some confusion in Rank nullity theorem. Let $ U,V$ and $ W$  be finite dimensional   real  vector spaces,  $T: U→V,\;S:V→W$ and $P:W→U $ be  linear transformations. If $\operatorname{range}(ST) = \operatorname{null space} (P)$,  $\;\operatorname{null space}(ST)=\operatorname{range} P\;$  and $\operatorname{rank}(T)=\operatorname{rank}(S)$. Then  which of the following is true?
Option 1) nullity of T= nullity of S
option 2) $\dim U \neq \dim W$
option 3) if $\dim V= 3$, $\;\dim U=4$ ,then $P$ is not identically zero
option 4) If  $\dim V= 4$, $\;\dim U= 3$ and $T$ is one-one ,then $P$ is identically zero.
My thinking : By rank nullity theorem,  I got only option  1 is correct  as I don't know the other options...
Pliz help me and  I'm very confused as any hints or solution  can be appreciated.
Thank  u 

What I can say:


*

*Option 2) is false. On the contrary, $\dim U=\dim W$, since
$$\dim U=\dim (\ker ST)+\dim(\operatorname{Im}ST)=\dim(\operatorname{Im}P)+\dim (\ker P)=\dim W.$$

*Option 3) is true: if $\dim U>\dim V$, $T$ cannot be injective, nor $ST$.  Hence $$\operatorname{Im}P = \ker ST\ne\{0\}.$$

