Inverse is Mapping implies Mapping is Surjection

Theorem

Let $S$ and $T$ be sets.

Let $f: S \to T$ be a mapping.

Let the inverse $f^{-1} \subseteq T \times S$ itself be a mapping.

Then $f$ is a surjection.

Proof 1

Let $f^{-1}: T \to S$ be a mapping.

Aiming for a contradiction, suppose $f$ is not a surjection.

That is:

$\exists y \in T: \neg \exists x \in S: \tuple {x, y} \in f$

By definition of inverse of mapping:

$\exists y \in T: \neg \exists x \in S: \tuple {y, x} \in f^{-1}$

which would mean that $f^{-1}$ is not a mapping.

From this contradiction it follows that $f$ is a surjection.

$\blacksquare$

Proof 2

Let $f^{-1}: T \to S$ be a mapping.

We have:

\(\ds t\)

\(\in\)

\(\ds T\)

\(\ds \leadsto \ \ \)

\(\ds \map {f^{-1} } t\)

\(\in\)

\(\ds S\)

as $f^{-1}$ is a mapping

\(\ds \leadsto \ \ \)

\(\ds \map f {\map {f^{-1} } t}\)

\(=\)

\(\ds t\)

Definition 1 of Inverse Mapping

\(\ds \leadsto \ \ \)

\(\ds \exists s \in S: \, \)

\(\ds \map f s\)

\(=\)

\(\ds t\)

setting $s = \map {f^{-1} } t$

Thus $f$ is by definition a surjection.

$\blacksquare$

Also see

• Inverse is Mapping implies Mapping is Injection