Question

State the following statement is True or False

Two lines that are respectively perpendicular to two intersecting lines, always intersect each other.
If the mappings $f:A\to B$ and $g:B\to C$ are both bijective, then the mapping $gof:A\to C$ is also bijective

• A. True
• B. False

Answer 1

The correct option is A True

Given $f:A\to B$ is bijective
$\Rightarrow f$ is one-one and onto.
Also, given $g:B\to C$ is bijective
$\Rightarrow g$ is one-one and onto.
Now, we will check whether $gof$ is bijective

Let $x,y\in A$ such that

$\left(gof\right)\left(x\right)=\left(gof\right)\left(y\right)$

$\Rightarrow g\left[f\right(x\left)\right]=g\left[f\right(y\left)\right]$

$\Rightarrow f\left(x\right)=f\left(y\right)$ (Since,$g$ is one-one , so $g\left(x\right)=g\left(y\right)\Rightarrow x=y$

$\Rightarrow x=y$ ($\because f$ is one-one)

Hence, $gof$ is one-one.
Now, for surjective, let $z\in C$ be an arbitrary element

Since, g is onto , so for $z\in C$, there exists an element $r\in B$ such that $g\left(r\right)=z$

Also since, f is one so for every $x\in A$, there is an element $r\in B$ such that $f\left(x\right)=r$

$\Rightarrow g\left(f\right(x\left)\right)=z$

$\Rightarrow \left(gof\right)\left(x\right)=z$

Hence, for $z\in C$ , there is an element $x\in A$ .

Hence, $gof$ is onto.