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318 U. Bunke, T. Schick, M. Spitzweck, and A. ThomIn order to calculate this sheaf in terms of the cohomology sheaves H i .K/ of K wechoose an injective resolution T ! I . Then we haveRHom ShAb S .K; T/ Š Hom Sh Ab S .K; I /:We must calculate the first total cohomology of this double complex. We first take thecohomology in the K-direction, and then in the I -direction. The second page of theassociated spectral sequence has the following form.1 Ext 0 ShAb S .H 1 .P /; T/ Ext 1 ShAb S .H 1 .P /; T/ Ext 2 ShAb S .H 1 .P /; T/ Ext 3 ShAb S .H 1 .P /; T/0 Ext 0 ShAb S .H 0 .P /; T/ Ext 1 ShAb S .H 0 .P /; T/ Ext 2 ShAb S .H 0 .P /; T/ Ext 3 ShAb S .H 0 .P /; T/0 1 2 3The sequence (45) is exactly the edge sequence for the total degree-1 term.5.1.10 The appearance in (45) of the groups Ext i Sh Ab S .H 0 .P /; T/ for i D 1; 2 was themotivation for the introduction of the notion of an admissible sheaf in 4.1.Corollary 5.10. Let P 2 PIC.S/ be such that H 0 .P / is admissible. Then we haveH 0 .D.P // Š D.H 1 .P //; H 1 .D.P // Š D.H 0 .P //:Theorem 5.11. Let P 2 PIC.S/ and assume that1. H 0 .P / and H 1 .P / are dualizable;2. H 0 .P / and D.H 1 .P // are admissible.Then P is dualizable.Proof. In view of 2.14 it suffices to show that the evaluation map c W P ! D.D.P //induces isomorphismsH i .c/W H i .P / ! H i .D.D.P ///for i D 1; 0. Consider first the case i D 0. Then by Corollary 5.10 we have anisomorphismh 0 W H 0 .D.D.P /// ! D.H 1 .D.F // ! D.D.H 0 .P ///:One now checks that the maph 0 ı H 0 .c/W H 0 .P / ! D.D.H 0 .P ///is the evaluation map (44). Since we assume that H 0 .P / is dualizable this map is anisomorphism. Hence H 0 .c/ is an isomorphism, too.For i D 1 we use the isomorphismh 1 W H 1 .D.D.P /// ! D.H 0 .D.P /// ! D.D.H 1 .P ///and the fact that h 1 ı H 1 .c/W H 1 .P / ! D.D.H 1 .P /// is an isomorphism.