# Equivalence of Definitions of Bijection/Definition 1 iff Definition 4

## Theorem

The following definitions of the concept of Bijection are equivalent:

### Definition 1

A mapping $f: S \to T$ is a bijection if and only if both:

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

and:

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

### Definition 4

A mapping $f \subseteq S \times T$ is a bijection if and only if:

for each $y \in T$ there exists one and only one $x \in S$ such that $\tuple {x, y} \in f$.

## Proof

Let $f: S \to T$ be a bijection by definition 1.

Then by definition:

$f$ is an injection
$f$ is a surjection

By definition of injection:

every element of $T$ is the image of at most $1$ element of $S$.

By definition of surjection:

every element of $T$ is the image of at least $1$ element of $S$.

So:

for each $y \in T$ there exists one and only one $x \in S$ such that $\tuple {x, y} \in f$.

Thus $f$ is a bijection by definition 4.

$\Box$

Let $f: S \to T$ be a bijection by definition 4.

Then by definition:

for each $y \in T$ there exists one and only one $x \in S$ such that $\tuple {x, y} \in f$.

But:

every element of $T$ is the image of at most $1$ element of $S$

defines an injection

and:

every element of $T$ is the image of at least $1$ element of $S$

defines a surjection.

From Injection iff Left Inverse, $f$ is an injection if and only if $f$ has a left inverse mapping.

From Surjection iff Right Inverse, $f$ is a surjection if and only if $f$ has a right inverse mapping.

Putting these together, it follows that:

$f$ is an injection
$f$ is a surjection

Thus $f$ is a bijection by definition 1.

$\blacksquare$