Let A, B, C be finite sets such that A and B have the same number of elements, that is, |A| = |B|. Let f : A → B and g : B → C be functions.
(a) Suppose f is one-to-one. Show that f is onto.
(b) Suppose g ◦ f is one-to-one. Show that g is one-to-one.
The Attempt at a Solution
We can let n = |A| = |B|
Since f is injective, we know that for all b ∈ B there exists at most ONE a ∈ A, such that f(a)=b right? that is, given x,y ∈ A f(x) != f(y) => x != y right? I can just see this from the definition, not sure how useful it is or where to go from here.
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