Definition:Inverse Mapping/Definition 2

Theorem
Let $$f$$ be a bijection.

Then the mappings $$g_1$$ and $$g_2$$ as defined in Bijection iff Left and Right Inverse are both equal to $$f^{-1}$$.

$$f^{-1}$$ is known as the two-sided inverse of $$f$$.

Usually we dispense with calling it the two-sided inverse, and just refer to it as the inverse.

Proof
Let $$f: S \to T$$ be a bijection.

First, take the case where $$S = \varnothing$$.

Then $$T = \varnothing$$ and the mapping $$\varnothing = \varnothing \times \varnothing$$ is a two-sided inverse for $$f$$.

There are clearly no other such mappings, as $$f: \varnothing \to \varnothing$$ is unique, by Null Mapping.

Now we assume $$S \ne \varnothing$$.

From Bijection iff Left and Right Inverse, $$f$$ is a bijection iff:


 * $$\exists g_1: T \to S: g_1 \circ f = I_S$$
 * $$\exists g_2: T \to S: f \circ g_2 = I_T$$

where both $$g_1$$ and $$g_2$$ are mappings.

Thus:

$$ $$ $$ $$ $$

Every right inverse $$g_2$$ is therefore the same as every left inverse $$g_1$$, so there has to be a unique inverse on each side.

Thus we can say that $$f^{-1} = g_1 = g_2$$ is a two-sided inverse for $$f$$ and it is unique.

As it is both an injection and a surjection, it is a bijection.