ok I think I've got a solution to **Puzzle 280**

> Suppose \\(f : I \to I\\) and \\(g : I \to I\\) are morphisms in a monoidal category going from the unit object to itself. Show that \\(fg = gf\\).

We've seen [earlier](https://forum.azimuthproject.org/discussion/comment/20720/#Comment_20720) how to prove

> \[(1\otimes f)\circ(k\otimes 1) = (k\otimes 1)\circ(1\otimes f)\]

I've switched \\(f\\) and \\(k\\) around to match the diagram below, sorry!

The other ingredient we need is the fact that \\(\lambda_I = \rho_I : I \otimes I \to I\\)

This is listed in Mac Lane as one of the coherence equations, although more recent treatments seem to omit it. I'm assuming this because it can be derived from the other two equations ("pentagon" and "triangle"). I'm at a loss as to how to go about this, however.

Putting these together we get the following commutative diagram:

The central diamond commutes because of the result proved earlier, and the wonky squares are all naturality diagrams.

The outer square, together with \\(\lambda_I = \rho_I\\), gives us what we're looking for: \\(f\circ k = k\circ f\\)

> Suppose \\(f : I \to I\\) and \\(g : I \to I\\) are morphisms in a monoidal category going from the unit object to itself. Show that \\(fg = gf\\).

We've seen [earlier](https://forum.azimuthproject.org/discussion/comment/20720/#Comment_20720) how to prove

> \[(1\otimes f)\circ(k\otimes 1) = (k\otimes 1)\circ(1\otimes f)\]

I've switched \\(f\\) and \\(k\\) around to match the diagram below, sorry!

The other ingredient we need is the fact that \\(\lambda_I = \rho_I : I \otimes I \to I\\)

This is listed in Mac Lane as one of the coherence equations, although more recent treatments seem to omit it. I'm assuming this because it can be derived from the other two equations ("pentagon" and "triangle"). I'm at a loss as to how to go about this, however.

Putting these together we get the following commutative diagram:

The central diamond commutes because of the result proved earlier, and the wonky squares are all naturality diagrams.

The outer square, together with \\(\lambda_I = \rho_I\\), gives us what we're looking for: \\(f\circ k = k\circ f\\)