Commutativity of Twisted Inverse with Restrictions

The proof consists in straightforward diagram drawings.  

Note that Ji is admissible by Lemma 18.9, 3, and hence 0(Ji,/ ) and Ji, /,) are defined. Since we have Ji, / ) is the conjugate of 0(Ji, / ) by definition, it suffices to show that 0 Ji,ft) is an isomorphism. It suffices to show that 

If i G Jo, then both hand sides are zero functors, and it is an isomorphism. On the other hand, if i Ji, then the map in question is equal to the composite isomorphism 

Note that U, has flat arrows, since Y has flat arrows and /, o g, is cartesian. 

We prove that g, is cartesian. Let t — j be a morphism in I. Then, the canonical map U, fjPj) Uj Ui Xy. Yj is an isomorphism by assumption. This map factors through U, gj) Uj — Ui Xj, and it is easy to see that (U, gj) is a closed immersion. On the other hand, it is an image dense open immersion, as can be seen easily, and hence it is an isomorphism. So p, is cartesian. 

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Commutativity of Twisted Inverse with Restrictions 2 The composite map... 

The fourth step is to prove various commutativities related to the well-definedness of the twisted inverse pseudofunctors, Chapters 16, 18, and 19. Among them, the compatibility with restrictions (Proposition 18.14) is the key to our construction. Given a separated G-morphism of finite t3rpe / X Y between noetherian G-schemes, the associated morphism BQ f) Bq X) Bq Y) is cartesian, see (4.2) for the definition. If we could find a compactification... 

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