Proof:Surjections have right inverses

If is surjective, then has a right inverse.

To demonstrate the proof, we start with an example.

Let [math]f := [/math]

We want to construct an inverse for [math]f [/math]; obviously such a function must map [math]1 [/math] to a, but we have a choice of where to map [math]2 [/math]. We can't map it to both [math]b [/math] and [math]c [/math] (because then [math]f [/math] would be ambiguous), but we can just pick one of them (say [math]b [/math]).

So we have [math]g := [/math]

Then we plug [math]g [/math] into the definition of right inverse and we see that and , so that [math]g [/math] is indeed a right inverse.

Note that this wouldn't work if [math]f [/math] was not surjective, (for example, if [math]2 [/math] had no pre-image) we wouldn't have any output for [math]g(2) [/math] (so that [math]g [/math] wouldn't be total).

Here is the general proof:

Proof: surjections have right inverses

Choose an arbitrary , , and a surjection . We want to show that there exists a right inverse of .

Define on input [math]y [/math] as follows: we know that there exists at least one with [math]f(x) = y [/math], since [math]f [/math] is surjective. Choose one of them and call it [math]g(y) [/math].

Note: it is not clear that there is an unambiguous way to do this; the assumption that it is possible is called the axiom of choice.

Now, we must check that [math]g [/math] is a right inverse of [math]f [/math], i.e. that .

Choose an arbitrary . We have by definition of [math]g [/math]. Therefore

as required.