Nicolas Boubarki, Algebra I, Chapter 1, § 2, Ex. 12 
Nicolas Boubarki, Algebra I, Chapter 1, § 2, Ex. 12:

($E$ is a Semigroup with associative law (represented multiplicatively), $\gamma_a(x)=ax$.)
Under a multiplicative law on $E$, let $ a \in E $  be such that $\gamma_a $ is surjective.
(a) Show that, if there exists $u$ such that $ua=a$, then $ux=x$ for all $x\in E$.
(b) For an element $b\in E$ to be such that $ba$ is left cancellable, it is necessary and sufficient that $\gamma_a$ be surjective and that $b$ be left cancellable.
For those interested in part (a), simple proof is that for every $x\in E$ there exists $x^\prime \in E$ such that $ax^\prime=x$, consequently $ua=a \Rightarrow uax^\prime=ax^\prime \Rightarrow ux=x$.
In (b), surjectivity of $\gamma_a$ and left cancellability of $b$ is required.
However, I am concerned with "sufficiency" portion of part (b). When $E$ is infinite set there can always be a surjective function $\gamma_a$ which need not be injective, and left translation by $b$ is cancellable, however $ba$ need not be left cancellable.

I have a Russian translation of Bourbaki. In it Ex.12 looks as follows:
"For $\gamma_{ba}$ to be an one-one mapping of $E$
into  $E$,  it is necessary and sufficient that $\gamma_{a}$ be an one-one mapping of
$E$ onto  $E$ and $\gamma_{b}$ be an one-one mapping of
$E$ into  $E$."
So I guess that there is a misprint in English  translation. I wonder how it looks in the French original?
