Inverse is Mapping implies Mapping is Injection

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 an injection.

Proof 1

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

Let $\map f {x_a} = y$ and $\map f {x_b} = y$.

Then:

\(\ds \tuple {x_a, y}\)

\(\in\)

\(\ds f\)

Definition of Mapping

\(\, \ds \land \, \)

\(\ds \tuple {x_b, y}\)

\(\in\)

\(\ds f\)

\(\ds \leadsto \ \ \)

\(\ds \tuple {y, x_a}\)

\(\in\)

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

Definition of Inverse of Mapping

\(\, \ds \land \, \)

\(\ds \tuple {y, x_b}\)

\(\in\)

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

\(\ds \leadsto \ \ \)

\(\ds x_a\)

\(=\)

\(\ds x_b\)

Definition 4 of Mapping: $f^{-1}$ is many-to-one

Thus, by definition, $f$ is an injection.

$\blacksquare$

Proof 2

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

\(\ds \map f x\)

\(=\)

\(\ds \map f y\)

\(\ds \leadsto \ \ \)

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

\(=\)

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

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

\(\ds \leadsto \ \ \)

\(\ds x\)

\(=\)

\(\ds y\)

Definition 1 of Inverse Mapping: as $\map {f^{-1} } {\map f x}$ and so on

Thus $f$ is by definition an injection.

$\blacksquare$

Also see

• Inverse is Mapping implies Mapping is Surjection