Proving composition of functions

Question

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!

Answer 1

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.....

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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?