Choose $\vect{x}\in U$, then $\lt{T}{\vect{x}}\in V$ and we can form $\preimage{T}{\lt{T}{\vect{x}}}$. Almost trivially, $\vect{x}\in\preimage{T}{\lt{T}{\vect{x}}}$, so every vector in $U$ is in some preimage. For (b), suppose that $\preimage{T}{\vect{v}_1}$ and $\preimage{T}{\vect{v}_2}$ are two different preimages, and the vector $\vect{u}\in U$ is an element of both. Then $\lt{T}{\vect{u}}=\vect{v}_1$ and $\lt{T}{\vect{u}}=\vect{v}_2$. But because $T$ is a function, we conclude that $\vect{v}_1=\vect{v}_2$. It then follows that $\preimage{T}{\vect{v}_1}=\preimage{T}{\vect{v}_2}$, contrary to our assumption that they were different. So there cannot be a common element $\vect{u}$.