0

I am trying to prove the following theorems:

Let A, B, and C be nonempty sets and let $f : A \rightarrow B$ and $g : B \rightarrow C$.

  • If $g \circ f : A \rightarrow C$ is an injection, then $f : A \rightarrow B$ is an injection.
  • If $g \circ f : A \rightarrow C$ is a surjection, then $g : B \rightarrow C$ is an surjection.

Any suggestions? Thanks!

Community
  • 1
mosca1337
  • 113
  • 3
  • For injective functions see e.g. http://math.stackexchange.com/questions/63552/composite-functions-and-one-to-one or http://math.stackexchange.com/questions/229065/if-f-circ-g-is-onto-then-f-is-onto-and-if-f-circ-g-is-one-to-one-then-g – Martin Sleziak May 26 '15 at 07:38
  • For surjective functions see http://math.stackexchange.com/questions/22572/injective-and-surjective-functions http://math.stackexchange.com/questions/229065/if-f-circ-g-is-onto-then-f-is-onto-and-if-f-circ-g-is-one-to-one-then-g – Martin Sleziak May 26 '15 at 07:40

1 Answers1

0

Suppose that $f(x_1) = f(x_2)$. Then we have $$g(f(x_1)) = g(f(x_2))$$ so the injectivity of $g \circ f$ implies.....

Suppose that $c \in C$, and choose an $a \in A$ such that $g(f(a)) = c$, by surjectivity of $g \circ f$. So can you conclude that $g$ is onto?