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
Set Theory: Functions (injectivity/surjectivity) Watch
- Thread Starter
- 11-12-2010 01:05
- 11-12-2010 01:22
Have you drawn the pictures? The wikipedia page bijections injections and surjections will give you an idea of how the pictures may look.
- 11-12-2010 01:39
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.