The Composition of Injective, Surjective, and Bijective Functions

Theorem 1: If $f : A \to B$ and $g : B \to C$ are injective then $g \circ f : A \to C$ is injective.

• Proof: Let $x, y \in A$ and suppose that $(g \circ f)(x) = (g \circ f)(y)$. Then $g(f(x)) = g(f(y))$. Since $g$ is injective we have that $g(f(x)) = g(f(y))$ implies that $f(x) = f(y)$. And since $f$ is injective we have that $f(x) = f(y)$ implies that $x = y$. So $g \circ f$ is injective. [[$ \blacksquare

Theorem 2: If $f : A \to B$ and $g : B \to C$ are surjective then $g \circ f : A \to C$ is surjective.

• Proof: Let $c \in C$. Since $g $] is surjective there exists a [[$ b \in B$ such that $g(b) = c$. For $b \in B$, since $f$ is surjective there exists an $a \in A$ such that $f(a) = b$. Therefore:

(1)

\begin{align} \quad (g \circ f)(a) = g(f(a)) = g(b) = c \end{align}

• So $g \circ f$ is surjective. $\blacksquare$

Theorem 3: If $f : A \to B$ and $g : B \to C$ are bijective then $g \circ f : A \to C$ is bijective.

• Proof: This follows immediately from theorem 1 and theorem 2. $\blacksquare$