Example 48.15.2. The base change map (48.5.0.1) is not an isomorphism if f is perfect proper and g is perfect. Let k be a field. Let Y = \mathbf{A}^2_ k and let f : X \to Y be the blowup of Y in the origin. Denote E \subset X the exceptional divisor. Then we can factor f as
This gives a factorization a = c \circ b where a, b, and c are the right adjoints of Lemma 48.3.1 of Rf_*, Rp_*, and Ri_*. Denote \mathcal{O}(n) the Serre twist of the structure sheaf on \mathbf{P}^1_ Y and denote \mathcal{O}_ X(n) its restriction to X. Note that X \subset \mathbf{P}^1_ Y is cut out by a degree one equation, hence \mathcal{O}(X) = \mathcal{O}(1). By Lemma 48.15.1 we have b(\mathcal{O}_ Y) = \mathcal{O}(-2)[1]. By Lemma 48.9.7 we have
Last equality by Lemma 48.14.2. Let Y' = \mathop{\mathrm{Spec}}(k) be the origin in Y. The restriction of a(\mathcal{O}_ Y) to X' = E = \mathbf{P}^1_ k is an invertible sheaf of degree -1 placed in cohomological degree 0. But on the other hand, a'(\mathcal{O}_{\mathop{\mathrm{Spec}}(k)}) = \mathcal{O}_ E(-2)[1] which is an invertible sheaf of degree -2 placed in cohomological degree -1, so different. In this example the hypothesis of Tor independence in Lemma 48.6.2 is violated.
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