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

(N/A) Define $f: N \rightarrow N$ by $f(x) = x + 1$.
Define $g: N \rightarrow N$ by $g(x) = \begin{cases} x - 1, & \text{if } x > 1 \\ 1, & \text{if } x = 1 \end{cases}$.
First,we show that $f$ is not onto. The range of $f$ is $\{2, 3, 4, \dots\}$,which is a proper subset of the codomain $N$. Specifically,the element $1 \in N$ has no preimage in the domain $N$ such that $f(x) = 1$. Thus,$f$ is not onto.
Now,consider the composition $g \circ f: N \rightarrow N$ defined by $(g \circ f)(x) = g(f(x))$.
Since $f(x) = x + 1$,we have $(g \circ f)(x) = g(x + 1)$.
Since $x \in N$,$x \geq 1$,so $x + 1 \geq 2$. Thus,$x + 1 > 1$.
Using the definition of $g$,we get $g(x + 1) = (x + 1) - 1 = x$.
Therefore,$(g \circ f)(x) = x$ for all $x \in N$.
Since for every $y \in N$,there exists $x = y \in N$ such that $(g \circ f)(x) = y$,the function $g \circ f$ is onto.