Injection iff Left Inverse

Theorem
A mapping $$f: S \to T, S \ne \varnothing$$ is an injection iff:
 * $$\exists g: T \to S: g \circ f = I_S$$

where $$g$$ is a mapping.

That is, iff $$f$$ has a left inverse.

In general, that left inverse is not unique.

Uniqueness occurs under either of two circumstances:


 * $$S$$ is a singleton;
 * $$f$$ is a surjection.

Proof 1

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

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

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

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


 * Now, assume $$f$$ is an injection.

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

As $$S \ne \varnothing$$, we choose $$x_0 \in S$$. Then we define:


 * $$g \left({y}\right) =

\begin{cases} x_0: & y \in T - \operatorname{Im} \left({f}\right) \\ f^{-1} \left({y}\right): & y \in \operatorname{Im} \left({f}\right) \end{cases} $$

Because $$f$$ is an injection, we know we can do this, as $$f^{-1}: \operatorname{Im} \left({f}\right) \to S$$ is a mapping, from Injection iff Inverse of Image Mapping.


 * InjectionIffLeftInverse.png

So, for all $$x \in S$$, $$g \circ f \left({x}\right) = g \left({f \left({x}\right)}\right)$$ is the unique element of $$S$$ which $$f$$ maps to $$f \left({x}\right)$$.

This unique element is $$x$$.

Thus $$g \circ f = I_S$$.

Comment
Notice that it does not matter what the elements of $$T - \operatorname{Im} \left({f}\right)$$ are - using the construction given, the equation $$g \circ f = I_S$$ holds whatever value (or values) we choose for $$g \left({T - \operatorname{Im} \left({f}\right)}\right)$$. The left-over elements of $$T$$ we can map how we wish and they will not affect the final destination of any $$x \in S$$ under the mapping $$g \circ f$$.

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.

Then:
 * $$\forall x, y \in A: f \left({x}\right) = f \left({y}\right) \implies g \left({x}\right) = g \left({y}\right)$$

iff:
 * $$\exists h: B \to C$$ 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 $$h \circ f = g$$

iff:
 * $$\forall x, y \in S: f \left({x}\right) = f \left({y}\right) \implies I_S \left({x}\right) = I_S \left({y}\right)$$

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


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

iff:
 * $$\forall x, y \in S: f \left({x}\right) = f \left({y}\right) \implies x = y$$

which is our result.

Proof of Non-Uniqueness

 * If $$f$$ is a surjection, then $$T - \operatorname{Im} \left({f}\right) = \varnothing$$, and we have that $$g = f^{-1}$$.

As $$f^{-1}$$ is uniquely defined $$g$$ is itself unique.


 * If $$S$$ is a singleton then there can only be one mapping $$g: T \to S$$:
 * $$\forall t \in T: g \left({t}\right) = s$$


 * If $$f$$ is not a surjection, then $$T - \operatorname{Im} \left({f}\right) \ne \varnothing$$.

Let $$t \in T - \operatorname{Im} \left({f}\right)$$.

We can now choose any $$x_0 \in S$$ such that $$g \left({t}\right) = x_0$$.

If $$S$$ is not a singleton, such an $$x_0$$ is not unique.

Hence the result.

Also see

 * Surjection iff Right Inverse