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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?

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You write that $W=\alpha^{-1}(X) = \beta^{-1}(Y)$, but this would only be true of $Y$ were equal to $\delta^{-1}(Z)$, which isn't true in general. –  Dustin Cartwright Jun 9 '12 at 16:01
    
Thanks for pointing this out -- I have corrected the question. –  crocodile Jun 9 '12 at 16:32
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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.

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I don't quite follow why the left square is cartesian -- could you explain? –  crocodile Jun 9 '12 at 16:34
    
Cartesian squares satisfy the following 2 out of 3 properties. Let L be the left square R the right square and LR the outside "composed" square. If L and R are Cartesian then so is LR. If LR and R are Cartesian then so is L. Both of these are checked using the universal property of pullbacks, and they are true in any category. As such your question is also true in any category if you generalise it as follows: $Z \to D$ is any morphism that $Y \to D$ factors through, $X = Z \times_D B$ and $W = Y \times_C A$. –  name Jun 9 '12 at 17:15
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