I have been trying to understand why the ground monoid in a monoidal category is commutative and every proof I have seen essentially uses the same thing to prove it namely by using the fact(apparently) that if every other subdiagram of a diagram is commutative then the remaining subdiagram is commutative. They all leave the proof of this fact and I don't understand why this should hold. I know that it has something to do with the fact that all the arrows in this diagram are isomorphisms but I don't know the whole picture.
I have added this picture from Vladimir's book on monoidal categories the emphasis here is on the 2nd last paragraph where it states "consequently the lower right triangle commutes" I don't understand this point.
$\text{End}(1)$ has two monoid operations. One is composition and the other is tensor product, by which I mean: if $f, g : 1 \to 1$ are two morphisms they have a tensor product $f \otimes g : 1 \otimes 1 \to 1 \otimes 1$, and then we use the unit maps $1 \otimes 1 \cong 1$ to turn this into another endomorphism of $1$. These two monoid operations satisfy the interchange law and so you can run the Eckmann-Hilton argument on them: see for example this blog post.