Basic properties of the categorical trace

Sliding a morphism. The defining mate identities for $f:U\to V$ are $(f\otimes 1_{U^{\vee }}){\mathrm{coev}}_U=(1_V\otimes f^{\vee }){\mathrm{coev}}_V$ and ${\mathrm{ev}}_U(f^{\vee }\otimes 1_U)={\mathrm{ev}}_V(1_{V^{\vee }}\otimes f)$. Apply the first identity to $f=a:X\to X^{\vee \vee }$. Postcomposing with ${\mathrm{ev}}_{X^{\vee }}$ gives ${\mathrm{Tr}}_L(a)={\mathrm{ev}}_{X^{\vee }}(1_{X^{\vee \vee }}\otimes a^{\vee }){\mathrm{coev}}_{X^{\vee \vee }}={\mathrm{Tr}}_R(a^{\vee })$, where the last trace is taken on $X^{\vee }$. This proves clause 1 without identifying a double dual with the original object.

Biproducts. In an additive rigid category, tensoring by a fixed object has both adjoints, furnished by its two duals, and hence preserves finite biproducts; duality carries a biproduct to the corresponding biproduct of duals. Write $i_U,p_U$ for the injections and projections of $Z=X\oplus Y$. Relative to $Z^{\vee }=X^{\vee }\oplus Y^{\vee }$, its coevaluation is ${\sum }_{U=X,Y}(i_U\otimes p_U^{\vee }){\mathrm{coev}}_U$, and its evaluation pairs matching summands by ${\mathrm{ev}}_U$ and different summands by zero. These formulas satisfy the zig-zags since $p_U{i}_V$ is the identity for $U=V$ and zero otherwise, so uniqueness of duals permits their use for the chosen dual. Inserting $a\oplus b$ leaves exactly the two diagonal trace composites. Their sum is ${\mathrm{Tr}}_L(a)+{\mathrm{Tr}}_L(b)$, proving clause 2.

Nested tensor trace. Use $(X\otimes Y{)}^{\vee }=Y^{\vee }\otimes X^{\vee }$. The coevaluation of $X\otimes Y$ is $(1_X\otimes {\mathrm{coev}}_Y\otimes 1_{X^{\vee }}){\mathrm{coev}}_X$. The evaluation of its dual $Y^{\vee }\otimes X^{\vee }$ first pairs the middle $Y^{\vee \vee },Y^{\vee }$ and then the outer $X^{\vee \vee },X^{\vee }$. Substitute $a\otimes b$ into this composite. Evaluating the middle pair gives $s={\mathrm{Tr}}_L(b)$ situated between $X^{\vee \vee }$ and $X^{\vee }$. Thus the result is exactly ${\mathrm{ev}}_{X^{\vee }}(r_{X^{\vee \vee }}(s)a\otimes 1_{X^{\vee }}){\mathrm{coev}}_X={\mathrm{Tr}}_L(r_{X^{\vee \vee }}(s)a)$. If $r_{X^{\vee \vee }}(s)={\mathrm{\ell }}_{X^{\vee \vee }}(s)$, the inserted scalar instead acts on the left of the whole evaluation input. Naturality of that left action at ${\mathrm{ev}}_{X^{\vee }}$ moves it to the output unit, giving $s{\mathrm{Tr}}_L(a)$. Endomorphisms of the unit commute under composition by interchange, so this equals the asserted product. No equality of left and right actions is assumed in the general nested formula.

Cyclicity, including different objects. More generally, for $f:U\to V$ and $h:V\to U^{\vee \vee }$, the first mate identity in step 1.1 and then the second identity applied to $f^{\vee }$ give ${\mathrm{ev}}_{U^{\vee }}(hf\otimes 1){\mathrm{coev}}_U={\mathrm{ev}}_{U^{\vee }}(h\otimes f^{\vee }){\mathrm{coev}}_V={\mathrm{ev}}_{V^{\vee }}(f^{\vee \vee }h\otimes 1){\mathrm{coev}}_V$. Thus ${\mathrm{Tr}}_{L,U}(hf)={\mathrm{Tr}}_{L,V}(f^{\vee \vee }h)$. Taking $U=V=X$, $f=c$ and $h=a$ proves clause 4. The mirrored mate identities give the corresponding right-trace cyclicity.

Sufficient hypotheses and right trace. In a braided category the hexagon at $\mathbf{1}\otimes \mathbf{1},Z$, with unit constraints inserted, makes the unit-transported $c_{\mathbf{1},Z}$ an idempotent automorphism of $Z$, hence the identity. Naturality of $c_{\mathbf{1},Z}$ at $s:\mathbf{1}\to \mathbf{1}$ gives $c_{\mathbf{1},Z}(s\otimes 1_Z)=(1_Z\otimes s)c_{\mathbf{1},Z}$. Compatibility with the unit identifies $c_{\mathbf{1},Z}$ with the composite $\mathbf{1}\otimes Z\to Z\to Z\otimes \mathbf{1}$, so ${\mathrm{\ell }}_Z(s)=r_Z(s)$. Alternatively, in a $k$-linear category with bilinear tensor and $\mathrm{End}(\mathbf{1})=k$, each $s=\lambda 1_{\mathbf{1}}$ has both actions $\lambda 1_Z$. For the right tensor trace, use the coevaluation of $Y^{\vee }\otimes X^{\vee }$: first close the middle $X^{\vee },X^{\vee \vee }$ pair with $p$, leaving ${\mathrm{Tr}}_R(p)$ immediately to the left of $Y^{\vee \vee }$. The remaining composite is exactly ${\mathrm{Tr}}_R(q{\mathrm{\ell }}_{Y^{\vee \vee }}({\mathrm{Tr}}_R(p)))$. When left and right actions agree at this scalar, move it to the outer right of the evaluation and hence to the output unit; this proves the right multiplicative formula. The mirrored mate and biproduct calculations prove the other right-trace identities. Together with steps 1.1–1.3 and 2.1 this proves all assertions.