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I found the following argument in more than one article: Let $X_0$ be a complex space and define the functor $F:Art \to Sets$ s.t. $F(A)=\{\text{Isomorphism classes of deformations of X_0 over A}\}$. Let $(X",A"),(X',A'), (X,A)$ deformations of $X$ and consider 2 maps $A"\to A$ and $A' \to A$ such that the first one is surjective, so we can consider $A' \times_{A} A"$. Then a deformation over $A' \times_{A} A"$ is $Y=(X_0,\mathcal{O}_{X'} \times_{\mathcal{O}_X} \mathcal{O}_{X"})$. Can someone explain to me what is Y? Suppose that $X=V(f)$ for some $f\in \mathbb{C}[x_1, \dots, x_n]$, how I compute Y?

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Sorry, I don't know why it is not compiling properly. – Michele Torielli Jun 11 '10 at 14:18
I fixed your formatting. – David Speyer Jun 11 '10 at 14:21
Can't compute $Y$ unless specify $X'$ and $X''$. Meaning of $Y$ is that underlying top. space is $X$, and ring of functions over open $U$ is fiber product ring $O_ {X'}(U) \times_ {O_ X(U)} O_ {X''}(U)$ (makes sense since $X'$ and $X''$ have same underlying top. space as $X$). Good exercise to check $Y$ really is a complex-analytic space and that natural $A' \times_A A''$-algebra structure on its structure sheaf makes it flat over $A' \times_A A''$. Good behavior of flatness under "fiber product" construction is explained in Schlessinger's first paper on def. theory (from his thesis). – BCnrd Jun 11 '10 at 14:52
I think of $Y$ as the gluing of the two deformations $X'$ and $X''$ along $X$. Simplistic example: $A'=k[\epsilon],A''=k[\delta]$ then the fibered product is $k[\epsilon,\delta]/(\epsilon.\delta)$ an infinitesimal coordinate cross. – Heinrich Hartmann Jun 11 '10 at 16:14
This is really a question about fiber products, not deformation theory. – Kevin H. Lin Jun 11 '10 at 22:44

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