**Proof.**
Part (1) follows from the definition of $U(\psi )$ and the fact that $D_{+}(f) = D_{+}(f^ n)$ for any $n > 0$. For $f \in A_{+}$ homogeneous we see that $A_{(f)} \to B_{(\psi (f))}$ is surjective because any element of $B_{(\psi (f))}$ can be represented by a fraction $b/\psi (f)^ n$ with $n$ arbitrarily large (which forces the degree of $b \in B$ to be large). This proves (2). The same argument shows the map

\[ A_ f \to B_{\psi (f)} \]

is surjective which proves the surjectivity of $\theta $. For an example where this map is not an isomorphism consider the graded ring $A = k[x, y]$ where $k$ is a field and $\deg (x) = 1$, $\deg (y) = 2$. Set $I = (x)$, so that $B = k[y]$. Note that $\mathcal{O}_ Y(1) = 0$ in this case. But it is easy to see that $r_\psi ^*\mathcal{O}_ X(1)$ is not zero. (There are less silly examples.)
$\square$

## Comments (5)

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