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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) = {(x-1, ifx >1),(1, if x = 1):}`
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Solution
Define f: N → N by,
f(x) = x + 1
And, g: N → N by,
`g(x) = {(x -1, if x>1), (1, if x = 1):}`
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 → N is defined by,
`gof(x) = g(f(x)) =g(x + 1) = (x +1) - 1` [x in N => (x + 1) > 1]
= x
Then, it is clear that for y ∈ N, there exists x = y ∈ N such that gof(x) = y.
Hence, gof is onto.
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