Question

Show that the inverse mapping f1 of a mapping f exists iff f is one-one onto

Answer 1

Theorem: A mapping $f:A\to B$ is invertible if and only if $f$ is a bijection.

Proof:

Necessary part.

Firstly we consider that $f:A\to B$ is invertible, and thus there exists a mapping $g:B\to A$ such that

$\quad\quad g\:\small{o}\:f=i_{A}$

$\quad\quad f\:\small{o}\:g=i_{B}$

where $i_{A}$ = identity mapping on $A$

$\quad\quad\quad i_{B}$ = identity mapping on $B$

Since $i_{A}$ is an identity mapping, it is injective and $g\:\small{o}\:f=i_{A}$

$\implies f$ is an injective mapping.

Since $i_{B}$ is an identity mapping, it is injective and $f\:\small{o}\:g=i_{B}$

$\implies f$ is a surjective mapping.

Therefore, $f$ is a bijective mapping.

Sufficient part.

Here we consider $f:A\to B$ be a bijective mapping.

Let, $y\in B$.

Since $f$ is a bijective mapping, the element $y$ has a unique pre-image $x\in A$.

Let us define a mapping $g:B\to A$ such that $g(y)=x$, where $x$ is the pre-image of $y$ under the mapping $f$ and $y\in B$.

$\implies g\:\small{o}\:f=g(y)=x,\:x\in A$ and

$\quad\quad\: f\:\small{o}\:g=f(x)=y,\:y\in B$.

We have both $g\:\small{o}\:f$ and $f\:\small{o}\:g$ being identity mapping, and therefore $f$ is invertible.

This completes the proof.

Related question:

Prove that the inverse of one-one onto mapping is unique.

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