 Solution

Let $n:k\to V^{\ast }\otimes V$ be the supplied coevaluation morphism, $n(1)=x\otimes u+y\otimes v$, and let $e:V^{\ast }\otimes V\to k$ be the supplied evaluation morphism with diagonal coefficients $A,B$. The tangle obtained by inserting the given crossing between the created strand and a through strand, then evaluating, is a removable curl. The first Reidemeister move gives the identityWe calculate this map on both basis vectors. The crossing matrix givesFor input $u$, applying $n$, then $c$, then $e$ givesFor input $v$, the off-diagonal evaluation vanishes and the result is $q^{-1}Bv$. Therefore the removable-curl identity requiresSolve the first equation to obtain $B=q$; substitution in the second gives $A=q^{-1}$. HenceFinally the closed identity strand is represented by $e\circ n$. Its value is $e(x\otimes u+y\otimes v)=A+B$, so the quantum dimension isIt is a weighted categorical contraction, rather than the ordinary vector space dimension $2$. Using a framed category without imposing the removable-curl identity would require additional twist data; the ordinary tangle category and the specified crossing normalization determine these coefficients.