A braided rigid category has a Drinfeld morphism

Bending a crossing. For any object $Y$, naturality at evaluation and coevaluation, followed by the two hexagons and the zig-zags, gives $c_{X^{\vee },Y}=({\mathrm{ev}}_X\otimes 1_Y\otimes 1_{X^{\vee }})(1_{X^{\vee }}\otimes c_{X,Y}^{-1}\otimes 1_{X^{\vee }})(1_{X^{\vee }}\otimes 1_Y\otimes {\mathrm{coev}}_X)\mathrm{.}$ Indeed expand $c_{Y,X\otimes X^{\vee }}=(1_X\otimes c_{Y,X^{\vee }})(c_{Y,X}\otimes 1_{X^{\vee }})$ in naturality at ${\mathrm{coev}}_X$, then tensor on the left with $X^{\vee }$ and apply ${\mathrm{ev}}_X$ to cancel the adjacent cup and cap. Replacing $c_{A,B}$ throughout by $c_{B,A}^{-1}$ gives the companion identity for $c_{Y,X^{\vee }}^{-1}$. These are algebraic identities, independent of a graphical calculus.

A second left dual of $X^{\vee }$. Put $e_X^{\mathrm{\prime }}={\mathrm{ev}}_X{c}_{X,X^{\vee }}:X\otimes X^{\vee }\to \mathbf{1}$ and $h_X^{\mathrm{\prime }}=c_{X^{\vee },X}^{-1}{\mathrm{coev}}_X:\mathbf{1}\to X^{\vee }\otimes X$. These give $X$ as a left dual of $X^{\vee }$. To check the first zig-zag, expand $(e_X^{\mathrm{\prime }}\otimes 1_X)(1_X\otimes h_X^{\mathrm{\prime }})$; move the crossing incident to the cup across that cup by step 1.1, cancel the resulting inverse crossing pair, and the remaining composite is $(1_X\otimes {\mathrm{ev}}_X)({\mathrm{coev}}_X\otimes 1_X)=1_X$. For the other zig-zag expand $(1_{X^{\vee }}\otimes e_X^{\mathrm{\prime }})(h_X^{\mathrm{\prime }}\otimes 1_{X^{\vee }})$; the companion bending identity cancels the crossing pair and leaves $({\mathrm{ev}}_X\otimes 1_{X^{\vee }})(1_{X^{\vee }}\otimes {\mathrm{coev}}_X)=1_{X^{\vee }}$.

Comparison and inverse. If $(A,e_A,h_A)$ and $(B,e_B,h_B)$ are left duals of an object $Z$, the maps $(e_A\otimes 1_B)(1_A\otimes h_B):A\to B$ and $(e_B\otimes 1_A)(1_B\otimes h_A):B\to A$ are inverse: substitution of either composite permits first the $Z,B$ zig-zag and then the $A,Z$ zig-zag, leaving the identity. They are the unique maps intertwining the evaluations, since tensoring that intertwining equation with $h_B$ recovers the displayed formula. Apply this to $Z=X^{\vee }$, $A=X$ from step 2.1, and $B=X^{\vee \vee }$ with its chosen duality. The comparison is precisely the displayed $u_X$, and its inverse is $v_X=({\mathrm{ev}}_{X^{\vee }}\otimes 1_X)(1_{X^{\vee \vee }}\otimes c_{X^{\vee },X}^{-1})(1_{X^{\vee \vee }}\otimes {\mathrm{coev}}_X)\mathrm{.}$ Thus $v_X{u}_X=1_X$ and $u_X{v}_X=1_{X^{\vee \vee }}$. For $f:X\to Y$, the mate equations for $f,f^{\vee }$, and naturality of $c$ show that both $u_{Y}f$ and $f^{\vee \vee }{u}_X$ intertwine the same evaluation pairing; uniqueness of the comparison gives $u_{Y}f=f^{\vee \vee }{u}_X$.

Tensor relation. Work first with the canonical tensor dual $(X\otimes Y{)}^{\vee }=Y^{\vee }\otimes X^{\vee }$. Its evaluation is ${\mathrm{ev}}_Y(1_{Y^{\vee }}\otimes {\mathrm{ev}}_X\otimes 1_Y)$, and the coevaluation of its dual is $(1_{Y^{\vee }}\otimes {\mathrm{coev}}_{X^{\vee }}\otimes 1_{Y^{\vee \vee }}){\mathrm{coev}}_{Y^{\vee }}$. Insert these into the formula for $u_{X\otimes Y}$. Expand its middle braiding using, successively, $c_{A\otimes B,C}=(c_{A,C}\otimes 1_B)(1_A\otimes c_{B,C})$ and $c_{A,B\otimes C}=(1_B\otimes c_{A,C})(c_{A,B}\otimes 1_C)$. This produces the four crossings of $X,Y$ with $Y^{\vee },X^{\vee }$. Bend the $X,Y^{\vee }$ crossing across its $Y$ evaluation and the $Y,X^{\vee }$ crossing across its $X$ evaluation by the two identities of step 1.1. The remaining $X,X^{\vee }$ and $Y,Y^{\vee }$ crossings and their cups and caps are respectively $u_X$ and $u_Y$, while the two bent crossings on the input are $c_{X,Y}^{-1}{c}_{Y,X}^{-1}$. Consequently the expanded composite is $u_{X\otimes Y}=(u_X\otimes u_Y)c_{X,Y}^{-1}{c}_{Y,X}^{-1}$. Reinsert the chosen-dual comparison $d_{X,Y}$ from [L2] and multiply on the right by $c_{Y,X}{c}_{X,Y}$ to obtain exactly the stated identity. Steps 3.1 and 4.1 prove the natural-isomorphism and obstruction claims.