,b)\to a\times b$ and on arrows, the unique morphism induced by the UMP of the product.
As long as $C$ has products and small hom-sets, this will work. 
The obvious unit is $\eta_c=\langle 1_c,1_c\rangle .$
All that remains is to check that $G\Delta f\circ \eta_c=f$ for $f:c\to a\times b.$
I guess you can take it from here. 
