I am tryng to prove a theorem in McLane without looking at his proof and I am stuck on one point. Suppose $F:X\times P\rightarrow A$ is a bifunctor, that for each $p\in P $, the functor $F(-,p):X\rightarrow A$ has a right adoint $G(p,-):A\rightarrow X$. Then $G$ fills out to a bifunctor $:P^{op}\times A\rightarrow X$ s.t. the arrow assignment is unique and the isomorphism of hom sets in the adjunction is natural in $x,a \textit{and}$ p.
final edit: The adjunction isomorpism is $\left [ F(x,p),a \right ]\cong \left [x,G(p,a) \right ]$. We know this is natural in $x$ and $a$. Naturality in $p$ means that the following square must commute, for $f:p\rightarrow p'$
$$\begin{matrix}[F(x,p'),a] &\stackrel{\phi '}{\rightarrow}&[x,G(p',a)]\\\downarrow{\tau_{x}^{*} }&&\downarrow{(\sigma_{*})_{a} }\\ [F(x,p),a]& \stackrel{\phi }{\rightarrow}&[x,G(p,a)]\end{matrix}$$
where $\tau_{x}$ is an arrow:$F(x,p)\rightarrow F(x,p')$ , and $\sigma_{a}$ is an arrow: $G(p',a)\rightarrow G(p,a)$. So if the square commutes, we get the arrow assigmment for $G(-,a)$, namely $G(f,a)=\sigma_{a}$.
On the other hand, since $F$ is a bifuntor, $F(x,f)$ is the $x$ component of the natural transformation $F(-,f):F(-,p)\overset{\cdot }{\rightarrow}F(-,p')$ and now, since $F$ and $G$ are adjoint, we get a unique natural transformation $\sigma :G(p',-)\overset{\cdot }{\rightarrow}G(p,-)$ which makes the above square commute. and we just saw that from this we get an arrow assignment for $G(-,a)$, which is evidently unique. Functoriality in the first argument $p$ follows because the choice of conjugates is unique, so that $G(f,\cdot f',-)$ = $G(f',-)\cdot G(f,-)$ just because each side is a natural transformation which corresponds to the same naturality square.
So $G$ is functorial in each argument. Now I use the following easy fact: suppose $G$ is functorial in both arguments. Let $f:p\rightarrow p'$ and $g:a\rightarrow a'$. Then $G$ is a bifunctor iff $G(f,a')\cdot G(p',g)=G(p,g)\cdot G(f,a)$. But this follows immediately from the fact that $G(f,-)$ is a natural transformation. Indeed, we have the naturality square
$$\begin{matrix}G(p',a') &\stackrel{G(f,a')}{\rightarrow}&G(p,a')\\\downarrow{G(p,g) }&&\downarrow{G(p',g)}\\ G(p',a)& \stackrel{G(f,a)}{\rightarrow}&G(p,a)\end{matrix}$$