Following Eisenbud-Harris The Geometry of Schemes, and I'm having trouble understanding a specific part of their proof that fibered products exist in the category of schemes.
The affine case is okay, no problems there. However, then we go to the general case, where we consider $X, Y$ and $S$ to be schemes with morphisms $\varphi \colon X \to S$ and $\psi \colon Y \to S$. The idea is to cover the diagram
$\require{AMScd}$ \begin{CD} &&X\\ & @VVV{\varphi}\\ Y @>{\psi}>> S \end{CD} with diagrams \begin{CD} &&\mathrm{Spec}\, A_{\rho\alpha}\\ & @VVV{\varphi_{\rho\alpha}}\\ \mathrm{Spec}\, B_{\rho\beta} @>{\psi_{\rho\beta}}>> \mathrm{Spec}\, R_{\rho} \end{CD} and apply the affine case to obtain a family of fibered products of affine schemes and glue them together.
The problem is how they define that. So the $\mathrm{Spec}\, R_{\rho}$ are an affine cover of $S$. Then, $\mathrm{Spec}\, A_{\rho\alpha}$ and $\mathrm{Spec} \, B_{\rho\beta}$ are the coverings of the preimages of $\mathrm{Spec}\, R_{\rho}$ in $X$ and $Y$ respectively.
This last sentence is what I don't get. The only time the word preimage has appeared before in the book is in the definitions of the preimage of a closed subscheme $X'$ defined by an ideal $I$ of $R$ where $\psi \colon \mathrm{Spec}\, T \to \mathrm{Spec}\, R$ is a morphism of affine schemes.
How do I make sense of the sentence in bold?