Characterization of Bijectivity¶

Problem¶

Let \(X\) and \(Y\) be sets and \(f:X\to Y\) a function. Show that \(f\) is bijective if and only if there exists a function \(g:Y\to X\) with \(f\circ g = \operatorname{id}_Y\) and \(g\circ f = \operatorname{id}_X\). Furthermore, show that in this case the function \(g\) is unique.

Proof¶

\(\implies\): Let \(f\) be bijective. The typical characterization of bijectivity is given as follows.

For all \(y\in Y\), let \(x\in X\) be the unique element such that \(f(x)=y\) and define the function \(g\) to be the following.

\(\Longleftarrow\):

Let \(g:Y\to X\) with the following properties.

First, we show that \(f\) is injective which is defined by the following statement.

Let \(a,b\in X\) be arbitrary. Then the following to equations hold.

This shows that the function \(f\) is injective.

Now, we show that \(f\) is surjective which is characterized by the following statement.

Let \(y\in Y\) be arbitrary.

Choose \(x = g(y)\).

This shows that the function \(f\) is surjective and as a consequence bijective.

Uniqueness:

Let \(g_1,g_2:Y\to X\) be two functions with the following properties for all \(i\in\{1,2\}\).

We have already proven that the function \(f\) is bijective. Now, let \(y\in Y\) be arbitrary.

Because \(f\) is injective this implies the uniqueness.