= Solution
Let $n:k\to V^*\otimes V$ be the supplied <coevaluation morphism>, $n(1)=x\otimes u+y\otimes v$, and let $e:V^*\otimes V\to k$ 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 identity
$$
(e\otimes1_V)(1_{V^*}\otimes c)(n\otimes1_V)=1_V.
$$
We calculate this map on both <basis> vectors. The crossing matrix gives
$$
\begin{aligned}
c(u\otimes u)&=q^{-1}u\otimes u,\\
c(u\otimes v)&=q^{-2}v\otimes u,\\
c(v\otimes u)&=q^{-2}u\otimes v+q^{-2}(q-q^{-1})v\otimes u,\\
c(v\otimes v)&=q^{-1}v\otimes v.
\end{aligned}
$$
For input $u$, applying $n$, then $c$, then $e$ gives
$$
\left[q^{-1}A+q^{-2}(q-q^{-1})B\right]u.
$$
For input $v$, the off-diagonal evaluation vanishes and the result is $q^{-1}Bv$. Therefore the removable-curl identity requires
$$
q^{-1}B=1,\qquad q^{-1}A+q^{-2}(q-q^{-1})B=1.
$$
Solve the first equation to obtain $B=q$; substitution in the second gives $A=q^{-1}$. Hence
$$
\boxed{A=q^{-1},\qquad B=q.}
$$
Finally 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> is
$$
\boxed{\dim_q(V)=q^{-1}+q=[2].}
$$
It 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.
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