(or mappings) f: A -%26gt; B and g: B -%26gt; C.
i. show that if f and g are both injective then so is their composition g*f
ii. show that if f and g are surjective then so is their composition g*f
recall that (g*f)(a)=g(f(a))
Suppose A, B, C are sets and consider the function?
(i) if f and g are injective then f(a) = f(b) iff a=b and g(c) = g(d) iff c=d
hence g(f(a)) = g(f(b)) iff f(a) = f(b) iff a=b
so g*f is injective
(ii) f and g are surjective iff for all b in B there exists a in A such that f(a)=b and similarly the exists b in B such that g(b) = c for all c in C
so for all c in C there exists b in B such that g(b)=c, and for all b in B there exists a in A such that f(a) = b
therefore, for all c in C there exists a in A such that g(f(a)) = c, and g*f is surjective
and as g*f is both surjective and injective it is also bijective!
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