Is it true that if multiplication on the right by a element of the monoid is a bijection, then the element is a invertible element?
In other words, is it true that for the homomorphism $M \to \mathrm{End}(M): m \mapsto (x \mapsto xm)$ the inverse image of the group of invertible elements is subset the group of invertible elements?
Best Answer
Yes. Trivially, if multiplication by $a$ on the right is surjective, then $a$ has a left inverse $b$. Now here's the trick to show that $b$ is also a right inverse: consider $aba$. We have $a(ba)=a1=1a$. But since right multiplication by $a$ is injective, $aba=1a$ implies $ab=1$.
From a much fancier point of view, this is a corollary of the Yoneda lemma. Considering your monoid as a one-object category, multiplication by $a$ can be considered as a natural transformation from the functor represented by the one object to itself. If this natural transformation happens to be pointwise bijective, then its pointwise inverse is also a natural transformation. Yoneda then says that this inverse transformation comes from a morphism of the category, which will be an inverse for $a$.