I want to solve exercise 3.10 (a) of Hartshorne's book, chapter II, which asks to prove the following:
Let $f\colon X\to Y$ be a morphism of schemes and let $y\in Y$, then $X_y=X\times_Y \operatorname{Spec}k(y)$ is homeomorphic to $f^{-1}(y)$ with the induced topology.
The idea is clear to me and I proved the statement for affines. Now, choosing some open affine $V=\operatorname{Spec}A$ of $Y$ containing $y$, it follows that $X_y\cong f^{-1}(V) \times_V \operatorname{Spec}k(y)$. Thus, it suffices to prove that the projection of the latter one to $f^{-1}(V)$ induces a homeomorphism with $f^{-1}(y)$.
Cover $f^{-1}(V)$ by open affines $U_i=\operatorname{Spec}B_i$. By the construction of the fiber product and in particular of the projection by gluing, the projection $p\colon f^{-1}(V) \times_V \operatorname{Spec}k(y) \to f^{-1}(V)$ is obtained by gluing the projections $p_i\colon \operatorname{Spec}(B_i \otimes_A k(y)) \to \operatorname{Spec}B_i \hookrightarrow f^{-1}(V)$.
$\textbf{The problem is:}$ I don't see why $p$ should remain injective. Let $z,z'$ be two points of $f^{-1}(V) \times_V \operatorname{Spec}k(y)$, then $z \in \operatorname{Spec}B_i \otimes_A k(y)$ and $z'\in \operatorname{Spec}B_j \otimes_A k(y)$ for some $i,j$. Now, $p(z)=p(z')$ implies that $z \in p^{-1}(f^{-1}(y)\cap \operatorname{Spec}B_i \cap \operatorname{Spec}B_j)=p^{-1}(\operatorname{Spec}B_i \cap \operatorname{Spec}B_j)$. Thus, I $\textbf{need to prove}$ that $p^{-1}(\operatorname{Spec}B_i \cap \operatorname{Spec}B_j)= \operatorname{Spec}(B_i \otimes_A k(y)) \cap \operatorname{Spec}(B_j \otimes_A k(y))$, where I fail to see that "$\subseteq$" holds.
I appreciate any kind of help.
Best Answer
The required equality $p^{-1}(\operatorname{Spec}B_i \cap \operatorname{Spec}B_j)=\operatorname{Spec}(B_i \otimes_A k(y)) \cap \operatorname{Spec}(B_j \otimes_A k(y))$ follows from the following:
Applying this to the above situation gives the desired result.