As in my [other question](https://mathoverflow.net/questions/99173/commutativity-of-pullback-and-pushforward), 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?



"Restriction" of a fibre product to a subvariety
I believe this should be true: If you factor the map $\delta\mid\_Y$ via the inclusion $Z \hookrightarrow D$ as
$$
Y \to Z \hookrightarrow D
$$
and pull back the map $\gamma$ along each of these maps, you should get your desired diagram as the left-hand side. i.e. you should obtain
$$
\begin{matrix}
W & \to & X & \to & B \\\\
\downarrow & & \downarrow & & \downarrow \\\\
Y & \to & Z & \hookrightarrow & D
\end{matrix}
$$
with each of the squares cartesian by definition.


