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

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.