Mapping is Injection and Surjection iff Inverse is Mapping/Proof 2

Theorem

Let $S$ and $T$ be sets.

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

Then:

$f: S \to T$ can be defined as a bijection in the sense that:

$(1): \quad f$ is an injection

$(2): \quad f$ is a surjection.

if and only if:

the inverse $f^{-1}$ of $f$ is such that:

for each $y \in T$, the preimage $\map {f^{-1} } y$ has exactly one element.

That is, such that $f^{-1} \subseteq T \times S$ is itself a mapping.

Proof

Necessary Condition

Let $f: S \to T$ be a mapping such that:

$(1): \quad f$ is an injection

$(2): \quad f$ is a surjection.

Let $t \in T$.

Then as $f$ is a surjection:

$\exists s \in S: t = \map f s$

As $f$ is an injection, there is only one $s \in S$ such that $t = \map f s$.

Define $\map g t = s$.

As $t \in T$ is arbitrary, it follows that:

$\forall t \in T: \exists s \in S: \map g t = s$

such that $s$ is unique for a given $t$.

That is, $g: T \to S$ is a mapping.

By the definition of $g$:

$(1): \quad \forall t \in T: \map f {\map g t} = t$

Let $s \in S$.

Let:

$(2): \quad s' = \map g {\map f s}$

Then:

\(\ds \map f {s'}\)

\(=\)

\(\ds \map f {\map g {\map f s} }\)

\(\ds \)

\(=\)

\(\ds \map f s\)

from $(1)$

\(\ds \leadsto \ \ \)

\(\ds s\)

\(=\)

\(\ds s'\)

as $f$ is an injection

\(\ds \)

\(=\)

\(\ds \map g {\map f s}\)

from $(2)$

Thus $f: S \to T$ and $g: T \to S$ are inverse mappings of each other.

$\blacksquare$

$\Box$

Sufficient Condition

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

By Inverse Mapping is Bijection, both $f$ and $f^{-1}$ are bijections.

Hence, in particular, $f$ is a bijection.

$\blacksquare$