Injection iff Left Inverse

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A mapping $f: S \to T, S \ne \O$ is an injection if and only if:

$\exists g: T \to S: g \circ f = I_S$

where $g$ is a mapping.

That is, if and only if $f$ has a left inverse.

Proof 1


$\exists g: T \to S: g \circ f = I_S$

From Identity Mapping is Injection, $I_S$ is injective, so $g \circ f$ is injective.

So from Injection if Composite is Injection, $f$ is an injection.

Note that the existence of such a $g$ requires that $S \ne \O$.


Now, assume $f$ is an injection.

We now define a mapping $g: T \to S$ as follows.

As $S \ne \O$, we choose $x_0 \in S$.

By definition of injection:

$f^{-1} {\restriction_{\Img f} } \to S$ is a mapping

so it is possible to define:

$\map g y = \begin{cases}

x_0: & y \in T \setminus \Img f \\ \map {f^{-1} } y: & y \in \Img f \end{cases}$


It does not matter what the elements of $T \setminus \Img f$ are.

Using the construction given, the equation $g \circ f = I_S$ holds whatever value (or values) we choose for $g \sqbrk {T \setminus \Img f}$.

The remaining elements of $T$ can be mapped arbitrarily, and will not affect the image of any $x \in S$ under the mapping $g \circ f$.

So, for all $x \in S$:

$\map {g \circ f} x = \map g {\map f x}$

is the unique element of $S$ which $f$ maps to $\map f x$.

This unique element is $x$.

Thus $g \circ f = I_S$.


Proof 2

Take the result Condition for Composite Mapping on Left:

Let $A, B, C$ be sets.

Let $f: A \to B$ and $g: A \to C$ be mappings.


$\forall x, y \in A: \map f x = \map f y \implies \map g x = \map g y$

if and only if:

$\exists h: B \to C$ such that $h$ is a mapping and $h \circ f = g$.

Let $C = A = S$, let $B = T$ and let $g = I_S$.

Then the above translates into:

$\exists h: T \to S$ such that $h$ is a mapping and $h \circ f = g$

if and only if:

$\forall x, y \in S: \map f x = \map f y \implies \map {I_S} x = \map {I_S} y$

But as $\map {I_S} x = x$ and $\map {I_S} y = y$ by definition of identity mapping, it follows that:

$\exists h: T \to S$ such that $h$ is a mapping and $h \circ f = g$

if and only if:

$\forall x, y \in S: \map f x = \map f y \implies x = y$

which is our result.


Proof 3

Let $f: S \to T$ be an injection.

Then $f$ is a one-to-many relation.

By Inverse of Many-to-One Relation is One-to-Many, $f^{-1}: T \to S$ is many-to-one.

By Many-to-One Relation Extends to Mapping, there is a Mapping $g: T \to S$ such that $f^{-1} \subseteq g$.

Let $\tuple {x, y} \in g \circ f$.


$\exists z \in T: \tuple {x, z} \in f, \tuple {z, y} \in g$

Since $\tuple {x, z} \in f$:

$\tuple {z, x} \in f^{-1} \subseteq g$

Since $\tuple {z, y} \in g$, $\tuple {z, x} \in g$ and $g$ is a mapping:

$x = y$


$\tuple {x, y} \in I_S$

So we see that:

$g \circ f \subseteq I_S$

Let $x \in S$.


$\tuple {x, \map f x} \in f$


$\tuple {\map f x, x} \in f^{-1} \subseteq g$


$\tuple {x, x} \in g \circ f$


$I_S \subseteq g \circ f$

By definition of set equality:

$I_S = g \circ f$


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