If `f and g` are one-one functions, then (a)`f+g` is one one (b)`fg` is one one (c)`fog` is one one (d)`n on e of these`

Explanation

(i) Suppose that `f` is onto and g is not one-one. We can fnd elements in B to witness the fact that g is not one-to-one, namely we can find `x, y in B` with g(x)=g(y). Since `f` is onto, we can then find `a,b in A` with `f(a)=x " and " f(b)= y.` Since `x ne y`, the definition of a function tells us that `a ne b.` Therefore, we have that `(gof)(a)=f(f(a))=g(x)=g(y)=g(f(b))=(gof)(b), " but " a ne b` and so `gof` is not one-one.
(ii) Suppose that `f` is not onto and g is one-one . We can find an `x in B ` which witnesses the fact that `f` is not onto, i.e., `x notin`Ranger`(f)`. In order to prove that `gof` is not onto, we now want to show that for `a in A, (gof) (a) ne g(x),` demonstrating that Range `(gof) subset C.` Given any ` a in A,` we know that `f(a) ne x` since x is not in the range of `f`. Since g is one-one, `f(a) ne x` tells us that `g(f(a)) ne g(x)` as well, which is what we were trying to prove.