The endomorphisms of the tensor unit form a commutative monoid

On ${\mathrm{End}}_{\mathcal{C}}(\mathbf{1})$ let $f\circ g$ be ordinary composition and define $f\ast g:={\lambda }_{\mathbf{1}}\circ (f\otimes g)\circ {\lambda }_{\mathbf{1}}^{-1}$. Composition is associative with unit $1_{\mathbf{1}}$. Naturality of $\lambda$ at $f$ gives ${\lambda }_{\mathbf{1}}\circ (1_{\mathbf{1}}\otimes f)=f\circ {\lambda }_{\mathbf{1}}$, so $1_{\mathbf{1}}\ast f=f$. Naturality of $\rho$ at $f$ gives ${\rho }_{\mathbf{1}}\circ (f\otimes 1_{\mathbf{1}})=f\circ {\rho }_{\mathbf{1}}$, and [L2] identifies ${\rho }_{\mathbf{1}}$ with ${\lambda }_{\mathbf{1}}$. Hence $f\ast 1_{\mathbf{1}}=f$ as well, so $\ast$ has the same unit $1_{\mathbf{1}}$.

For $f,g,h,k\in {\mathrm{End}}_{\mathcal{C}}(\mathbf{1})$, one has $((f\circ g)\ast (h\circ k))={\lambda }_{\mathbf{1}}\circ ((f\circ g)\otimes (h\circ k))\circ {\lambda }_{\mathbf{1}}^{-1}={\lambda }_{\mathbf{1}}\circ ((f\otimes h)\circ (g\otimes k))\circ {\lambda }_{\mathbf{1}}^{-1}=(f\ast h)\circ (g\ast k)$, so the interchange law holds.

By [L4], the operations $\circ$ and $\ast$ coincide and their common value is commutative. Since composition is already associative and unital, [L3] shows that composition makes ${\mathrm{End}}_{\mathcal{C}}(\mathbf{1})$ into a commutative monoid.

If $\mathcal{C}$ is strict and has one object, then tensor and composition on endomorphisms are the same literal operation. Thus its endomorphism monoid is commutative; [L5] supplies the converse comparison with an ordinary one-object category.

Therefore the endomorphisms of the tensor unit form a commutative monoid.