Bijection iff Inverse is Bijection

Theorem

Let $f: S \to T$ be a mapping.

Then $f$ is a bijection iff its inverse $f^{-1}$ is a mapping from $T$ to $S$.

It follows that $f^{-1}$ is itself a bijection.

Proof

First we establish the definition of $f^{-1}$ in this context.

Let $f: S \to T$ be a mapping.

Then $f^{-1} \subseteq T \times S$ is a relation defined as:

$f^{-1} = \left\{{\left({t, s}\right): t = f \left({s}\right)}\right\}$

Sufficient Condition

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

As $f$ is a bijection, it is by definition an injection.

So, by Inverse of Injection is Functional Relation‎, $f^{-1}$ is functional.

Also, as $f$ is a bijection, it is by definition a surjection.

From Surjection iff Image equals Codomain, $\operatorname{Im} \left({f}\right) = T$.

Thus from the definition of inverse relation, the domain of $f^{-1}$ is:

$\operatorname {Dom} \left({f^{-1}}\right) = T$

Thus we have established that:

• $\forall y \in T: \exists x \in S: f^{-1} \left({y}\right) = x$ as $\operatorname {Dom} \left({f^{-1}}\right) = T$;
• $\forall y_1, y_2 \in T: f^{-1} \left({y_1}\right) \ne f^{-1} \left({y_2}\right) \implies y_1 = y_2$ as $f^{-1}$ is functional.

Hence by definition $f^{-1}$ is a mapping.

$\Box$

Necessary Condition

Now suppose $f^{-1}$ is a mapping.

Then by definition:

$\forall y_1, y_2 \in T: f^{-1} \left({y_1}\right) \ne f^{-1} \left({y_2}\right) \implies y_1 = y_2$

which implies that:

$\forall y_1, y_2 \in T: f \left({x_1}\right) = f \left({x_2}\right) \implies x_1 = x_2$

and so $f$ is an injection.

From Preimage of Mapping equals Domain we have that:

$\operatorname{Im}^{-1} \left({f^{-1}}\right) = \operatorname{Dom} \left({f^{-1}}\right)$

From the definition of inverse relation, the domain of $f^{-1}$ is:

$\operatorname {Dom} \left({f^{-1}}\right) = T$

From Surjection iff Image equals Codomain:

$\operatorname{Im} \left({f}\right) = T$

So $f$ is a surjection.

Being both an injection and a surjection, it follows by definition that $f$ is a bijection.

$\Box$

Now, from Inverse of Inverse Relation, we have $\left({f^{-1}}\right)^{-1} = f$.

So we have that the inverse of $f^{-1}$ is $f$.

But then $f$, being a bijection, is by definition a mapping.

So if the inverse of $f^{-1}$ is a mapping, then $f^{-1}$ must be a bijection.

$\blacksquare$

Also see

It also follows, from Bijection Composite with Inverse, that $f^{-1}$ is the two-sided inverse of $f$, i.e.:

• $f \circ f^{-1} = I_S$
• $f^{-1} \circ f = I_T$

where $I_S$ and $I_T$ are the identity mappings on $S$ and $T$.

Sources

• J.A. Green: Sets and Groups (1965)... (previous)... (next): $\S 3.3$
• Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 5$: Theorem $5.5$
• Ian D. Macdonald: The Theory of Groups (1968): Appendix
• John F. Humphreys: A Course in Group Theory (1996): $\S 2$: Proposition $2.15$
• H. Jerome Keisler and Joel Robbin: Mathematical Logic and Computability (1996): Appendix $\text{A}.7$: Proposition $\text{A}.7.5 \ (2)$

• For a video presentation of the contents of this page, visit the Khan Academy.