Assume $(\mathcal{C},\otimes,\mathbf{1},\phi,\psi)$ (let's denote this as $\mathcal{C}$) is a tensor-category in the sense of Deligne/J.S. Milne (https://www.jmilne.org/math/xnotes/tc2018.pdf).
Assume furthmore that for each object $X$ in $\mathcal{C}$, we have an adjunction $- \otimes X \dashv \underline{\text{Hom}}(X,-)$.
Let's define the dual of an object $X$ as $X^{\vee} := \underline{\text{Hom}}(X,\mathbf{1})$.
It is claimed in ibid. that by diagram $1.6.6$ we get the following
where $\text{ev}_{X}:X^{\vee} \otimes X \to \mathbf{1}$, and we say that ${}^{t}f:Y^{\vee} \to X^{\vee}$ is induced from a morphism $f$, and is the unique morphism so that the diagram above commutes. How do we actually get this unique arrow?
My guess is that by $1.6.5$ in ibid., we have $\text{Hom}(Y^{\vee} \otimes X,\mathbf{1}) \cong \text{Hom}(Y^{\vee},X^{\vee})$.
So any morphism $Y^{\vee} \otimes X \to \mathbf{1}$ corresponds to a morphism ${}^{t}f:Y^{\vee} \to X^{\vee}$.
But why is the induced ${}^{t}f$ such that the diagram commutes. I suppose uniqueness follows from the bijection I laid out, but how can I specifically see that this diagram commutes from earlier results in the article?
Best Answer
By definition, $X^{\vee}$ represents the (contravariant) functor $F = \text{Hom}(- \otimes X, \mathbf{1})$. This means that there is a natural isomorphism $\eta : \text{Hom}(-,X^{\vee}) \to F$. By the Yoneda Lemma, $\eta$ is determined by $\eta_{X^{\vee}}(\text{id}_{X^{\vee}}) = \text{ev}_X$. Namely, for any $\varphi : T \to X^{\vee}$ we have $$\eta_T(\varphi) = F(\varphi)(\text{ev}_X) = \text{ev}_X \circ (\varphi \otimes \text{id}_X).$$ Given $f : X \to Y$, note that $\text{ev}_Y \circ (\text{id}_{Y^{\vee}} \otimes f) \in F(Y^{\vee})$. Since $\eta_{Y^{\vee}}$ is a bijection, there is a unique $\varphi : Y^{\vee} \to X^{\vee}$ such that $\text{ev}_Y \circ (\text{id}_{Y^{\vee}} \otimes f) = \text{ev}_X \circ (\varphi \otimes \text{id}_X)$. This $\varphi$ is what the authors call ${}^t\!f$.