Isomorphism (Category Theory) is Epic

Theorem
Let $\mathbf C$ be a metacategory.

Let $f: C \to D$ be an isomorphism.

Then $f: C \twoheadrightarrow D$ is epic.

Proof
Let $g: D \to C$ be the inverse of $f$.

Suppose that $x, y: D \to E$ are morphisms such that:


 * $x \circ f = y \circ f$

Then necessarily also:


 * $x \circ f \circ g = y \circ f \circ g$

and hence, since $f \circ g = \operatorname{id}_D$, it follows that:


 * $x \circ \operatorname{id}_D = y \circ \operatorname{id}_D$

which yields the result by the definition of identity morphism.

The situation is illustrated by the following commutative diagram:


 * $\begin{xy}

<0em,0em> *+{C} = "C", <0em,-4em>*+{D} = "D", <4em,0em> *+{D} = "D2", <8em,0em> *+{E} = "E",

"D2"+/r.5em/+/^.25em/;"E"+/l.5em/+/^.25em/ **@{-} ?>*@{>} ?*!/_.6em/{x}, "D2"+/r.5em/+/_.25em/;"E"+/l.5em/+/_.25em/ **@{-} ?>*@{>} ?*!/^.6em/{y},

"C";"D2" **@{-} ?>*@{>} ?*!/_.6em/{f}, "D";"D2" **@{-} ?>*@{>} ?*!/^.6em/{\operatorname{id}_C}, "D";"C" **@{-} ?>*@{>} ?*!/_.6em/{g}, \end{xy}$