# Inverse Mapping is Bijection

 It has been suggested that this page or section be merged into Mapping is Injection and Surjection iff Inverse is Mapping/Sufficient Condition/Proof 2. (Discuss)

## Theorem

Let $S$ and $T$ be sets.

Let $f: S \to T$ and $g: T \to S$ be inverse mappings of each other.

Then $f$ and $g$ are bijections.

## Proof

 $\displaystyle f \paren x$ $=$ $\displaystyle f \paren y$ $\displaystyle \leadsto \ \$ $\displaystyle g \paren {f \paren x}$ $=$ $\displaystyle g \paren {f \paren y}$ as $g$ is a mapping $\displaystyle \leadsto \ \$ $\displaystyle x$ $=$ $\displaystyle y$ Definition 1 of Inverse Mapping

Thus $f$ is by definition an injection.

 $\displaystyle t$ $\in$ $\displaystyle T$ $\displaystyle \leadsto \ \$ $\displaystyle g \paren t$ $\in$ $\displaystyle S$ as $g$ is a mapping $\displaystyle \leadsto \ \$ $\displaystyle f \paren {g \paren t}$ $=$ $\displaystyle t$ Definition 1 of Inverse Mapping $\displaystyle \leadsto \ \$ $\, \displaystyle \exists s \in S: \,$ $\displaystyle f \paren s$ $=$ $\displaystyle t$ setting $s = g \paren t$

Thus $f$ is by definition a surjection.

Similarly:

 $\displaystyle g \paren x$ $=$ $\displaystyle g \paren y$ $\displaystyle \leadsto \ \$ $\displaystyle f \paren {g \paren x}$ $=$ $\displaystyle f \paren {g \paren y}$ as $f$ is a mapping $\displaystyle \leadsto \ \$ $\displaystyle x$ $=$ $\displaystyle y$ Definition 1 of Inverse Mapping

Thus $g$ is by definition an injection.

 $\displaystyle s$ $\in$ $\displaystyle S$ $\displaystyle \leadsto \ \$ $\displaystyle f \paren s$ $\in$ $\displaystyle T$ as $f$ is a mapping $\displaystyle \leadsto \ \$ $\displaystyle g \paren {f \paren s}$ $=$ $\displaystyle s$ Definition 1 of Inverse Mapping $\displaystyle \leadsto \ \$ $\, \displaystyle \exists t \in T: \,$ $\displaystyle g \paren t$ $=$ $\displaystyle s$ setting $t = f \paren s$

Thus $g$ is by definition a surjection.

So $f$ and $g$ are both injections and surjections.

The result follows by definition of bijection.

$\blacksquare$