Suppose that $(C,\otimes,I)$ is a [traced symmetric monoidal category](https://en.wikipedia.org/wiki/Traced_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](http://sci-prew.inf.ua/v119/3/S0305004100074338.pdf) 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 $T(f,g)$?



Does trace handle composition in a traced symmetric monoidal category?
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 $\mathrm{T}(f,g)=g \cdot \theta\_B \cdot f$. 


