# Functor $G$ has right adjoint iff $Y \mapsto Hom(X, G(Y))$ is corepresentable

My professor in his notes claims that a functor $$G$$ has a right adjoint iff the functor $$Y \mapsto Hom(X, G(Y)$$ is corepresentable, i.e for each $$X$$ there is an object $$F(X)$$ and a nutural by $$X$$ bijection

$$Hom(F(X), Y) \cong Hom(X, G(Y))$$.

Is that true? Of course, from corepresentability follows that $$X \mapsto F(X)$$ is indeed a functor (functoriality by $$X$$ plus Yoneda lemma gives us a unique arrow $$F \phi : F(X) \to F(X')$$ for each $$\phi: X \to X'$$).But I don't see how do we get functoriality by Y.

If it was true it would be enough to check if two given functors is adjoint pair to verify functoriality just by one argument. I never heard of it and always checked both arguments.

• Didn't you mean to write and a bijection $\hom(F(X),Y)\cong \hom(X,G(Y))$ natural in $Y$''? – Nex Sep 30 '19 at 10:50
• I guess I did, but it makes everything even more confusing... – Vladislav Sep 30 '19 at 10:56
• After that don't your observations show that there is a unique way to make $F$ into a functor such that the above becomes natural in $X$? – Nex Sep 30 '19 at 11:01
• Not yet, but does it show that in general it is sufficient to check naturality in just one argument whenever I am checking if functors are adjoint? – Vladislav Sep 30 '19 at 11:17
• Yes, because here the isomorphism is a huge constraint. As Nex pointed out, the isomorphism is natural in $Y$ by definition of corepresentability, and natural in $X$ by definition of $F$ as a functor (that is, its definition on maps). Now you just have to prove that if $F,G : C\times D\to E$ are functors and $\psi_{x,y} : F(x,y) \to G(x,y)$ is natural in $x$ and in $y$ (separately) then it is natural in $(x,y)$ – Maxime Ramzi Sep 30 '19 at 16:51