Why are certain restrictions morphisms of prevarieties?

Question

At page 30 the author of this book states that the restrictions $f_{\mid f^{-1}(U_i)}$ are morphisms of prevarieties. Can you explain why this is the case?

Answer 1

By definition, $f^{-1}(U_i)$ is the distinguished open set $D(x_i) \subset \mathbb{A}^{n+1}(k)$. By Lemma 1.50 on p. 24, $f^{-1}(U_i)$, as an open subprevariety of $\mathbb{A}^{n+1}(k)$, is isomorphic to the affine variety corresponding to the $k$-algebra $k[x_0,\ldots,x_n]_{x_i}$ (the localisation of $k[x_0,\ldots,x_n]$ at $x_i$). Let $g : U_i \to \mathbb{A}^n(k)$ be the map defined by $g(x_0:\ldots:x_n) = (x_0/x_i,\ldots,x_{i-1}/x_i,x_{i+1}/x_i,\ldots,x_n/x_i).$ By Corollary 1.60 on p. 28, this map is an isomorphism of prevarieties. The composite map $h = g\circ f|_{f^{-1}(U_i)} : f^{-1}(U_i) \to \mathbb{A}^n(k)$ induces the homomorphism of $k$-algebras $\phi : k[y_1,\ldots,y_n] \to k[x_0,\ldots,x_n]_{x_i}$ given by $$ \phi(y_j) = \begin{cases} x_{j-1}/x_i & \text{for} \quad 1\leq j \leq i\\ x_j/x_i & \text{for} \quad i < j \leq n\end{cases}$$ Therefore, $h$ is a morphism of affine varieties, by Proposition 1.44 on p. 22. Since $g$ is an isomorphism, we conclude that the composite map $f|_{f^{-1}(U_i)} = g^{-1}\circ h : f^{-1}(U_i) \to U_i$ is a morphism of prevarieties (in fact, of affine varieties).