Question about Example 1.27

[jiwhiz]

In Page 8, Example 1.27 to prove that a group in the category of groups is an abelian group, I don't understand the last step:

Furthermore, taking into account that m and μ agree, we also obtain m(x, y) = m(m(u, x), m(y, u)) = m(m(u, y), m(x, u)) = m(y, x).

How do you get the middle equation after switching x and y?

Maybe we should use the inverses i and ι agree, and ι is a group homomorphism: m(x, y) = i(i(m(x, y))) = i(m(i(y), i(x))) = ι(m(i(y), i(x))) = m(ι(i(y)), ι(i(x))) = m(y, x)