Answer:
Let
f(x) = x + 1 and g(x) =
Let
be any
element
x + 1
f(x)
Hence f is not
onto.
Also gof : N
N is such
that
(gof) (x) =
g(f(x)) =g(x +1)
= (x + 1) ? 1 =x
Hence gof is onto.
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