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Proposition Let $M$ be a nonunital monoid, let $e$ be an element of $M$, and let $\ell _{e}: M \rightarrow M$ denote the function given by the formula $\ell _{e}(x) = ex$. The following conditions are equivalent:


The element $e$ is a left unit of $M$: that is, $\ell _{e}$ is the identity function from $M$ to itself.


The element $e$ is idempotent (that is, it satisfies $ee = e$) and the function $\ell _{e}: M \rightarrow M$ is a bijection.


The element $e$ is idempotent and the function $\ell _{e}: M \rightarrow M$ is a monomorphism.

Proof. The implications $(a) \Rightarrow (b) \Rightarrow (c)$ are immediate. To complete the proof, assume that $e$ satisfies condition $(c)$ and let $x$ be an element of $M$. Using the assumption that $e$ is idempotent (and the associativity of the multiplication on $M$), we obtain an identity $\ell _{e}(x) = ex = (ee)x = e(ex) = \ell _{e}( ex )$. Since $\ell _{e}$ is a monomorphism, it follows that $x = ex$. $\square$