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$\newcommand{\Spec}{\operatorname{Spec}}$ $\require{AMScd}$

Let $f:X\rightarrow Y$ be a morphism of schemes, and suppose $y\in Y$. Let $X_y=\Spec k_y\times_YX$, where $k_y$ is the residue field at $y$. I am trying to identify this scheme with the fibre $f^{-1}(y)$. I know there are some other questions on this site that deal with this, but I am trying to stay firmly in the realm of scheme theory, as I think this should fall out naturally from the Cartesian square and appealing to affine opens (perhaps I am wrong though).

In particular, I am trying to see that the image of $\pi_X$ in the following diagram is $f^{-1}(y)$: \begin{CD} X_y@>\pi_X>> X \\ @V\pi_yVV @VV f V \\ \Spec k_y@>>g> Y \end{CD} It is clear that $g(\pi_Y(X_s)=f(\pi_X(X_s)$, so $f(\pi_X(X_s))=y$, so $f^{-1}(f(\pi_X(X_s))=f^{-1}(y)$. This only gives me that $\pi_X(X_s)\subset f^{-1}(y)$, and I cannot seem to figure out to show the other inclusion. If $x\in f^{-1}(y)$, how do I show that there is some $z\in X_s$ such that $\pi_{X}(z)=x$?

I am also not totally sure how to show this is a homeomorphism, but I figured that that might be easier once I can actually show this.

Semiclassical
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Chris
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    Have you looked through this answer? I think this does what you want. (If it doesn't, please let me know what's not in there that you are looking for.) – KReiser May 28 '24 at 17:56
  • @KReiser , this is mostly what I was looking for, and I understand everything up to reducing to the affine case. Personally, I am not sure what it means to localize $A$ at the prime ideal $\mathfrak q\subset B$. Is this localization as a $B$-algebra? – Chris May 28 '24 at 22:53
  • Yes, it is localization as a $B$-algebra. It's also equivalent to localizing the ring $A$ at the multiplicatively closed subset which is the image of $B\setminus\mathfrak{q}$ (multiplicatively closedness is preserved by homomorphisms). – KReiser May 28 '24 at 23:50

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