Composition of injections is injective

Statement

Let $f \colon A \to B$ and $g \colon B \to C$ be injective functions. Then the composition $g \circ f \colon A \to C$ is injective.

Remarks

Proof: if $g(f(a)) = g(f(a'))$, then $f(a) = f(a')$ because $g$ is injective, and $a = a'$ because $f$ is injective. A partial converse holds: if $g \circ f$ is injective, then $f$ is injective. The function $g$ need not be injective.

Explanations

L1 human

Think of $f$ and $g$ as machines, and of $g \circ f$ as the two machines in a chain. If neither machine merges two inputs, the chain does not merge two inputs either.