Mapping is Injection and Surjection iff Inverse is Mapping/Proof 1

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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.

iff:

the inverse $f^{-1}$ of $f$ is such that:
for each $y \in T$, the preimage $f^{-1} \left({y}\right)$ has exactly one element.
That is, such that $f^{-1} \subseteq T \times S$ is itself a mapping.


Proof

Recall the definition of the inverse of $f$:

$f^{-1} \subseteq T \times S$ is the relation defined as:

$f^{-1} = \left\{{\left({t, s}\right): t = f \left({s}\right)}\right\}$


Necessary Condition

Let $f: S \to T$ be a bijection in the sense that:

$(1): \quad f$ is an injection
$(2): \quad f$ is a surjection.


By Inverse of Injection is Functional Relation‎, $f^{-1}$ is functional.

That is:

$\forall y_1, y_2 \in T: f^{-1} \left({y_1}\right) \ne f^{-1} \left({y_2}\right) \implies y_1 = y_2$

Hence the preimage $f^{-1} \left({y}\right)$ has at most one element.


By Surjection iff Image equals Codomain:

$\operatorname{Im} \left({f}\right) = T$

That is:

$\forall y \in T: \exists x \in S: f^{-1} \left({y}\right) = x$

Hence the preimage $f^{-1} \left({y}\right)$ has at least one element.


Thus the preimage $f^{-1} \left({y}\right)$ has exactly one element.


Hence, by definition, $f^{-1}$ is a mapping.

$\Box$


Sufficient Condition

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

It is necessary to show that $f$ is both an injection and a surjection.


Let $f \left({x_a}\right) = y$ and $f \left({x_b}\right) = y$.

Then:

\(\displaystyle \left({x_a, y}\right)\) \(\in\) \(\displaystyle f\) $\quad$ $\quad$
\(\, \displaystyle \land \, \) \(\displaystyle \left({x_b, y}\right)\) \(\in\) \(\displaystyle f\) $\quad$ by definition of mapping as a relation $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle \left({y, x_a}\right)\) \(\in\) \(\displaystyle f^{-1}\) $\quad$ $\quad$
\(\, \displaystyle \land \, \) \(\displaystyle \left({y, x_b}\right)\) \(\in\) \(\displaystyle f^{-1}\) $\quad$ by definition of inverse of mapping $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle x_a\) \(=\) \(\displaystyle x_b\) $\quad$ as $f^{-1}$ is a mapping $\quad$

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

$\Box$


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

That is:

$\exists y \in T: \neg \exists x \in S: \left({x, y}\right) \in f$

By definition of inverse of mapping:

$\exists y \in T: \neg \exists x \in S: \left({y, x}\right) \in f^{-1}$

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

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

$\Box$


So, being both an injection and a surjection, it follows by definition that $f$ is a bijection.

$\blacksquare$