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# Set Theory: Functions (injectivity/surjectivity) Watch

1. Suppose f : A -> B and g : B -> C are functions.
(a) Show that if f is surjective and g is not injective then g o f is not injective.
(Hint: draw pictures to get an idea.)
(b) Show that if f is not surjective and g is injective then g o f is not surjective.

hey, I have a general idea how to solve questions like this, but these ones seem more challenging than any ive done so far, so any help is is really appreciated & rep of course
2. Have you drawn the pictures? The wikipedia page bijections injections and surjections will give you an idea of how the pictures may look.
3. part (a). if f is surjective then f maps onto the entire codomain i.e. for any b in B there exists an a in A s.t. f(a) = b. g not injective means there exists some b1 and b2 in B, with b1 not equal to b2, such that g(b1) = g(b2). but b1 = f(a1) and b2 = f(a2), and a1 cannot equal a2 by the definition of a function, so there exist a1 and a2 in A such that a1 is not equal to a2, and g(f(a1)) = g(f(a2)) => g o f is not injective.

Now try part (b) yourself.

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