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.