Representable functor is left-adjoint $\newcommand{\catname}[1]{{\textbf{#1}}}
\newcommand{\Set}{\catname{Set}}$
Let $\mathcal{C}$ be a locally small category.
I am trying to show that if all coproducts exist in $\mathcal{C}$ and if $F:\mathcal{C}\rightarrow \Set$ is representable, then $F$ has a 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.
 A: 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


*

*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).

*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 $L\dashv F$.
