Why are invertible objects reflexive in a tensor category? I am reading Deligne and Milne's notes on Tannakian categories. I'm only just getting used to the idea of an abstract tensor category, and I've encountered a very believable statement that I'm struggling to verify myself in the abstract setting. Although it's not really particularly important for the rest of the notes I would be really pleased to see a proof of it for completeness and for my own interest. Let me define some terms (all of which are lifted from the notes) first, and then ask my question at the end:
Let $(\mathsf{C}, \otimes, \phi, \psi)$ be a tensor category; here $\phi$ is an associativity constraint (shuffles brackets around tensor products) and $\psi$ is a commutativity constraint (swaps order of tensor products) and they are compatible (pentagon/hexagon axioms). Let $1$ denote the unique (up to isomorphism) identity object of $\mathsf{C}$ i.e. an object $1$ and an isomorphism $u: 1\to 1\otimes 1$ such that the functor $1\otimes - :\mathsf{C}\to\mathsf{C}$ is an equivalence.
An object $X\in\mathsf{C}$ is invertible if $X\otimes-:\mathsf{C}\to\mathsf{C}$ is an equivalence. Hence the identity object is invertible, but also comes with a canonical isomorphism $1\to 1\otimes 1$. If $X$ is invertible then there is an object $X^{-1}$ (unique up to isomorphism) and an isomorphism $\delta: X^{-1}\otimes X\to 1$, and the converse also holds. The pair $(X^{-1}, \delta)$ is called the inverse of $X$.
If $X$ and $Y$ are objects, the internal hom-object $Hom(X,Y)\in\mathsf{C}$, if it exists, is the object representing the functor 
$$\text{Hom}(-\otimes X,Y): \mathsf{C}^{op}\to\mathsf{Set}$$
The dual of an object $X$ is $X^\vee = Hom(X,1)$. There is a canonical morphism $ev_X: X^\vee \otimes X\to 1$ called "evaluation" coming from the identity map $Hom(X,1)\to Hom(X,1)$.
There is a canonical map $i_X : X\to X^{\vee \vee} = (X^\vee)^\vee$ corresponding to the composition
$$X\otimes X^\vee \xrightarrow{\psi_{X,X^\vee}} X^\vee \otimes X \xrightarrow{ev_X} 1$$
$X$ is reflexive if $i_X$ is an isomorphism.

On page 9 of the linked notes the authors claim that if an object $X\in\mathsf{C}$ is invertible then it is reflexive. Intuitively I understand why this is so by analogy with vector spaces over a field $K$/modules over a ring $R$ - invertibility is a strong "finiteness condition" and being isomorphic to the double dual only requires some (weaker) finiteness hypotheses (e.g. dimension $=1$ vs dimension $<\infty$) . However when I tried to check this claim myself in this abstract setting I got quite confused with all the different morphisms/adjunctions involved. I would be grateful if anyone can show/suggest a proof for this.
 A: Note, that $X^{-1} = X^\vee$ and $ev_X$ corresponds to $\delta$. To see this, notice that by the Yoneda Lemma we have a bijection
$$
\operatorname{Nat}(h_{X^{-1}},\operatorname{Hom}(-\otimes X,1)) \longrightarrow \operatorname{Hom}(X^{-1}\otimes X,1),
$$
where $h_{X^{-1}}(Y):= \operatorname{Hom}(Y,X^{-1})$. So $\delta\in \operatorname{Hom}(X^{-1}\otimes X,1)$ gives a natural transformation $\tau\colon h_{X^{-1}}\rightarrow \operatorname{Hom}(-\otimes X,1)$. More explicitly, for $Y\in \mathsf C$ we have
$$
\tau_Y\colon \operatorname{Hom}(Y,X^{-1}) \xrightarrow{f\mapsto \operatorname{Hom}(f\otimes X,1)(\delta)} \operatorname{Hom}(Y\otimes X,1),
$$
i. e. given $f\colon Y\rightarrow X^{-1}$, then
$$
\tau_Y(f)\colon Y\otimes X\xrightarrow{f\otimes1_X} X^{-1}\otimes X \xrightarrow\delta 1.
$$
It still needs to be checked that $\tau$ is an isomorphism, which is clear since 
$$
\operatorname{Hom}(Y,X^{-1}) \xrightarrow{f\mapsto f\otimes1_X} \operatorname{Hom}(Y\otimes X, X^{-1}\otimes X)
$$
is bijective as $-\otimes X$ is an equivalence of categories (and in particular fully faithful) and $\delta$ is an isomorphism.
With this, $ev_X = \tau_{X^{-1}}(1_{X^{-1}}) = \delta$ is immediate.
From this, it follows that $X = (X^{-1})^{-1} = X^{\vee\vee}$. Now, $i_X\colon X\rightarrow (X^{-1})^{-1}$ corresponds to
$$
X\otimes X^{-1} \xrightarrow{\psi_{X,X^{-1}}} X^{-1}\otimes X \xrightarrow\delta 1,
$$
i. e. it is the composition of isomorphisms
$$
X\xrightarrow{1_X\otimes\delta^{-1}} X\otimes(X^{-1})^{-1}\otimes X^{-1}\xrightarrow{1_X\otimes\psi} X\otimes X^{-1}\otimes (X^{-1})^{-1}\\
\xrightarrow{\psi\otimes1_{(X^{-1})^{-1}}} X^{-1}\otimes X\otimes (X^{-1})^{-1} \xrightarrow{\delta\otimes1_{(X^{-1})^{-1}}} 1\otimes (X^{-1})^{-1} = (X^{-1})^{-1}
$$
and thus itself an isomorphism.
A: A more conceptual reason for this takes the view that a tensor category is simply a 2-category which happens to have only one object. Then duals in the tensor category become adjoints in the 2-category, and inverses of objects in the tensor category are equivalences in the 2-category. Since every equivalence can be refined to an adjoint equivalence, it follows that an inverse is a dual.
