# Representable functor is left-adjoint


My proof so far is:

Let us assume that all coproducts exist in $$\mathcal{C}$$ and that $$F:\mathcal{C}\rightarrow\Set$$ is representable, say $$h^X$$ and $$F$$ are naturally isomorphic for some object $$X$$ in $$\mathcal{C}$$. Let us define $$G:\Set\rightarrow \mathcal{C}: S\rightarrow \bigsqcup_{s\in S} X$$. We will show $$G$$ is left-adjoint to $$F$$. By constructing a natural isomorphism $$\alpha: \operatorname{Hom}_\mathcal{C}(G(-),-)\tilde{\Longrightarrow}\operatorname{Hom}_\Set(-,F(-))$$. So for every set $$S$$ and every object $$Y$$ of $$\mathcal{C}$$, we need a bijection $$\alpha_{S,Y}:\operatorname{Hom}_\mathcal{C}(GS,Y)\rightarrow\operatorname{Hom}_\Set(S,FY)$$ Note that since $$F$$ is representable we have an isomorphism from $$\operatorname{Hom}_\mathcal{C}(X,Y)$$ to $$FY$$, say $$\psi$$. We see that there exists an isomorphism (bijection) $$\phi$$ from $$\operatorname{Hom}_\mathcal{C}(GS,Y) =\operatorname{Hom}_\mathcal{C}(\bigsqcup_{s\in S}X,Y)$$ to $$\operatorname{Hom}_\Set(S,\operatorname{Hom}_\mathcal{C}(X,Y))$$. Indeed, let $$\phi$$ map $$f$$ to a function $$f'$$ in $$\operatorname{Hom}_\Set(S,\operatorname{Hom}_\mathcal{C}(X,Y))$$ such that $$f'(s)$$ is a morphism in $$\operatorname{Hom}_\mathcal{C}(X,Y))$$ for which $$f'(s)(x) = f(x^{(s)})$$. The latter notation meaning that $$x$$ belongs to the set $$X$$ corresponding to $$s$$ in the disjoint union. $$\phi$$ is an isomorphism of sets (bijection). It is injective, let $$\phi(f_1) = \phi(f_2)$$, then $$\phi(f_1)(s)(x) = \phi(f_2)(s)(x)$$ for all $$s\in S$$ and $$x\in X$$, so $$f_1(x^{(s)}) = f_2(x^{(s)})$$ for all $$s\in S$$ and $$x\in X$$, which means $$f_1 = f_2$$. $$\phi$$ is also surjective, take $$f'\in \operatorname{Hom}_\Set(S,\operatorname{Hom}_\mathcal{C}(X,Y))$$, then $$f'$$ is the image of a map $$f$$ for which $$f(x^{(s)}) = f'(s)(x)$$. So we conclude the isomorphism $$\alpha_{S,Y}$$ exists, it is the composition of $$(\psi\circ -)\circ \phi$$.

Now take a morphism $$f: S\rightarrow S'$$ in $$\Set$$. Then for $$a\in \operatorname{Hom}_\mathcal{C}(GS',Y)$$ we get $$(\alpha_{S,Y}\circ (-\circ Gf))(a) = \alpha_{S,Y}(aGf)= ((\psi\circ -)\circ \phi_S)(a(Gf)) = \psi\phi_S(aGf)$$ and also that $$((-\circ f)\circ\alpha_{S',Y})(a) = \alpha_{S',Y}(a)f=((\psi\circ -)\circ \phi_{S'})(a)f=\psi\phi_{S'}(a)f$$ where $$\phi_S,\phi_{S'}$$ is the map $$\phi$$ from above acting on $$S$$ and $$S'$$ respectively.

But at this point I am getting a bit lost in the notation and not sure what I can use. I am trying to show that $$\psi\phi_S(aGf)=\psi\phi_{S'}(a)f$$ such that the condition for $$\alpha$$ being a natural isomorphism is satisfied. (I also need to show that the left compositions commute of course). But I am unsure how to proceed. I also don't really know why we require that the coproducts exist.

Are you familiar with the unit-counit formalism? If yes, it's way easier to argue that way :-)

Let $$F=\hom(A,\_)$$ be representable; your putative left adjoint $$L$$ is the functor $$\_\odot A$$, sending a set $$X$$ to the $$X$$-indexed coproduct $$\coprod_{x\in X}A$$. You can

1. Define a natural transformation $$L\circ F \Rightarrow 1$$ whose components are the maps $$\epsilon_C : \hom(A,C)\odot C \to C$$ obtained from the universal property of the coproduct by the $$\hom(A,C)$$-indexed family of morphisms $$\hom(A,C)$$ (this should seem tautological to you).
2. Define a natural transformation $$\eta : 1 \Rightarrow F\circ L$$ whose components are the maps $$\eta_X : X \to \hom(A, X\odot A)$$ obtained sending $$x\in X$$ to the $$x^\text{th}$$ coproduct injection $$A\to \coprod_{x\in X} A$$.

Now prove that $$\epsilon$$ and $$\eta$$ satisfy the zig-zag identities: $$F \overset{\eta*F}\Rightarrow FLF \overset{F*\epsilon}\Rightarrow F$$ is the identity natural transformation of the functor $$F$$, and $$L \overset{L*\eta}\Rightarrow LFL \overset{\epsilon*L}\Rightarrow L$$ is the identity natural transformation of the functor $$G$$. This entails that $$L\dashv F$$.

• I have seen that an adjunction gives rise to a unit and a counit transformation. But not that these transformations satisfying those identities is equivalent to L and F being adjoint. It feels like proving this fact seperately would be the same as proving it in this special case directly? – Jarne Renders Dec 24 '19 at 14:50
• Sure it takes a certain effort to prove the equivalence; but Mac Lane does it for you, just follow him :) and once you know it is true, you can just choose the most convenient proof each time you need it. It's better to work hard once, and never again, than work a little, one thousand times. – Fosco Dec 24 '19 at 17:05
• I'm unsure how the components in $1.$ are defined. They should be maps from the $\operatorname{Hom}_\mathcal{C}(A,C)$-indexed coproduct $\bigsqcup_{f\in \operatorname{Hom}_\mathcal{C}(A,C)}A$ to $C$, right? But how do you obtain a map using the universal property of coproduct? $C$ is not necessarily a coproduct, is it? – Jarne Renders Dec 25 '19 at 11:03
• For every set $X$ there is a bijection between $\hom(\coprod_{x\in X} E_x, F)$ and $\prod_{x\in X}\hom(E_x,F)$; if now $X =\mathcal{C}(A,C)$.... – Fosco Dec 25 '19 at 11:15