$\dim \ker (T\circ S) \ge \dim \ker T$.

Question

Suppose we have two transformations $T\colon V\to W$ and $S\colon W\to V$ where $W,V$ are finite dimensional vector spaces.
And also suppose that $\ker T \subset \operatorname{Im}~S$ then is it true that $\dim \ker T \le \dim \ker (T\circ S)$?.

I thought this to be true but I don't see how to prove it.
Let $v\in \ker T$ then I want to show that $v\in \ker (T\circ S)$ which is sufficient to prove this claim.
So we have $w\in W$ s.t $v=Sw$ and $w\in \ker (T\circ S)$, but I want show that $v\in \ker (T\circ S)$.

So I guess this claim is false, but how to find a counterexample, I am clueless with this question, any help is appreciated.

Answer 1

$\dim\ker T\circ S=\dim W-\dim TS(W)\ge\dim W-\dim T(V)=\dim\ker T$