It is well known that, for square matrix $x$ and $y$, we have $\operatorname{tr}(xy)=\operatorname{tr}(yx)$. Here of course the trace of a matrix is just the sum of the elements of the diagonal.


The notion of trace has a lot of generalization. As I know, the most general definition is the following: let $(\mathcal C, \otimes, 1, ^\vee)$ be a rigid symmetric monoidal category, $X$ an object of $\mathcal C$ and $f$ an endomorphism of $X$. Then $\operatorname{tr}(f) \in \operatorname{End}(1)$ is defined by the following composition
$$ 1 \longrightarrow X^\vee \otimes X \stackrel{\operatorname{id}\_{X^\vee} \otimes f}{\longrightarrow} X^\vee \otimes X \longrightarrow X \otimes X^\vee \longrightarrow 1 $$
So my questions is: it is true, in this generality, that $\operatorname{tr}(f\circ g)=\operatorname{tr}(g \circ f)$, for $f$ and $g$ in $\operatorname{End}(X)$?


Ricky



trace(xy)=trace(yx) in full generality
Yes. The string diagram chase can be found on page 8 of [Ponto and Shulman - Traces in symmetric monoidal categories](https://arxiv.org/abs/1107.6032).


