Question

Let $f:[0,1]\to [–1,1]$ and $g:[–1,1]\to [0,2]$ be two functions such that $g$ is injective and $g\circ f:[0,1]\to [0,2]$ is surjective. Then

• A. $f$ must be injective but need not be surjective
• B. $f$ must be surjective but need not be injective
• C. $f$ must be bijective
• D. $f$ must be a constant function

Answer 1

The correct option is B $f$ must be surjective but need not be injective

$g\circ f$ is surjective.
$\Rightarrow$ For each $g\left(f\right(x\left)\right)\in [0,2],\exists f\left(x\right)\in \text{Domain of}g$
$\Rightarrow f\left(x\right)\in [-1,1]=R_f$
$\Rightarrow f$ is surjective.

Given, $g$ is injective.
So, given $x_1,x_2\in Dom(g\circ f)$
$g\left(f\right(x_1\left)\right)=g\left(f\right(x_2\left)\right)\Rightarrow f\left(x_1\right)=f\left(x_2\right)$
We don't have enough evidence to conclude $x_1=x_2$ further.
So, we can't conclude about injectivity of $f$ from the given conditions.