Let $C$ be a strict monoidal category. Take objects, $a,b\in C$ and morphisms $T\in C(a,1)$, $S\in C(1,b)$. Does it hold then that $ST=S\otimes T=T\otimes S$? In particular, does the identity morphism of the identity object act as an identity also with respect to the tensor product?
It surely holds for the category of Hilbert spaces and for some diagramatical categories I am working with, but it is not clear to me whether it follows from the axioms.
Best Answer
Yes, it follows from the axioms. In fact in any monoidal category the diagrams $$\require{AMScd} \begin{CD}A @>{T}>> I @>{S}>> B \\ @V{\rho_A}VV @V{\rho_I}V{=\lambda_I}V @VV{\lambda_B}V\\ A\otimes I@>>{T\otimes id_I}> I\otimes I @>>{id_I\otimes S}> I\otimes B \end{CD} $$ commute (this is simply the naturality of the unitors), and $$(id_I\otimes S)\circ (T\otimes id_I)=(id_I\circ T)\otimes(S\circ id_I)=T\otimes S $$ (this is is just the functoriality of $\otimes$). Thus in any monoidal category, you have $$ST=\lambda_B^{-1} \circ T\otimes S \circ \rho_A,$$ and by a similar argument $$ST=\rho_B^{-1} \circ S\otimes T \circ \lambda_A.$$ In a strict monoidal category, these identities simply become $ST=T\otimes S=S\otimes T$.