Question

If ‘\[f\]’ and ‘\[g\]’ are bijective functions and \[gof\] is defined then \[gof\] must be
A. Injective
B. Surjective
C. Bijective
D. Into only

Answer 1

Hint: Here we use the definition of bijective function and write the two functions in the form of mapping from one set to another where the domain in function \[g\] will be the co-domain of the function \[f\].
Using the concept of composition function we check if \[gof\]is one-one or onto or both and then decide from the given options.
* A bijective function is a mapping that is both one-one and onto.
* We check the function \[f\] is one-one if \[f(a) = f(b) \Rightarrow a = b\]
* We check the function \[f\] is onto if for every value \[f(x)\] in the co-domain there is an element\[x\] in the domain.

Complete step-by-step answer:
Let us assume the two functions such that the domain in function \[g\] will be the co-domain of the function\[f\].
\[f:A \to B\]
\[g:B \to C\]
Let us assume the elements \[{a_1},{a_2} \in A\]
Now we know that the composition function is defined.
We write \[gof = g[f(x)]\] where x is an element of the domain.
Now we will check if the composition function is one-one by using the concept\[f(a) = f(b) \Rightarrow a = b\].
We take \[gof({a_1}) = gof({a_2})\]
Opening the composition function we write
\[ \Rightarrow g[f({a_1})] = g[f({a_2})]\]
Since g is bijective function so we have \[g[f({a_1})] = g[f({a_2})] \Rightarrow f({a_1}) = f({a_2})\]
\[ \Rightarrow f({a_1}) = f({a_2})\]
Since, f is a bijective function so we have \[f({a_1}) = f({a_2}) \Rightarrow {a_1} = {a_2}\]
\[ \Rightarrow {a_1} = {a_2}\]
Therefore, the composition function is one-one.
Now we check if the composition function is onto.
Since,
We know that the function f is onto, therefore we can write for every \[b \in B,\exists a \in A\] such that \[f(a) = b\].
Since,
Also, we know that the function g is onto, therefore we can write for every \[c \in C,\exists b \in B\] such that \[g(b) = c\].
We can substitute the value of \[f(a) = b\]in \[g(b) = c\]
\[ \Rightarrow g(f(a)) = c\]
LHS is the composition function \[gof\].
So, \[gof\]is onto function.
Since, \[gof\] is one-one and onto, we can say \[gof\] is a bijective function.

So, option C is correct.

Note: Students might get confused if they take a set of elements and try to prove one-one and onto for each element which will be a long process, so we use the concepts and basic definitions of bijective functions to find the nature of the function.
* Composition function is the function where we apply one function to the value obtained from another function. So we can write \[gof = g[f(x)]\] where x is an element from the domain.