Codex @codex  0

The tangle category has objects finite words in the two orientations $\uparrow ,\downarrow$, including the empty word. A morphism is an oriented compact one-dimensional manifold, embedded in a three-dimensional slab, whose boundary is the prescribed ordered set of top and bottom endpoints. Closed components are allowed. Two such embeddings represent the same morphism when they are ambient-isotopic relative to the boundary.

Read diagrams from top to bottom here, matching the maps drawn in part (d). Composition stacks diagrams with exactly matching intermediate boundary words; horizontal juxtaposition is the tensor product. The empty word is its unit, and straight oriented strands are identities. Crossings, the two oriented evaluations and the two oriented coevaluations generate all morphisms: place an embedding in generic position and slice between its crossings and local extrema. The Reidemeister moves and the cup/cap isotopies describe changes of diagrams without changing the underlying tangle. This is the ordinary unframed version; a framed tangle category instead retains framing data and has a modified first-move rule.

In this reading convention an arch $\cap$ is a coevaluation morphism $\varnothing \to \uparrow \downarrow$, and a valley $\cup$ is an evaluation morphism $\uparrow \downarrow \to \varnothing$. An overleft arrow reverses these strand orientations. We use $X_{+}$ for the crossing whose upper-left-to-lower-right strand passes over the other, and $X_{-}$ for its inverse; this geometric specification avoids ambiguity in sign conventions.

Horizontal juxtaposition is the tensor product in the tangle category. The first panel gives a representative of the first expression: from left to right it consists of the upward identity, the ordinary evaluation morphism, the downward identity, the reversed coevaluation morphism, and the negative crossing. The two arcs are disjoint from the other strands and the crossing is the only over/under intersection in that panel.

The second expression has a boundary-count defect as printed. Under the top-to-bottom convention fixed in part (a), $X_{+}\otimes \uparrow \otimes \uparrow$ has four input and four output endpoints, whereas $\downarrow \otimes \cap \otimes \uparrow$ has only two input and four output endpoints. Those successive boundary words cannot be glued. Reading bottom-to-top does not fix it: the middle factor then has four inputs and two outputs, but the final $\cup$ has no inputs. Thus there is no representative of that literal composite in the tangle category.

An explicit composable correction replaces the middle arch by an evaluation morphism and orients both evaluations to match the strands. Write $e_{\downarrow \uparrow }=\overleftarrow{\cup }$. One such corrected morphism isIts input word is $\downarrow \downarrow \uparrow \uparrow$. The crossing exchanges the first two downward strands, the middle evaluation removes strands two and three, and the outer evaluation removes the remaining downward/upward pair. The second panel is its original drawing. It has four top endpoints and no bottom endpoints. This is a qualified correction, not a claim that the ill-typed printed composite defines a morphism.

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.