= Solution
{c}
The <image of a linear map> $\operatorname{im}(ST)=S(\operatorname{im}T)$ lies inside $\operatorname{im}S$. Therefore $r(ST)\leq r(S)$. Also $S$ restricted to $\operatorname{im}T$ cannot have <rank of a linear map> greater than its domain's <dimension>, so $r(ST)\leq\dim\operatorname{im}T=r(T)$. Thus
$$
\boxed{r(ST)\leq\min\{r(S),r(T)\}.}
$$
Equivalently, apply the <rank-nullity theorem> to that restriction: $r(ST)=r(T)-\dim(\operatorname{im}T\cap\ker S)\leq r(T)$. This proof keeps track of the intermediate <vector subspace>, rather than assuming an invertible factor.
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