Let $(\mathcal{C},\otimes)$ and $(\mathcal{C}',\otimes')$ be rigid tensor categories (in the sense of Deligne/Milne; see https://www.jmilne.org/math/xnotes/tc2022.pdf). My question is asked in the context of page $11$ in ibid. Let $\text{ev}_{X,Y}:\underline{\text{Hom}}(X,Y) \otimes X \to Y$ correspond to $\text{id}_{\underline{\text{Hom}}(X,Y)}$ under the natural isomorphism $$\text{Hom}(\underline{\text{Hom}}(X,Y),\underline{\text{Hom}}(X,Y)) \simeq \text{Hom}(\underline{\text{Hom}}(X,Y) \otimes X,Y).$$
Then if $(F,c)$ is a pair where $F$ is a tensor functor $F:\mathcal{C} \to \mathcal{C}'$ and $c_{X,Y}:F(X) \otimes F(Y) \xrightarrow[]{\simeq} F(X \otimes Y)$ are natural isomorphisms, then it is claimed that $F(\text{ev}_{X,Y}):F(\underline{\text{Hom}}(X,Y)) \otimes F(X) \to F(Y)$ gives rise to morphisms $F_{X,Y}:F(\underline{\text{Hom}}(X,Y)) \to \underline{\text{Hom}}(FX,FY)$.
Now, I assume that they by $F(\text{ev}_{X,Y})$ they really mean $F(\text{ev}_{X,Y})$ composed with some natural isomorphism $c_{X,Y}$, since the domain of $F(\text{ev}_{X,Y})$ naively interpreted is not the same as their definition.
I am interested in understanding in which way we get morphisms $F_{X,Y}:F(\underline{\text{Hom}}(X,Y)) \to \underline{\text{Hom}}(FX,FY)$?
Best Answer
They write that you get a morphism $F(\underline{\mathrm{Hom}}(X,Y))\to\underline{\mathrm{Hom}}(FX,FY)$, not a morphism with target $\underline{\mathrm{Hom}}(X,Y)$. This morphism $F(\underline{\mathrm{Hom}}(X,Y))\to\underline{\mathrm{Hom}}(FX,FY)$ is obtained via the tensor-hom adjunction from the morphism $F(\mathrm{ev}_{X,Y})\colon F(\underline{\mathrm{Hom}}(X,Y))\otimes FX\to FY$. (It is by the way standard to omit the isomorphisms $c_{X,Y}$ from the notation, as they are canonical and otherwise just cluttering up the notation.)
Edit: I will write out why the diagram $$ \require{AMScd} \begin{CD} F(\underline{\mathrm{Hom}}(X,Y)) @>{F_{X,Y}}>> \underline{\mathrm{Hom}}(FX,FY)\\ @A{\alpha}AA @AA{\beta}A\\ F(X^\vee\otimes Y) @>{\gamma}>> (FX)^\vee\otimes FY \end{CD} $$ commutes, where the vertical maps are the canonical isomorphisms of this type, and the bottom map is the map you have shown to be an isomorphism. By the tensor-hom adjunction, it suffices to show that the diagram $$ \require{AMScd} \begin{CD} FX\otimes F(\underline{\mathrm{Hom}}(X,Y)) @>{F(\mathrm{ev})}>> FY\\ @A{FX\otimes\alpha}AA @AA{\varepsilon_{FX}\otimes FY}A\\ FX\otimes F(X^\vee\otimes Y) @>{FX\otimes \gamma}>> FX\otimes (FX)^\vee\otimes FY \end{CD} $$ commutes, where $\varepsilon_{-}\colon (-)\otimes(-)^\vee\to \mathbf{1}$ is the evaluation map that a duality datum comes with. You have checked that $\varepsilon_{FX}$ is the map $FX\otimes(FX)^\vee\cong F(X\otimes X^\vee)\xrightarrow{F\varepsilon_X}F(\mathbf{1})\cong\mathbf{1}$. Therefore, the map in the square given by first going right and then going up is just the map $$ FX\otimes F(X^\vee\otimes Y)\cong FX\otimes F(X^\vee)\otimes FY\cong F(X\otimes X^\vee)\otimes FY\xrightarrow{F\varepsilon_X\otimes FY}FY. $$ On the other hand, the diagram $$ \require{AMScd} \begin{CD} X\otimes\underline{\mathrm{Hom}}(X,Y) @>{\mathrm{ev}}>> Y\\ @AAA @AA{\mathrm{id}}A\\ X\otimes X^\vee \otimes Y @>{\varepsilon_X\otimes Y}>> Y \end{CD} $$ commutes, because $X^\vee\otimes -\cong\underline{\mathrm{Hom}}(X,-)$ as functors. Therefore, both are adjoint to $X\otimes -$, and in particular the counits agree under the natural isomorphism. This is exactly what the above diagram says. If I apply $F$ to the commutative diagram above, I find that in the second square, the map $F(\mathrm{ev})\circ(FX\otimes\alpha)$ is given by $$ FX\otimes F(X^\vee\otimes Y)\cong F(X\otimes X^\vee\otimes Y)\xrightarrow{F(\varepsilon_X\otimes Y)} FY. $$ This is the same as what we wrote down for the map $(\varepsilon_{FX}\otimes FY)\circ(FX\otimes \gamma)$, because of the ways the isomorphisms $F(A\otimes B)\cong FA\otimes FB$ and $A\otimes(B\otimes C)\cong (A\otimes B)\otimes C$ are required to interact.