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# Give examples of two functions $f:N\to N$ and $g:N\to N$ such that gof is onto but f is not onto.  Answer Verified
Hint: Use the conditions for surjectivity for taking the right examples.

Complete step-by-step answer:
Consider examples of $f\left( x \right)=x+1$and $g\left( x \right)=\left\{ \begin{matrix} x-1,x>1 \\ 1,x=1 \\ \end{matrix} \right.$.
First prove that f is not onto by proving the digit formed is not a natural number. As mentioned in $f:N\to N$. Then prove gof is onto. Find the composite of g and f$\Rightarrow \left( gof \right)\left( x \right)=g\left( f\left( x \right) \right)$. Thus prove gof is onto.
Given to us two functions of $f:N\to N$and $g:N\to N$.
We need to prove that gof is onto and f is not onto.
Onto function could be explained by considering two sets, set A and set B which consist of elements. If for every element of B there is at least one or more than one element matching with A, then the function is said to be onto.
Let us consider $f:N\to N$be $f\left( x \right)=x+1$.
And $g:N\to N$be taken as $g\left( x \right)=\left\{ \begin{matrix} x-1,x>1 \\ 1,x=1 \\ \end{matrix} \right.$.
We will first show that f is not onto.
Let $f:N\to N$be $f\left( x \right)=x+1$
Let us consider$y=f\left( x \right)$, where $y\in N$.
\begin{align} & f\left( x \right)=x+1\Rightarrow y=x+1 \\ & \therefore x=y-1 \\ \end{align}
Let us put $y=1$,
$x=1-1=0$
We get x = 0, which is not a natural number.
It is said that $f:N\to N$i.e. natural numbers.
But zero is not a natural number. So f is not onto and it doesn’t belong in the elements of natural numbers.
If given functions f and g with domains and target sets consisting sets consisting of real numbers, the composite of g and f is the new function gof defined as,
$\left( gof \right)\left( x \right)=g\left( f\left( x \right) \right)$
Here, we need to find gof.
Let us consider $f\left( x \right)=x+1$and $g\left( x \right)=\left\{ \begin{matrix} x-1,x>1 \\ 1,x=1 \\ \end{matrix} \right.$.
$f\left( x \right)=x+1,g\left( x \right)=1$
Since, $g\left( x \right)=1$, for x = 1.
\begin{align} & g\left( f\left( x \right) \right)=1 \\ & \therefore gof=1 \\ \end{align}
Since, $g\left( x \right)=x-1,x>1$
$g\left( f\left( x \right) \right)=f\left( x \right)-1$, put, $x=f\left( x \right)$.
\begin{align} & gof=\left( x+1 \right)-1 \\ & \Rightarrow gof=x \\ & \therefore gof=\left\{ \begin{matrix} x,x>1 \\ 1,x=1 \\ \end{matrix} \right. \\ \end{align}
Let us consider, $gof=y$, where $y\in N$.
$\therefore y=\left\{ \begin{matrix} x,x>1 \\ 1,x=1 \\ \end{matrix} \right.$
Here, y is a natural number, as y = x.
So, x is also a natural number.
$\therefore$gof is onto, as y is a natural number and y = gof.
Hence, we proved that gof is onto and f is not onto.

Note: Remember what is onto function. We can only prove a function is not onto, if the elements found don’t belong to it. In the case of $f:N\to N$, the element was zero, which is not a natural number and doesn’t belong to f. If it was any number greater than zero, then it would belong to $f:N\to N$and f would be an onto function.
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