Give examples of two surjective functions $f_{1}$ and $f_{2}$ from $Z$ to $Z$ such that $f_{1}+f_{2}$ is not surjective.
We know that $f_{1}: R \rightarrow R$, given by $f_{1}(x)=x$, and $f_{2}(x)=-x$ are surjective functions.
Proving $f_{1}$ is surjective :
Let $y$ be an element in the co-domain $(R)$, such that $f_{1}(x)=y$.
$f_{1}(x)=y$
$\Rightarrow x=y$, which is in $R$.
So, for every element in the co-domain, there exists some pre-image in the domain.
So, $f_{1}$ is surjective.
Proving $f_{2}$ is surjective :
Let $y$ be an element in the co domain $(R)$ such that $f_{2}(x)=y$.'
$f_{2}(x)=y$
$\Rightarrow x=y$, which is in $R .$
$\Rightarrow x=y$, which is in $R$.
So, for every element in the co-domain, there exists some pre-image in the domain.
So, $f_{2}$ is surjective.
Proving $\left(f_{1}+f_{2}\right)$ is not surjective :
Given:
$\left(f_{1}+f_{2}\right)(x)=f_{1}(x)+f_{2}(x)=x+(-x)=0$
So, for every real number $x,\left(f_{1}+f_{2}\right)(x)=0$
So, the image of every number in the domain is same as 0 .
$\Rightarrow$ Range $=\{0\}$
Co-domain $=R$
So, both are not same.
So, $f_{1}+f_{2}$ is not surjective.