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
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