Question

Consider the functions
$f:X\to Y$ and $g:Y\to Z$
then which of the following is/are incorrect?

• A. If $f$ and $g$ are both injective then $gof:X\to Z$ is injective
• B. If $f$ and $g$ are both surjective then $gof:X\to Z$ is surjective
• C. If $gof:X\to Z$ is bijective then $f$ is injective and $g$ is surjective.
• D. none

Answer 1

The correct option is D none

Option A)
Let $x_1,x_2\in X$ be two distinct elements, then $f\left(x_1\right)\ne f\left(x_2\right)$ as $f$ is injective.
Hence,$g\left(f\left(x_1\right)\right)\ne g\left(f\left(x_2\right)\right)$ as $g$ is also an injective function.
Hence,$g\circ f$ is injective.

Option B)
For, $z\in Z,\exists y\in Y$, such
that $g\left(y\right)=z$, as $g$ is surjective. For $y\in Y,\exists x\in X$ such that $f\left(x\right)=y$ as $f$
is surjective. Combining the two we have $z\in Z,\exists x\in x$,
such that $g\left(f\left(x\right)\right)=z$.
Hence,$g\circ f$ is surjective.
Option C)
Let $x_1,x_2\in X$ be two distinct
elements. Since $g\circ f$ is injective,
$g\left(f\left(x_1\right)\right)\ne g\left(f\left(x_2\right)\right)$ implies that $f\left(x_1\right)\ne f\left(x_2\right)$. Hence $f$ is injective.
For, $z\in Z,\exists x\in X$, such
that $g\left(f\left(x\right)\right)=z$, as $g\circ f$ is surjective. let $y=f\left(x\right)$ for some $y\in Y$,
then we have $g\left(y\right)=g\left(f\left(x\right)\right)=z$. Hence$g$ is surjective.
Hence none of the three options are incorrect.