= Solution
{c}
For $u\in U$, the <vector> $Tu$ already belongs to $\operatorname{im}T$, so
$$
STu=0\iff S'(Tu)=0\iff Tu\in\ker S'.
$$
Therefore $\ker(ST)=T^{-1}(\ker S')$. Since $\ker S'=\ker S\cap\operatorname{im}T$, its intersection with $\operatorname{im}T$ is itself. Applying <3/a> with $X=\ker S'$ yields
$$
\boxed{n_{ST}=n_{S'}+n_T\leq n_S+n_T.}
$$
The inequality follows from the inclusion $\ker S'\subseteq\ker S$ and monotonicity of <dimension> for <vector subspaces>.
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