Equality of Frobenius Inequality 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!
 A: If $T:V\to W$ is a linear mapping, then dimension theorem says
$$
\dim V =\dim{N}(T)+\dim R(T)
$$ where ${N}(T)$ is the null space of $T$ and $R(T)$ is the range of $T$.
Given linear maps $T:V\to W$ and $S:W\to X$, we can apply dimension theorem to
$$
S|_{R(T)}:R(T)\to X
$$ and get
$$
\dim R(T) = \dim N(S|_{R(T)}) +\dim R(S|_{R(T)})=\dim \left[N(S)\cap R(T)\right] +\dim R(ST)
$$or equivalently
$$
\dim R(T)-\dim R(ST)=\dim \left[N(S)\cap R(T)\right].
$$
The given equation can be written as
$$
\DeclareMathOperator{\Rank}{Rank}\Rank(B)-\Rank(AB)  =  \Rank(BC) -\Rank(ABC)
$$ or 
$$
\dim R(B)-\dim R(AB) =\dim R(BC)-\dim R(ABC).
$$ As we've already seen, it is equivalent to
$$
\dim [N(A)\cap R(B)]=\dim [N(A)\cap R(BC)].
$$ Since $N(A)\cap R(B)\ge N(A)\cap R(BC)$, it says that the two spaces are equal.
Assume $v\in N(AB)$. Then $ABv=0$ implies that $$Bv\in N(A)\cap R(B)=N(A)\cap R(BC).$$ Thus there exists $w$ such that $Bv = BCw$. Now, since $B(v-Cw)=0$, there exists $x\in N(B)$ such that $v-Cw =x$. This gives
$$
v= x +Cw \in N(B)+R(C),
$$ proving $$N(AB)\le N(B)+R(C)$$ as desired.
