Question

Give examples of two surjective functions ${f_1}$ and ${f_2}$ from $Z$ to $Z$, such that ${f_1} + {f_2}$ is not surjective.

Answer 1

Hint: - Assume first function be $x$ and second function be $ - x$
A function $f:A \to B$ is said to be onto function or surjective if every element of $A$ i.e. $f\left( A \right) = B$ or range of $f$ is the co-domain of $f$.

Complete step-by-step solution -
So, $f:A \to B$ is surjective for each $b \in B$, there exists $a \in B$ such that $f\left( a \right) = b$
Let ${f_1}:Z \to Z$ and ${f_2}:Z \to Z$ be two functions given by
$
   \Rightarrow {f_1}\left( x \right) = x \\
   \Rightarrow {f_2}\left( x \right) = - x \\
$
For $1 \in R,{\text{ }}x \in R$ such that ${f_1}\left( x \right) = 1$
So for $x = 1$
The value of ${f_1}\left( x \right) = x = 1$, so ${f_1}\left( x \right) = x$ is a surjective function.
For $1 \in R,{\text{ }}x \in R$ such that ${f_2}\left( x \right) = 1$
For $x = - 1$
The value of ${f_2}\left( x \right) = - x = - \left( { - 1} \right) = 1$, so ${f_2}\left( x \right) = - x$ is a surjective function.
Now,
\[
   \Rightarrow {f_1} + {f_2}:Z \to Z \\
   \Rightarrow \left( {{f_1} + {f_2}} \right)x = {f_1}\left( x \right) + {f_2}\left( x \right) \\
   \Rightarrow \left( {{f_1} + {f_2}} \right)x = x - x = 0 \\
\]
Therefore, \[{f_1} + {f_2}:Z \to Z\] is a function given by
\[ \Rightarrow \left( {{f_1} + {f_2}} \right)x = 0\]
Since, \[{f_1} + {f_2}\] is a constant function, hence it is not an onto or surjective function, because for any value of $x$ expect zero the function \[\left( {{f_1} + {f_2}} \right)x\] value remains same which is zero.

Note: - In such type of question always remember the property of an onto or surjective function which is stated above, then assume two simple surjective functions as above, then add them we will get the required result.