Exercise 3, § 2

In the set F of mappings of E into E, for the relation $f \circ g = f \circ h$ to imply $g = h$, it is necessary and sufficient that $f$ be injective; for the relation $g \circ f = h \circ f$ to imply $g = h$, it is necessary and sufficient that $f$ be surjective; for $f$ to be cancellable (under the law $\circ$), it is necessary and sufficient that $f$ be bijective.