Question

If $f:A\to B$ and $g:B\to C$ then prove that $gof$ is a one-one function. $$$$

Answer 1

We recall the definition of composite function and the condition for a function to be one-one. . We assume any two elements ${{x}_{1}},{{x}_{2}}$ in $A$ and use the condition of one-one for $f:A\to B$ and $g:B\to C$ to prove $gof\left( {{x}_{1}} \right)=gof\left( {{x}_{2}} \right)\Rightarrow {{x}_{1}}={{x}_{2}}$.

Complete step-by-step solution
We know that a function is a relation from the domain set to range set where all the elements of the domain are mapped to an element of range and one element of domain cannot be mapped to two element of range. If we have the single valued functions (takes only one element at a time) $f:A\to B$ and $g:B\to C$ then the composite function from $gof$ is defined for some element $x\in A$ as
$gof:A\to C,gof\left( x \right)=g\left( f\left( x \right) \right)$
We also know that one-one function is a function which maps exactly one element of the domain set to exactly one element of the range set. We have first $f:A\to B$ as a single valued one-one function. Here the domain set is $A$ and the range set is $B.$ Let us we have any two elements in the range and domain as ${{x}_{1}},{{x}_{2}}\in A$. Then by the condition of one-one function we have,
$f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)\Rightarrow {{x}_{1}}={{x}_{2}}....\left( 1 \right)$
The second single valued one-one function is given to us $g:B\to C$. Here the domain set is $B$ and the range set is $C.$ Let us have any two elements in the domain and range as ${{y}_{1}},{{y}_{2}}\in B$ and Then by the condition of one-one function we have,
$g\left( {{y}_{1}} \right)=g\left( {{y}_{2}} \right)\Rightarrow {{y}_{1}}={{y}_{2}}....\left( 2 \right)$
The composite function $gof$ sends elements form set $A$ to $C$. Let us take elements ${{x}_{1}},{{x}_{2}}\in A$. Let us assume $gof\left( {{x}_{1}} \right)=gof\left( {{x}_{2}} \right)$. If we prove ${{x}_{1}}={{x}_{2}}$ indirectly we prove $gof$ as one-one functon. Let us proceed with our assumption,

& gof\left( {{x}_{1}} \right)=gof\left( {{x}_{2}} \right) \\\ & \Rightarrow g\left( f\left( {{x}_{1}} \right) \right)=g\left( f\left( {{x}_{2}} \right) \right) \\\ \end{aligned}$$ Since $g$ is a one-one function we have $$\begin{aligned} & f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right) \\\ & \Rightarrow {{x}_{1}}={{x}_{2}}\left( \because f\text{ is one-one} \right) \\\ \end{aligned}$$ **Hence $gof$ is one-one function.** **Note:** The other name of one-one function is injective function. If all the elements from the range set are mapped then the function is called surjective or onto. If a function is both one-one and onto then it is called bijective. If $f,g$ are bijective then $fog,gof$ are also bijective.