Let $f: X \to Y$ be a morphism of schemes. Then for open $U \subseteq Y$, I believe it is true that $X \times_Y U = f^{-1}(U)$, but I am not sure how to make rigorous the following proof sketch:
Just as in the construction of the fiber product, cover $Y = \cup \mathrm{Spec} ~B$, and for each $B$, cover $\pi^{-1}(\mathrm{Spec} ~B) = \cup \mathrm{Spec} ~A^B_{\alpha}$.
Finally, take a basic open cover (since $\mathrm{Spec} ~B \cap U$ is open) $\mathrm{Spec} ~B \cap U = \cup D(g)$.
Then we have the following commutative diagram:
$\require{AMScd}$
$
\begin{CD}
\mathrm{Spec}(A_{\alpha}^B \otimes_B B_g) = D(f^{\#}(g)) @>i>> \mathrm{Spec}(A_{\alpha}^B)\\@VV V @VV f V\\ D(f) = \mathrm{Spec}~B_f @>i>> \mathrm{Spec} ~B
\end{CD}
$
Thus, we have $\mathrm{Spec} ~(A_{\alpha}^B \otimes B_f) = \pi^{-1}(\mathrm{Spec} ~B \cap U)$. Is it then true that "glueing" will give us $U \times_Y X = f^{-1}(U)$? I have learned about fiber products very recently, so it is still difficult for me to fill in details.
Best Answer
Looking at the fiber product in the category of sets, the equality is clear.