Answer :

It is given that f: N N by, f(x) = x + 1

And, g : N N by,



Now, consider element 1 in co-domain N. So, it is clear that this element is not an image of any of the elements in domain N.


f is onto.


Now, gof: N N is defined as:


gof(x) = g(f(x)) = g(x+1) = (x + 1) – 1 [x ϵ N = > (x+1) > 1]


Then we can see that for y ϵ N, there exists x = y ϵ N such that gof(x) = y.


Therefore, gof is onto.


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