Question

Given examples of two functions f :N→N and g :N→N such that gof is onto but f is not onto.
(Hint: Consider f(x)=x+1 and $g(x) = \begin{cases} x-1 & \quad \text{if } x \geq 1\text{ }\\\ 1 & \quad \text{if } x \text{ = 1} \end{cases}$

Answer 1

Define f : N $\to$ N by,
f(x) = x + 1
And, g: N → N by,
$g(x) = \begin{cases} x-1 & \quad \text{if } x \geq 1\text{ }\\\ 1 & \quad \text{if } x \text{ = 1} \end{cases}$
We first show that g is not onto.
For this, consider element 1 in co-domain N.
It is clear that this element is not an image of any of the elements in domain N.
∴ f is not onto.
Now, gof : N $\to$ N is defined by,
gof (x)= g(f(x))= g(x+1)= (x+1)-1 [x ∈ N =>(x+1)>1]
Then, it is clear that for y ∈ N, there exists x = y ∈ N such that gof( x) = y.
Hence, gof is onto.