I'm new to linear algebra and I'm still quite shaky when it comes to all the new notations and concepts.
According to the dimension theorem of vector spaces $\dim(\ker(T)) + \operatorname{rank}(T) = \dim(U)$ for the linear transformation $T:U \mapsto V$. When I was thinking about that and tried to draw the domains and so one I thought about the following: Does the dimension theorem also imply that $\dim(\ker(T)\cup T^{-1}(V)) = \dim(U)$, whereby $T^{-1}$ is the inverse image of the range of $T$?
$T^{-1}(V)=U$, $\ker T\subseteq U$ hence $\ker(T)\cup T^{-1}(V)=U$ and therefore $\dim(\ker(T)\cup T^{-1}(V)) = \dim(U)$ is true, but the dimension theorem is not needed to prove this.