Can Linear Operators A and B Affect the Rank of AB in V?

[Zorba]

Studying old exam papers from my college I came across the following:

Given linear operators A,\,B: V\rightarrow V, show that:

\textrm{rk}AB\le \textrm{rk}A

My solution:
Since all v \in \textrm{Ker}B are also in \textrm{Ker}AB (viz ABv=A(Bv)=A(0)=0) and potentially there are w \in \textrm{Ker}A such that w \in \textrm{Im}B which implies w must be also in \textrm{Ker}AB (viz ABv=A(Bv)=A(w)=0 - either Bv is inconsistant then we still have Bv=0, or we have Bv=w and follows as earlier) thus we have \textrm{Ker}A \subset \textrm{Ker}AB and thus from the rank-nullity theorem it follows that \textrm{rk}A \ge \textrm{rk}AB.

Is everything I've done there legal? I'm a bit iffy about the step where I conclude \textrm{Ker}A \subset \textrm{Ker}AB...

 

[adriank]

Having w in Ker A and w in I am B does not imply that w is in Ker AB (although if w = Bv then v is in Ker AB, since ABv = Aw = 0).

Hint: You can solve this by considering only images, not kernels.