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gbean
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Homework Statement
a) Let g: A => B, and f: B => C. Prove that f is one-to-one if f o g is one-to-one.
b) Let g: A => B, and f: B => C. Prove that f is onto if f o g is onto.
Homework Equations
a) Since f o g is onto, then (f o g)(a) = (f o g)(b) => a = b.
b) Since f o g is onto, every element in C is an image under f o g.
(f o g)(a) = f(g(a)) = f(b) = c
The Attempt at a Solution
a) Proof by contradiction. Assume that g is not one-to-one.
(f o g)(a) = (f o g)(b)
f(g(a)) = f(g(b))
I don't know where to go from here.
I'm just wondering if this explanation is rigorous enough.
b) Proof by contradiction: Assume that f is not onto.
(f o g)(a) = f(g(a)) = f(b) = c.
However, f is not onto, thus there exists a c in C that is not mapped to by f. Thus by definition, f o g is not onto if not every c in C is an image under f o g. This is a contradiction of the given, hence f must be onto.