# Trace in monoidal category acts multiplicatively

Every paper I've found claims this is true but nobody actually proves it.

Let $(C,\otimes)$ be a strict monoidal category with duals and for every object $a\in C$ an evaluation $\epsilon_a:a \otimes a^* \longrightarrow \mathbb{F}$ and a coevaluation $\tilde{\epsilon}_a:\mathbb{F}\longrightarrow a\otimes a^*$ (i.e. $C$ is strict pivotal). These maps give a notion of (left and right but assume they are equal) trace of endomorphisms. Let $f \in$ End$(a)$, then we can define its trace as the endomorphism of the field given by the composition $$\mathbb{F}\xrightarrow{\tilde{\epsilon}_a} a\otimes a^*\xrightarrow{f\otimes id}a\otimes a^* \xrightarrow{\epsilon_a} \mathbb{F}.$$ Now the claim is that if $f\in$End$(a), g \in$End$(b)$ then $$tr(f\otimes g) = tr(f)\cdot tr(g)$$ But I have no idea how to show this. I know that this is true for the usual matrix trace but in this general context I don't know how to approach it. Any advice would be greatly appreciated.

The only non-obvious equality is the first: it follows from the monoidality of the duality functor $(X \otimes Y)^* \cong Y^* \otimes X^*$ (with the corresponding isomorphisms for evaluation and coevalutation). Then the definition of the trace of $g$ is used, equality of left and right traces, and again the definition of the trace of $g$.
To prove this equality without the graphical calculus would be a feat into itself, but it should be possible to convert the picture above to a symbolic computation. But be prepared to write things like $$\operatorname{ev}_{Y \otimes X} \circ (\operatorname{id}_{(Y \otimes X)^*} \otimes (g \otimes f)) \circ \operatorname{\widetilde{coev}}_{Y \otimes X} = (\operatorname{ev}_X) \circ (\operatorname{id}_{X^*} \otimes \operatorname{ev}_Y \otimes \operatorname{id}_{X^*}) \circ (\operatorname{id} \otimes \operatorname{id} \otimes g \otimes f) \circ \dots$$ and this is just the beginning of the first equality!