Let $C$ be a symmetric monoidal category. I am interested in objects $X \in C$ such that the symmetry


$S\_{X,X} : X \otimes X \cong X \otimes X$


is equal to the identity. There are many examples of such objects, e.g. invertible sheaves. My first question is: How would you call such an object?


Now assume that $X$ has a dual $Y$, i.e. we have morphisms $e: Y \otimes X \to 1$ and $c : 1 \to X \otimes Y$ such that the triangular identities are satisfied.


**Question**. Assuming $S\_{X,X}$ is the identity, can we conclude that $S\_{Y,Y}$ is the identity? If not, does it suffice to assume that $e$ (and thus $c$) is an isomorphism?


Edit: I am still interested how objects with $S\_{X,X}=\mathrm{id}$ are called in the literature or which terminology you would suggest.



Does the dual of an object with trivial symmetry also have trivial symmetry?
I believe the answer to your question is yes, without a further assumption that e is an isomorphism. The symmetry S\_{Y,Y} can be obtained from the symmetry S\_{X,X}


as follows


$Y\otimes Y \xrightarrow{c\circ c} Y\otimes Y \otimes X\otimes X \otimes Y \otimes Y \xrightarrow{id\_Y^{\otimes 2} S\_{X,X}\otimes id\_Y^{\otimes 2}} Y\otimes Y \otimes X\otimes X \otimes Y \otimes Y \xrightarrow{e\circ e}Y\otimes Y$.


Here, $c\circ c$ is shorthand for $(id\_Y^{\otimes 2}\otimes c \otimes id)\circ(id\_Y^{\otimes 2}\otimes c)$, and similarly for $e\circ e$.


In pictures, all I'm doing (which I would draw if I knew an easy way) is:


Take $Y \otimes Y$ up, and then bend them around to the right and back down (they become X's on the downward strand, apply $S\_{X,X}$, then bend the X's back around and up to the right (where they become Y's again.


here is a [pdf of the computation](http://math.mit.edu/~djordan/2011-10-25-Note-11-11.pdf)


I am just really repating a proof here that $S\_{U^\*,V^\*}=S\_{U,V}^\*$, which holds for the braiding in any rigid braided monoidal categetory.


Since $S\_{X,X}$ is the identity, you will get a diagram which is recognizable as the identity for $Y\otimes Y$.


