# Left inverse in monoid, left dual in monoidal category, and uniqueness

abstract-algebracategory-theorymonoidmonoidal-categories

The notion of monoidal category is a categorification of the notion of monoid. If $$M$$ is a monoid, consider the monoidal category $$Vec_M$$ (of $$M$$-graded vector spaces over a field $$\mathbb{k}$$).

If an object of a monoidal category admits a left dual then it is unique up to isomorphism [EGNO, Proposition 2.10.5]. Now, let $$M$$ be a monoid, and let $$m$$ be an element admiting left inverses $$m'$$ and $$m''$$. Then the objects $$\delta_{m'}$$ and $$\delta_{m''}$$ of $$Vec_M$$ should be left duals of $$\delta_{m}$$, so should be isomorphic by previous proposition, and then $$m'$$ and $$m''$$ should be the same element.

But in a monoid, an element can have two distinct left inverses (see the answers here), which contradicts the previous paragraph. Where is the mistake?

Reference

[EGNO] Etingof, Pavel; Gelaki, Shlomo; Nikshych, Dmitri; Ostrik, Victor. Tensor categories. Mathematical Surveys and Monographs, 205. American Mathematical Society, Providence, RI, 2015. xvi+343 pp.

It is not true that $$\delta_{m'}$$ is a left dual of $$\delta_m$$ unless $$m'$$ is inverse to $$m$$ on both sides. Indeed, for $$\delta_{m'}$$ to be left dual to $$\delta_m$$, you need a map $$\delta_{m'}\otimes \delta_m\to 1$$ and a map $$1\to\delta_m\otimes\delta_{m'}$$ that make certain diagrams commute. If $$mm'$$ and $$m'm$$ are not both equal to the identity element, then one of these maps will be forced to be $$0$$, and so the required diagrams will not be able to commute (they will end up saying the identity maps on $$\delta_m$$ and $$\delta_{m'}$$ have to be equal to $$0$$).