# Simple question on proving naturality of an adjunction

Say you want to prove that there exists an adjunction between the functors $F:\mathcal{C}\to\mathcal{D}$, $G:\mathcal{D}\to\mathcal{C}$ with $F$ left adjoint to $G$, and suppose that the bijection on hom-sets $$\hom_{\mathcal{D}}(FA,B)\cong\hom_{\mathcal{C}}(A,GB)$$ has been established. Is it enough to prove naturality of $A$ and $B$ using the map $$\hom_{\mathcal{D}}(FA,B)\to\hom_{\mathcal{C}}(A,GB)$$ (or the opposite) or does one need to check both directions?

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Natural isomorphisms are natural transformations that are objectwise isomorphisms. – Pece Mar 29 '14 at 10:32
...and isomorphisms preserve commutative diagrams (?) – Noah12 Mar 29 '14 at 11:19
No, I meant you have the following (not hard) result : if there is a natural transformation $\alpha \colon F \to G$ between two functors $F,G \colon \mathsf A \to \mathsf B$ such that for all object $a$ of $\mathsf A$ the morphism (in $\mathsf B$) $\alpha(a) \colon F(a) \to G(a)$ is an isomorphism, then $\alpha$ is a natural isomorphism. Apply to your situation ! Hint : $\mathsf A$ must be $\mathcal C \times {\mathcal D}^{\rm op}$ and $\mathsf B$ must be $\mathsf{Set}$. – Pece Mar 29 '14 at 11:50
If you have two bifunctors $K,L:X^{op}×A\to C$ (which is really just a functor where the domain category is a product), then naturality can be checked separately in both coordinates. That means if you fix $a\in A$, and $f:x'\to x$ is an arrow, then the square $$\begin{array}{ccc} K(x,a) & \to & L(x,a)\\ \downarrow & & \downarrow\\ K(x',a) & \to & L(x',a) \end{array}$$ commutes. And similarly if you fix the object $x$, and $g:a\to a'$. – Stefan Hamcke Mar 29 '14 at 15:15