Give example of two functions f : N $\rightarrow$ N and g : N $\rightarrow$ N such that gof is onto but f is not onto.

(Hint : Consider f(x) $=$ x + 1 and g(x)$=$ $\left\{ {\begin{array}{llllllllllllllllllll}{x - 1,}&{if}&{x > 1}\\{1,}&{if}&{x = 1}\end{array}} \right.$

Consider, f(x)$=$x + 1 and g(x)$=$ $\left\{ {\begin{array}{llllllllllllllllllll}{x - 1,}&{if}&{x > 1}\\{1,}&{if}&{x = 1}\end{array}} \right.$

$f(x) = x + 1 \ge 1 + 1\forall x \in N$
$\Rightarrow$ $f(x) \ge 2\forall x \in N.$

Clearly, range of $f \ne N$ [1$\notin$Range of f]

$\therefore$ f is not onto.

Now, (gof) : N $\rightarrow$ N such that (gof) (x)$=$g(f(x))$=$g(x + 1)
$=$ (x + 1)$-$1 [ x + $1 > 1$ for all x $\in$N]
$=$ $x\forall x \in N$

$\therefore$Range of (gof) $=$ N [gof is identity function]
Hence, gof is onto.