Condition for a symmetric monoidal category to be closed

Question

Let $\mathcal{C}$ be a symmetric monoidal category i.e. a category with a binary functor $\otimes$, a distinguished object $I$, and distinguished isomorphisms $A\otimes B\to B\otimes A$, $$(A\otimes B)\otimes C\to A\otimes (B\otimes C)$$ and $A\otimes I\to A$ for objects $A$, $B$ and $C$. These are subject to various coherence conditions. I am a little puzzled by the structure/properties required for $\mathcal{C}$ to be closed. For each object $B$, the functor $(-)\otimes B:\mathcal{C}\to\mathcal{C}$ has a right adjoint, which is unique up to natural isomorphism. Call it $F_B$, and we somehow form a functor $F:\mathcal{C}^{\text{op}}\times\mathcal{C}\to\mathcal{C}$ from this. Am I right in thinking there is no additional structure here? Is the isomorphism $$\text{Hom}(A\otimes B,C)\to\text{Hom}(A,F(B,C))$$ additional structure or just the statement that such isomorphisms exist? Basically, I just need to understand what is going on here, but neatly divided into structure and properties.

Answer 1

Let $C$ be any category and let $G:C\times C \to C$ be any functor such that $G(-,Y)$ has a right adjoint $F_Y$ for each $Y$. Then there is a unique way to turn the functors $F_Y$ into a functor $F:C^{op}\times C\to C$ such that \begin{align} C(G(X,Y),Z)=C(X,F(Y,Z)) \end{align} is natural in all three variables.

To see this fix some $Z$ and let $f:Y\to Y'$ be an arrow. Naturality means that the $X$th component of $C(-,F(f,Z))$ must be the following composite. \begin{align} C(X,F(Y',Z))&= C(G(X,Y'),Z)\\ &\to C(G(X,Y),Z) \\ &=C(X,F(Y,Z)) \end{align} The arrow is the $Z$th component of transformation $C(G(X,f),-)$.

Now that you know how $C(-,F(f,Z))$ must look like, you can find out $F(f,Z)$ via the Yoneda lemma. The arrows $F(f,g)$ are then determined by the equation \begin{align} F(f,g) = F(f,1)\circ F(1,g) \end{align} due to functoriality.