Ken Scambler wrote:

>The inverse operation takes x to -x, -x to x, and the identity 0 to itself. The group equations follow: -x + x = x + -x = 0 and 0 = -0

I don't think that will work since it just swaps inversion.

However, \\(-x \mapsto x\\), \\(x \mapsto x\\), and \\(0 \mapsto 0\\) I think will work.

The adjoint here takes a monoid and freely adds inverse elements, while it's right adjoint forgets what it means to be an inverse.

In the context of numbers, the adjoint is the multiplication of a natural number by \\(-1\\), \\[-n : \mathbb{N} \to \mathbb{Z},\\] it's right adjoint is the absolute value, \\[|z|: \mathbb{Z} \to \mathbb{N}.\\]

>The inverse operation takes x to -x, -x to x, and the identity 0 to itself. The group equations follow: -x + x = x + -x = 0 and 0 = -0

I don't think that will work since it just swaps inversion.

However, \\(-x \mapsto x\\), \\(x \mapsto x\\), and \\(0 \mapsto 0\\) I think will work.

The adjoint here takes a monoid and freely adds inverse elements, while it's right adjoint forgets what it means to be an inverse.

In the context of numbers, the adjoint is the multiplication of a natural number by \\(-1\\), \\[-n : \mathbb{N} \to \mathbb{Z},\\] it's right adjoint is the absolute value, \\[|z|: \mathbb{Z} \to \mathbb{N}.\\]