Duals are unique up to a unique compatible isomorphism

Define $\varphi :X_1^{\vee }\to X_2^{\vee }$ by the composite $X_1^{\vee }\stackrel{{\rho }^{-1}}{\to }{X}_1^{\vee }\otimes \mathbf{1}\stackrel{1\otimes {\mathrm{coev}}_2}{\to }{X}_1^{\vee }\otimes (X\otimes X_2^{\vee })\stackrel{{\alpha }^{-1}}{\to }(X_1^{\vee }\otimes X)\otimes X_2^{\vee }\stackrel{{\mathrm{ev}}_1\otimes 1}{\to }\mathbf{1}\otimes X_2^{\vee }\stackrel{\lambda }{\to }{X}_2^{\vee },$ and define $\psi :X_2^{\vee }\to X_1^{\vee }$ by the same formula with the subscripts interchanged.

Postcomposing the definition of $\varphi$ with ${\mathrm{ev}}_2$ and precomposing it with ${\mathrm{coev}}_1$, then using the zig-zag identities from [L1], yields the two compatibility equations in the statement. The same calculation with $\psi$ gives the analogous equations for $\psi$.

The composite $\psi \varphi$ is the unique morphism $X_1^{\vee }\to X_1^{\vee }$ compatible with ${\mathrm{ev}}_1$ and ${\mathrm{coev}}_1$, and the identity morphism has that same compatibility by [L1]. Expanding one copy of $\varphi$ and one copy of $\psi$ and then straightening with the zig-zag identities shows that $\psi \varphi =1_{X_1^{\vee }}$; similarly $\varphi \psi =1_{X_2^{\vee }}$. Thus $\varphi$ is an isomorphism with inverse $\psi$.

If $\chi :X_1^{\vee }\to X_2^{\vee }$ is any other morphism satisfying the two compatibility equations, insert $\chi$ into the formula of step 1.1 and use those compatibilities to collapse the same zig-zag composites; the result is $\chi =\varphi$. Hence the compatible isomorphism is unique. The right-dual statement is the mirror argument.