Does trace handle composition in a traced symmetric monoidal category?

Question

Suppose that $(C,\otimes,I)$ is a traced symmetric monoidal category (TSMC) with symmetrizor $\sigma$ and trace $Tr$. Given two morphisms $f\colon A\to B$ and $g\colon B\to C$, I can tensor them to get a morphism $$f\otimes g\colon A\otimes B\to B\otimes C.$$ I can then apply $\sigma$ to swap $B$ and $C$ in the codomain, and subsequently apply the trace to obtain some morphism $$Tr^{B}_{A,C}\Big(\sigma_{B,C}\;(f\otimes g)\Big)\colon A\to C,$$ which I'll denote $T(f,g)$.

The natural question is whether $T(f,g)\colon A\to C$ is, or is somehow related to, the composition $g\circ f\colon A\to C$. I'm not seeing an obvious proof that it is, nor did I see the issue mentioned explicitly in the seminal Joyal-Street-Verity paper. Does it follow from the axioms of TSMCs?

Trying to draw the relevant property with string diagrams seems to suggest I need $C$ to be compact closed or something. But then how can I interpret this mysterious $T(f,g)$?

Answer 1

I assume you use the trivial balanced monoidal structure where the twists $\theta_A \colon A \rightarrow A$ are identities? If so, then the result you are asking follows indeed from the axioms.

First you can use naturality of the symmetry to find that $\mathrm{T}(f,g)=\mathrm{Tr}(g \otimes B \cdot\sigma_{B,B} \cdot f \otimes B)$. By naturality of the trace (the axiom called "tightening" in Joyal-Street-Verity), this reduces the problem to computing the trace of $\sigma_{B,B}$. But the "yanking" axiom says precisely that the trace of the braiding is equal to the twist. In other words, if you have non-trivial balanced structure, then $\mathrm{T}(f,g)=g \cdot \theta_B \cdot f$.