Let $V$ be finite dimensional vector space. $T$ is a linear transformation from $V$ into $V$ and $E$ is a subspace of $V$. Shows, $T^{-1}(E)={\{u\in V|T(u)\in E}\}$
Prove dim$(T^{-1}(E))$=dim($ker(T)$)+dim($E\cap Im(T)$)
I know eventually I have to prove that dim$(T^{-1}(E))$= dim($ker(T^{-1}(E)$)+dim($Im(T^{-1}(E)$).
but I can't figur out the relationship between them. 

Consider the restiriction $T|_{T^{-1}E}: T^{-1}(E) \to E$, then the ordinary dimension formula for morphism gives preciswely the desired formula, since $Im(T|_{T^{-1}E})=E \cap Im(T) $
