Composition of linear mappings is $0$ if and only if $ \operatorname{im}(g) \subseteq \ker(h)$

Question

Let $U,V,W$ be vector spaces and $g:U\rightarrow V,h:V \rightarrow W$ be linear mappings. Prove that the composite linear mapping $h\circ g:U\rightarrow W$ satisfies $h\circ g=0$ if and only if $ \operatorname{im}(g) \subseteq \ker(h)$.

This statement seems obvious but I am terrible at proving things so I have no idea how to even begin.

Answer 1

I will show you one direction: Let $\text{im}(g) \subseteq \text{ker}(h)$. For $ u \in U$, we have $g(u) \in \text{im}(g)$, so $h(g(u)) = 0$ since $\text{im}(g) \subseteq \text{ker}(h)$. Try the other direction, you can do it.