Existence of Mappings Creating Identity Mappings

Table of Contents

Let’s introduce a fundamental theorem regarding the existence of mappings that create identity mappings.

Theorem

Let $f$ be a mapping from $A$ to $B$ .

If $f$ is surjective, then and only then there exists a mapping $s: B \rightarrow A$ such that $f \circ s = I_B$ .
If $f$ is injective, then and only then there exists a mapping $r: B \rightarrow A$ such that $r \circ f = I_A$ .

Proof

If $f \circ s = I_B$ , then for any $b \in B$ , $f(s(b)) = b$ . Therefore, $f$ is surjective.

Let $f: A \rightarrow B$ be surjective. Define $s(b)$ for $b \in B$ such that $s(b) \in f^{-1}(b)$ . By the Axiom of Choice, such $s(b)$ can be defined. Then, for any $b \in B$ , $f(s(b)) = b$ .
If $r \circ f = I_A$ , then if $f(a_1) = f(a_2)$ for any elements $a_1$ , $a_2$ of $A$ , it follows that $a_1 = r(f(a_1)) = r(f(a_2)) = a_2$ . Therefore, $f$ is injective.

Let $f: A \rightarrow B$ be injective. For $b \in B$ , if $b \in f(A)$ , there exists an element $a \in A$ such that $b = f(a)$ . Define $r$ using that $a$ as $r(b) = a$ . If $b \not\in f(A)$ , pick an element $a_0$ from $A$ and define $r(b) = a_0$ . Then, for any $a \in A$ , $r(f(a)) = a$ .

Corollary

Let $A$ and $B$ be two sets. The necessary and sufficient condition for the existence of an injective mapping from $A$ to $B$ is the existence of a surjective mapping from $B$ to $A$ .

Proof

If there exists an injective mapping $\varphi: A \rightarrow B$ , then by Theorem 2, a mapping $\psi: B \rightarrow A$ exists such that $\psi \circ \varphi = I_A$ . $\psi$ is surjective according to Theorem 1.

If there exists a surjective mapping $\psi: B \rightarrow A$ , then by Theorem 1, a mapping $\varphi: A \rightarrow B$ exists such that $\psi \circ \varphi = I_A$ . $\varphi$ is injective according to Theorem 2.