left inverse is not equal to right inverse

Question

Is it possible to have a function which has both left and right inverse but they are unequal ?

A left inverse means the function should be one-to-one whereas a right inverse means the function should be onto.

How can both of these conditions be valid simultaneously without being equal ?

An example will be really helpful. Thanks in advance

Answer 1

Suppose that $f(g(x))=x$ and $h(f(x))=x$, then $h(x)=h(f(g(x))=g(x)$. So they have to be the same.

Answer 2

For functions they are equal whenever they exist. Let $f$ be a function with left inverse $g$ and right inverse $h$, then $$g=g\cdot id=g(fh)=(gf)h=h$$

Answer 3

You can't have mismatched left and right inverses.

If $f$ is a left inverse for $g$ and $h$ is a right inverse for $g$ (denote the identity function $\mathrm{id}(x)=x$) we have $f \circ g = \mathrm{id}$ and $g \circ h = \mathrm{id}$ so $f = f \circ \mathrm{id} = f \circ (g \circ h) = (f \circ g) \circ h = \mathrm{id} \circ h = h$. So $f=h$ is a double sided inverse.

This is true whenever you have an associative operation. Only non-associative operations allow one to have mismatched left and right inverses.

By the way...the equivalence of "existence of a right inverse" and "being onto" assumes the axiom of choice (for those who care about such things).