The Composition of Injective, Surjective, and Bijective Functions

# 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} • Sog \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\$