Existence of adjunction

Question

I found some references saying that a functor $G:\mathcal{D}\to\mathcal{C}$ admitting a left adjoint iff for any object $A\in\mathrm{Ob}~\mathcal{C}$, the functor $\mathrm{hom}_{\mathcal{C}}(A,G(-))$ is representable. (like the dual https://stacks.math.columbia.edu/tag/0A8B) But when I was proving/verifying the naturality of the isomorphisms, I felt like I cannot give the naturality in $\mathcal{C}$, i.e. for any morphism $k:C\to A$ in $\mathcal{C}$ we have the following commutative diagram $\require{AMScd}$ \begin{CD} {\mathrm{hom}_{\mathcal{C}}(A,G(B))} @>{\eta_{A,B}^{-1}}>> {\mathrm{hom}_{\mathcal{D}}(F(A),B)}\\ @V{k^*}VV @VV{F(k)^*}V\\ {\mathrm{hom}_{\mathcal{C}}(C,G(B))} @>{\eta_{C,B}^{-1}}>> {\mathrm{hom}_{\mathcal{D}}(F(C),B)} \end{CD}

And there are also some references stating more are required (like https://ncatlab.org/nlab/show/adjoint+functor#AdjointFunctorFromObjectwiseRepresentingObject). So I'm a little confused. Which one is really correct? (My personal thought was like that in nLab, in which more natuality is required.)

Answer 1

Let $G:\mathcal{D} \to \mathcal{C}$ a functor such that for every $A$ in $\mathcal{C}$, the functor $X \mapsto {Hom}_\mathcal{C}(A,G(X))$ is representable. This means by definition that for every $A$, there is an object $F(A)$ in $\mathcal{D}$ and an isomorphism $\eta^A_X :{Hom}_\mathcal{C}(A,G(X)) \to {Hom}_\mathcal{D}(F(A),X)$ which is natural in X.

So now we have to show :

1. That we have a functor $X \mapsto F(X)$
2. For a fixed $X$ in $\mathcal{D}$, $\eta^A_X$ is natural in $A$.

Let's adress the first point : to a map $f : A \to B$ in $\mathcal{C}$, Define $\tilde{F(f)}_X: {Hom}_\mathcal{D}(F(B),X) \to {Hom}_\mathcal{D}(F(A),X)$ as the composition $\eta^A_X \circ f^*_X \circ (\eta^B_X)^{-1}$. It is natural in $X$ since it is the compostion of natural transformations in $X$. Now Yoneda says this corresponds to some morhism $F(f) : F(A) \to F(B)$, and $\tilde{F(f)}_X = F(f)^*_X$. This defines indeed a functor since for $f : A \to B$ and $g : B \to C$,$$\tilde{F(gf)}_X = \eta^A_X \circ (gf)^*_X \circ (\eta^C_X)^{-1} = \eta^A_X \circ f^*_X \circ g^*_X\circ (\eta^C_X)^{-1} = \eta^A_X \circ f^*_X \circ(\eta^B_X)^{-1} \circ \eta^B_X\circ g^*_X\circ (\eta^C_X)^{-1}$$ So $\tilde{F(gf)}_X = \tilde{F(f)}_X \circ \tilde{F(g)}_X$ and again by Yoneda, this gives the equality $F(gf) = F(g)\circ F(f)$. Identity is very easy again using Yoneda.

For the second point we have to show that for any $f : A\to B$, $$F(f)^*_X \circ \eta^B_X = \eta^A_X \circ f^*_X.$$ But remembering that $F(f)^*_X = \tilde{F(f)}_X = \eta^A_X \circ f^*_X \circ (\eta^B_X)^{-1}$, we get the desired equality.

Indeed we get that $G$ has a left adjoint.