X \to Y should be Y \to X (2x)

THanks to Neeraj Deshmukh
https://stacks.math.columbia.edu/tag/0271#comment-3580

examples.tex

@@ -1475,9 +1475,9 @@ \section{Non-quasi-affine variety with quasi-affine normalization}
 ring map $A \otimes_k B \to B_{x + y}$ is surjective. To see
 this use that $A \otimes_k B$ contains the element
 $xy/(x + y) \otimes 1/xy$ which maps to $1/(x + y)$.
-The morphism $X \to Y$ is given by the natural maps
+The morphism $Y \to X$ is given by the natural maps
 $D(x + y) \to \Spec(A)$ and $D(xy) \to \Spec(B)$.
-Since these are both finite we deduce that $X \to Y$ is finite
+Since these are both finite we deduce that $Y \to X$ is finite
 as desired. We omit the verification that $X$ is indeed the
 coequalizer of the displayed diagram above, however, see
 (insert future reference for pushouts in the category of schemes