Which of the following statements are definitely correct?

Solution 1:

Hint about C). Assume $f\circ g$ is surjective. Then for any $y$ in the codomain of $f$ you can find $x$ in the domain of $g$ such that $f(g(x)) = y$. This means $f$ is surjective, because $g(x)$ is a preimage of $y$ under $f$.

To see that $g$ is not necessarily surjective, take e.g. $$A = \{1,2,3,4\}, \quad B = \{1,2,3\}, \quad C = \{1,2\},$$ and define $g \colon A \to B$ via $$g(1) = 1, \quad g(2)=2, \quad g(3) = 2, \quad g(4) = 2,$$ and $f\colon B \to C$ via $$f(1) = 1, \quad f(2) = 2, \quad f(3) = 2.$$ Now $f$ is surjective, $f(g(1)) = f(1) = 1$ and $f(g(n)) = f(2) = 2$ for $n \neq 1$, so $f\circ g$ is surjective, but $g$ is not surjective.

You can work out D) with a similar example.