Let ${\mathcal{C}}$ be a monoidal category—I don’t actually want to define what that means, so I refer you to  John Armstrong’s post–with ${\otimes}$ as the monoidal operation. Suppose ${1}$ is a unital object.

The following is well-known:

Theorem 1 ${End(1)}$ is a commutative monoid.

Oftentimes, one has an additive structure on ${\mathcal{C}}$ as well, and one actually wants ${End(1)}$ to be a field. The result is interesting, because it strikes a parallel with the following:

Proposition 2 The endomorphisms of the identity functor in a category ${\mathcal{C}}$ form a commutative monoid.

The proof is different though.   In some places, it’s not even properly mentioned; in others, it’s always seemed extremely non-intuitive.

I learned of a neat proof of the first theorem in the first chapter of a book by Saavedra on Tannakian categories. It is as follows. By definition, ${1 \simeq 1 \otimes 1}$, so it is enough to prove ${End(1 \otimes 1)}$ commutative. Let ${f, g \in End(1 \otimes 1)}$. Since the functors ${- \rightarrow - \otimes 1}$ and ${- \rightarrow 1 \otimes -}$ are equivalences of categories, and in particular fully faithful, we can write ${f = u \otimes \mathrm{id}_1, g = \mathrm{id}_1 \otimes v}$ for appropriate ${u,v \in End(1)}$. But then

$\displaystyle f \circ g = u \otimes v = g \circ f ,$

which proves commutativity.