Consider the case when $$\DeclareMathOperator{\Rank}{Rank}\Rank(AB) + \Rank(BC) = \Rank(B) + \Rank(ABC)$$
where $A \in M_{m,k}(F)$ , $B \in M_{k,p}(F)$ and $C \in M_{p,n}(F)$ for any field $F$.
I wish to show that $$\DeclareMathOperator{\Ker}{Ker}\Ker(AB) \subseteq \Ker(B) + \operatorname{Range}(C)$$
I have been stuck on this problem for a while now. In general the problem statement doesn't even make sense to me, it seems to me that $\Ker(AB)$ is a subset of $F^m$, $\Ker(B)$ is a subset of $F^k$ and $ \operatorname{Range}(C)$ is a subset of $F^n$. So I don't see how we could have such a containment if the dimensions don't line up.
Any help is appreciated thanks!