As in my other question, suppose I have a Cartesian diagram of morphisms of algebraic varieties $$\begin{array}{ccc} A & \to^\alpha & B \\ \downarrow^\beta & & \downarrow^\gamma \\ C & \to^\delta & D \end{array}$$

This time, I'm going to suppose all the varieties are proper *surfaces*, and that the maps are (as before) finite and flat. Suppose I have a curve $Y \subset C$ (irreducible and nonsingular, if that helps), and I define $Z = \delta(Y)$, $X = \gamma^{-1}(Z)$, and $W = \beta^{-1}(Y)$. [EDIT: We have $W \subseteq \alpha^{-1}(X)$ but equality does not necessarily hold as I erroneously claimed -- thanks Dustin.]

Is it true that the diagram $$\begin{array}{ccc} W & \to^\alpha & X \\ \downarrow^\beta & & \downarrow^\gamma \\ Y & \to^\delta & Z \end{array}$$ obtained by restricting all the morphisms in the previous diagram is also Cartesian?