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

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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

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 Bijection has Inverse Mapping‎, $f^{-1}$ is a mapping.

$\Box$


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.

$\blacksquare$


Sources

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