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.