Question

Consider functions $f:A\to B$ and $g:B\to C$ $(A,B,C\subseteq R)$ such that $(gof{)}^{-1}$ exists, then

• A. $f$ is one-one and $g$ is onto
• B. $f$ is onto and $g$ is one-one
• C. $f$ and $g$ both are one-one
• D. $f$ and $g$ both are onto

Answer 1

The correct option is A $f$ is one-one and $g$ is onto

Given that $(gof{)}^{-1}$ exists. It means $gof=g\left(f\right(x\left)\right)$ is one-one and onto.
Let $h\left(x\right)=g\left(f\right(x\left)\right)$ and
$h:A\to C$
Since $h\left(x\right)$ is one-one, we have $h\left(x_1\right)=h\left(x_2\right)$ for $x_1,x_2\in A$
$\Rightarrow g\left(f\right(x_1\left)\right)=g\left(f\right(x_2\left)\right)\Rightarrow f\left(x_1\right)=f\left(x_2\right)$
It means $f$ must be one-one.

As $h\left(x\right)$ is onto it means range of $h\left(x\right)$ is $C$ that is possible only when range of $g\left(x\right)$ is also $C$.
Hence, $g\left(x\right)$ is onto.