Question

If $f:A\to B$ and $g:B\to C$ are one-one functions, show that gof is a one-one function.

Answer 1

In order to solve this problem, we need to know the definition of the one-one function. The one-one function denotes the mapping of two sets. A function f corresponds to exactly one element of the domain of f. It can also be explained as each element of one set is mapped with a unique element of another set. If $f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)$ then ${{x}_{1}}={{x}_{2}}$ if $f$ is the one-one function.

Complete step-by-step answer :
We are given that $f$ function is one to one function and $g$ is also a one to one function.
Let’s first understand what do we mean by one to one relation.
The one-one function denotes the mapping of two sets. A function f corresponds to exactly one element of the domain of f.
It can also be explained as each element of one set is mapped with a unique element of another set.
If $f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)$ then ${{x}_{1}}={{x}_{2}}$ if $f$ is the one-one function.
Similarly, we are given that $g$ is also a one-one function.
Therefore, if $g\left( {{x}_{1}} \right)=g\left( {{x}_{2}} \right)$ then ${{x}_{1}}={{x}_{2}}$ .
Now, let's understand by gof.
gof can be explained as $g$ is the function of $f$ of x.
Firstly, f is the function of x, then this function itself is a function of g.
It can be expressed as $gof=g\left( f\left( x \right) \right)$ .
Now, we need to prove that gof is also a one-one function.
$gof=g\left( f\left( x \right) \right)$.
We already know the g is a one-one function so we can write,
If $g\left( f\left( {{x}_{1}} \right) \right)=g\left( f\left( {{x}_{2}} \right) \right)$ , then $f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)$ .
Now we have that $f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)$.
But we also know the $f$ is also a one-one function. Therefore, ${{x}_{1}}={{x}_{2}}$ .
Therefore, we can show that if ${{x}_{1}}={{x}_{2}}$ then $gof$ is also a one-one function.

Note :We need to understand we are asked to prove that the function of gof and not fog.
fog of the function can be written as $f\left( g\left( x \right) \right)$ and gof is explained as $g\left( f\left( x \right) \right)$ . And both show different correlations. Also, we can show that whenever $f\left( {{x}_{1}} \right)\ne f\left( {{x}_{2}} \right)$ then ${{x}_{1}}\ne {{x}_{2}}$ if f is the one-one function