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math/9909019 | MATH . Let MATH, and let us consider the possible value of MATH: CASE: MATH. Impossible, similarly to case MATH in REF . CASE: MATH. Consider MATH such that MATH. Since MATH avoids MATH we have MATH for all MATH, hence if MATH then MATH contains MATH. So either MATH or MATH or MATH. Besides MATH for all MATH since otherwise MATH contains MATH, and MATH for MATH since otherwise MATH contains MATH. Hence in this case there are three possible permutations: MATH, MATH and MATH. CASE: MATH. Evidently MATH if and only if MATH. Since the above cases are disjoint we obtain MATH. Besides MATH, hence MATH. MATH . Let MATH, and let us consider the possible value of MATH: CASE: MATH. Let MATH, MATH, and MATH. Since MATH avoids MATH we have that MATH for all MATH. On the other hand MATH avoids MATH, so we have MATH, which means that MATH contains MATH, a contradiction. CASE: MATH. Evidently MATH if and only if MATH. CASE: MATH. Since MATH avoids MATH we have only one permutation MATH. CASE: MATH. Since MATH avoids MATH we have that MATH contains MATH. Since MATH avoids MATH we have MATH. So we have only one permutation MATH. CASE: MATH. Since MATH avoids MATH and MATH we have two permutations MATH or MATH, but MATH avoids MATH, so we we have only one permutation MATH. Since the above cases are disjoint we obtain MATH. Besides MATH, hence MATH. |
math/9909019 | MATH . Let MATH. Fix MATH such that MATH and let us consider the possible value of MATH: CASE: MATH. Evidently MATH if and only if MATH. CASE: MATH. If MATH then MATH contains MATH, so MATH. On the other hand MATH avoids MATH, so we have only one permutation MATH. CASE: MATH. Similarly to the above case, we have MATH or MATH, but the permutation MATH contains MATH, so we have only one permutation MATH. CASE: MATH. Since MATH avoids MATH we have that MATH and MATH contains MATH, a contradiction. Since the above cases are disjoint we obtain by REF that MATH. Besides MATH, hence MATH. MATH . Let MATH. Fix MATH such that MATH and let us consider the possible value of MATH: CASE: MATH or MATH. Similarly to cases MATH or MATH respectively, CASE: MATH. Since MATH avoids MATH we have only one permutation MATH. CASE: MATH. Since MATH avoids MATH we have that MATH; since MATH avoids MATH we have MATH and MATH for all MATH, so MATH contains MATH, a contradiction. Since the above cases are disjoint we obtain MATH. Besides MATH, hence MATH. MATH . Let MATH and MATH. By REF we get MATH. Let MATH and MATH. If MATH then MATH contains MATH, and if MATH then MATH contains MATH, so we have MATH or MATH. Let MATH; since MATH avoids MATH we have that MATH, anthere are MATH permutations of this type. Let MATH; since MATH avoids MATH we have that MATH conatins MATH. If MATH and MATH then MATH contains MATH, which means that MATH. Since MATH avoids MATH we have that MATH, and there are MATH permutations of this type. Hence MATH. MATH . Let MATH and MATH. If MATH then MATH contains MATH, and if MATH then MATH contains MATH, so we have that MATH or MATH. Let MATH; since MATH avoids MATH we get MATH, and there are MATH permutations of this type. Let MATH; if MATH then MATH contains MATH, and if MATH then MATH contains MATH, so we have that MATH. Since MATH avoids MATH we have that MATH, and there are MATH permutations of this type. Hence MATH. MATH . Let MATH and MATH. Similarly to case MATH we have that MATH or MATH. Let MATH; since MATH avoids MATH we have that MATH, and there are MATH permutations of this type. Let MATH. If MATH then, since MATH avoids MATH we get MATH. If MATH then, since MATH avoids MATH we have that MATH contains MATH. If there exist MATH such that MATH, then MATH contains MATH, so MATH. Since MATH avoids MATH we have that MATH, and there are MATH permutations of this type. Hence MATH. MATH . Let MATH, fix MATH such that MATH and let us consider the possible value of MATH: CASE: MATH. Since MATH avoids MATH we have that MATH for all MATH, so since MATH avoids MATH we get MATH for all MATH or MATH. Hence MATH, and there are MATH permutations of this type. CASE: MATH. Evidently MATH if and only if MATH. By CITE we have MATH. Since the above cases are disjoint we obtain MATH. |
math/9909019 | MATH . By CITE we get MATH, hence by REF we have that MATH for all MATH containing MATH or MATH. MATH . Let MATH and let us consider the possible value of MATH: CASE: MATH. Impossible, similarly to case MATH in REF CASE: MATH. Let MATH; since MATH avoids MATH we have for all MATH, MATH. Since MATH avoids MATH we have for all MATH, MATH. Since MATH avoids REF we have MATH, and there are MATH permutations of this type. CASE: MATH. Evidently MATH if and only if MATH. Since the above casses are disjoint we obtain MATH. Besides MATH, hence MATH for all MATH. It is easy to see that for MATH the same formula holds for all MATH. MATH . Let MATH and let us consider the possible value of MATH: CASE: MATH. Impossible, similarly to case MATH in REF . CASE: MATH. Consider MATH such that MATH and MATH. Since MATH avoids MATH we get MATH for all MATH, and since MATH avoids MATH we get MATH for all MATH, So MATH. Hence MATH if and only if MATH. By CITE we get MATH. If MATH then since MATH avoids MATH we have only one permutation MATH. CASE: MATH. Evidently MATH if and only if MATH. By CITE we have MATH. Since the above cases are disjoint we obtain MATH. MATH . Let MATH. Let us consider the possible value of MATH: CASE: MATH. Since MATH avoids MATH we have that MATH. CASE: MATH, MATH. If MATH then MATH contains MATH, and if MATH then MATH contains MATH, So MATH. Since MATH avoids MATH we have that MATH contains MATH. Fix MATH such that MATH. If there MATH such that MATH then MATH contains MATH which is order-isomorphic to MATH, so MATH. Since MATH avoids MATH we have that MATH, and there are MATH permutations of this type. CASE: MATH. Evidently MATH if and only if MATH. Since the above cases are disjoint we obtain MATH. Besides MATH, hence similarly to the second proof we get MATH. MATH . Let MATH and MATH. By REF we get MATH. Let MATH and let us consider the possible value of MATH: CASE: MATH. Since MATH avoids MATH we have that MATH contains MATH, a contradiction. CASE: MATH. Consider MATH such that MATH. Since MATH avoids MATH we have MATH for all MATH, and since MATH avoids MATH we have that MATH for all MATH. Therefore MATH. Hence MATH if and only if MATH. By CITE and corollary MATH we get that MATH contains MATH permutations for all MATH, and exactly one permutation for MATH. CASE: MATH. Similarly to REF, MATH if and only if MATH. By CITE and REF we have that MATH permutations. Since the above cases are disjoint we obtain MATH. MATH . Let MATH and let us consider the possible value of MATH: CASE: MATH. Fix MATH such that MATH. If MATH and MATH then MATH contains MATH, and if MATH and MATH then MATH contains MATH. Since MATH avoids MATH we have that MATH, and since MATH avoids MATH we get MATH, and there are MATH permutations of this type. CASE: MATH. Evidently MATH if and only if MATH. Since the above cases are disjoint we obtain MATH. Besides MATH, hence similarly to the second proof we get MATH. MATH . Let MATH and let us consider the possible value of MATH: CASE: MATH. Since MATH avoids MATH we get MATH, so MATH contains MATH, a contradiction. CASE: MATH. Since MATH avoids MATH the permutation MATH contains MATH which is order-isomorphic to MATH, a contradiction. CASE: MATH. Since MATH avoids MATH we have that MATH, and since MATH avoids MATH we get MATH and MATH. Since MATH avoids MATH we have that MATH for all MATH. Hence MATH. CASE: MATH. If MATH then MATH contains MATH, so MATH. Evidently MATH if and only if MATH. By CITE we have MATH. CASE: MATH. Evidently MATH if and only if MATH. Since the above cases are disjoint we obtain MATH. Besides MATH, hence similarly to the second proof we get MATH. MATH . Let MATH and let us consider the possible value of MATH: CASE: MATH. Since MATH avoids MATH we get MATH. CASE: MATH. Since MATH avoids MATH we get that MATH contains MATH and since MATH avoids MATH we get that MATH contains MATH which is order-isomorphic to MATH, a contradiction. CASE: MATH. Similarly to case MATH we get MATH. So MATH if and only if MATH. By CITE we have MATH. CASE: MATH. Evidently MATH if and only if MATH. Since the above cases are disjoint we obtain MATH. Besides MATH, hence similarly to the second proof we get MATH. MATH . Let MATH and let us consider the possible value of MATH: CASE: MATH. Similarly to case MATH we have only one permutation MATH. CASE: MATH. Since MATH avoids MATH we have that MATH, and since MATH avoids MATH we get MATH. CASE: MATH, MATH. Fix MATH such that MATH. Since MATH avoids MATH we have that MATH for all MATH and MATH for all MATH. Since MATH avoids MATH we have that MATH for all MATH. If MATH and MATH then MATH contains either MATH or MATH, which means that MATH contains MATH or MATH, a contradiction. Hence MATH, and there are MATH permutations of this type. CASE: MATH. Evidently MATH if and only if MATH. Since the above cases are disjoint we obtain MATH. Besides MATH, hence similarly to the second proof we get MATH. MATH . Let MATH and let us consider the possible value of MATH: CASE: MATH. Similarly to case MATH we have only one permutation MATH. CASE: MATH, MATH. Since MATH avoids MATH we have that MATH contains MATH. Fix MATH such that MATH, if MATH and MATH then MATH contains MATH. Hence MATH. Since MATH avoids MATH we have that MATH, and there are MATH permutations of this type. CASE: MATH. Evidently MATH if and only if MATH. Since th above cases are disjoint we obtain MATH. Besides MATH, hence similarly to the second proof we get MATH. MATH . Let MATH and let us consider the possible value of MATH: CASE: MATH. Evidently MATH if and only if MATH. CASE: MATH, MATH. If MATH then MATH contains MATH, and if MATH then MATH contains MATH or MATH which is order-isomorphic to MATH or MATH respectively, so MATH. Since MATH avoids MATH we have that MATH, and there are MATH permutations of this type. CASE: MATH. Since MATH avoids MATH we have that MATH. Since the above cases are disjoint we obtain by REF that MATH. Besides MATH, hence similarly to the second proof we get MATH. MATH . By CITE we have that MATH, hence by REF we get MATH when MATH. Proof REF : Let MATH and MATH. By REF we get MATH. Let MATH and let us consider the possible value of MATH: CASE: MATH. Since MATH avoids MATH we have that MATH. CASE: MATH, MATH. Similarly to case MATH we have that MATH. Evidently MATH if and only if MATH. By CITE we have MATH. CASE: MATH. Evidently MATH if and only if MATH. By CITE we have that MATH. Since the above cases are disjoint we obtain MATH. MATH . Let MATH and MATH. By REF we get MATH. Let MATH and let us consider the possible value of MATH: CASE: MATH. Since MATH avoids MATH we have that MATH. CASE: MATH, MATH. Fix MATH such that MATH, since MATH avoids MATH and MATH we have that MATH for MATH and MATH for MATH respectively, so MATH. Evidently MATH if and only if MATH. By CITE we have MATH. CASE: MATH. Since MATH avoids MATH we have that MATH. Evidently MATH if and only if MATH. By CITE we have that MATH. CASE: MATH. Evidently MATH if and only if MATH. By CITE we have that MATH. Since the above cases are disjoint we obtain MATH. |
math/9909019 | By CITE we have that MATH, hence by REF we get MATH. |
math/9909019 | By CITE. |
math/9909019 | MATH . By CITE we have that MATH, hence by REF we get MATH when MATH contains a permutation in MATH. MATH . Let MATH and let us consider the possible value of MATH: CASE: MATH. Since MATH avoids MATH we have that MATH contains MATH, which means that MATH contains MATH, a contradiction. CASE: MATH. Since MATH avoids MATH we get MATH. Since MATH avoids MATH we have only one permutation MATH. CASE: MATH. Evidently MATH if and only if MATH. Since the above cases are disjoint we obtain MATH. Besides MATH, hence MATH for all MATH. It is easy to see that for MATH the same formula holds. |
math/9909019 | By definitions. |
math/9909019 | By induction and REF . |
math/9909019 | By induction and REF . |
math/9909019 | By induction and REF . |
math/9909019 | By CITE we have that MATH, so by REF . |
math/9909019 | By induction and REF . |
math/9909019 | By CITE. |
math/9909024 | We may as well assume that our NAME category MATH is the localization of MATH with respect to a hereditary torsion theory MATH, for some ring MATH. Let MATH be the larger of MATH and the cardinality of MATH, let MATH be an ordinal with MATH, and let MATH be a colimit-preserving functor. We will first show that MATH, calculated in MATH, is still MATH-local, so is also the colimit in MATH. This proof will depend on the fact that both MATH and MATH are MATH-small in MATH CITE, for all (left) ideals MATH of MATH. To see this, first note that MATH is torsion-free. Indeed, MATH is generated by cyclic modules MATH, so it suffices to show that MATH. But we have chosen MATH so that MATH since each MATH is torsion-free. Hence MATH is torsion-free. It follows that the localization of MATH is MATH where the colimit is taken over ideals MATH such that MATH is torsion, as in CITE. But then we have MATH . Thus MATH is already local. Now suppose MATH is an arbitrary local module. There is a cardinal MATH such that MATH is MATH-small as a MATH-module, and we can choose MATH. It is then immediate from the argument above that MATH is MATH-small in MATH. |
math/9909024 | We have an exact sequence MATH for each object MATH of MATH. Since filtered colimits are exact, we find that MATH and smiilarly for MATH. Applying colimits to the short exact sequences MATH completes the proof. |
math/9909024 | By the small object argument CITE, every element of MATH is a retract of a transfinite composition of pushouts of elements of MATH. Injections are closed under retracts and pushouts in any abelian category; the ABREF condition guarantees that they are also closed under transfinite compositions. REF then shows that pushouts of maps of MATH are quasi-isomorphisms. REF shows that transfinite compositions of quasi-isomorphisms are quasi-isomorphisms. It is clear that retracts of injective quasi-isomorphisms are quasi-isomorphisms. |
math/9909024 | It is well-known that the category of sheaves on a ringed space is a NAME category CITE. The category of quasi-coherent sheaves on ringed space is an abelian subcategory of all sheaves, closed under colimits. So colimits are exact. It remains to show that the category of quasi-coherent sheaves on a quasi-compact, quasi-separated scheme has a generator. This is REF 's result in CITE, which asserts that every quasi-coherent sheaf is the colimit of finitely presented sheaves. Since there is only a set of finitely presented sheaves, the direct sum of all of them will serve as a generator. |
math/9909024 | We may as well assume MATH is torsion-free, since killing the torsion only decreases the cardinality of MATH, without changing MATH. Then MATH, where the colimit is taken over ideals MATH such that MATH lies in MATH. Thus MATH . In case MATH is finite, MATH, and one can easily see from this equation that MATH is countable when MATH is so. In case MATH is infinite, we have MATH as required. |
math/9909024 | The smallest subcomplex MATH of MATH-modules containing MATH is simply MATH in degree MATH and MATH in degree MATH, so MATH. Since localization is exact, the localization MATH of MATH will be a subcomplex in MATH containing MATH. REF guarantees that MATH. |
math/9909024 | We first show that any map in MATH is injective. Recall the disk functor MATH that takes an object MATH to the complex which is MATH in degrees MATH and MATH, and MATH elsewhere. The functor MATH is right adjoint to the exact functor MATH. Thus MATH is injective whenever MATH is injective in MATH. In particular, suppose MATH is a map of complexes with kernel MATH. Fix MATH, and embed MATH into an injective object MATH. This embedding extends to a map MATH, and so defines a map of complexes MATH, which is MATH in degree MATH. This map obviously cannot extend to a map MATH unless MATH is MATH. Since the map MATH is in MATH, this shows that every map in MATH is an injection. Conversely, suppose we have a commutative diagram in MATH as follows, MATH where MATH is an injection and MATH is a surjection with injective kernel MATH. Since MATH is injective in MATH, there is a splitting MATH of MATH. We have MATH, so, since MATH is injective, there is an extension MATH such that MATH. Then MATH is the desired lift. Hence MATH is in MATH. Now, we always have MATH. Conversely, suppose MATH has the right lifting property with respect to all injections. Consider the map MATH that is the identity in degree MATH. Since MATH has the right lifting property with respect to all injections, there is a lift MATH of this map. This shows that MATH is a split surjection in each dimension. Since the map MATH is a pullback of MATH, it too will have the right lifting property with respect to all injections, and so MATH is injective as an object of MATH. |
math/9909024 | Let MATH denote the class of surjections whose kernel is injective. Applying REF , we see that MATH, so MATH. Thus MATH consists of injections. Furthermore, if we can show MATH consists of all injections, REF will show that MATH is MATH, as required. So suppose MATH is an injection. To show that MATH, we will show that MATH has the left lifting property with respect to MATH. So suppose MATH is in MATH, and we have a commutative diagram as follows. MATH . Let MATH be the set of partial lifts of this diagram, so MATH is the set of all pairs MATH, where MATH is a subcomplex of MATH containing MATH and MATH is a chain map such that MATH and MATH. Then MATH is a partially ordered set, where MATH if MATH contains MATH and MATH extends MATH. The set MATH is nonempty and we claim that every chain in MATH has an upper bound. Indeed, given a chain MATH in MATH, the colimit MATH of the MATH is still a subcomplex of MATH, by the ABREF condition, and the union of the MATH defines a lift on MATH. Thus there is a maximal element MATH of MATH. Suppose that MATH is not all of MATH, and choose a homogeneous element MATH that is not in MATH. Let MATH be the smallest subcomplex (in MATH) of MATH containing MATH, so that MATH by REF . Let MATH denote the subcomplex of MATH generated by MATH and MATH, so that we have the pushout diagram below. MATH . Since the top horizontal map is in MATH, the bottom horizontal map is in MATH. Hence there is a lift MATH in the following diagram. MATH . This lift violates the maximality of MATH, so we must have MATH. Hence MATH, as required. |
math/9909024 | The second statement follows from the first. Indeed, a map in MATH has the right lifting property with respect to all injections, so in particular is an injective fibration. If MATH, then MATH is an injective object of MATH, so has no homology by the first statement. Thus MATH is a homology isomorphism, by the long exact sequence. Now suppose MATH is an injective object of MATH. Certainly MATH is injectively fibrant. To see that MATH has no homology, let MATH denote the complex defined by MATH with MATH. Then MATH is an inclusion of complexes, so since MATH is injective, has a retraction MATH. This retraction is equivalent to a contracting homotopy of MATH, so in particular MATH has no homology. |
math/9909024 | The failure of MATH to be a quasi-isomorphism is measured by MATH. Suppose for the moment that for every homogeneous element MATH of MATH, we can find a subcomplex MATH containing MATH such that MATH and the map MATH sends MATH to MATH. Since MATH, the MATH-module homology of MATH has size MATH. But then REF assures that MATH, so have MATH choices for MATH. We can therefore take the union of all the MATH to form a new subcomplex MATH with MATH (using REF again), such that the induced map MATH is the zero map. Now iterate this construction to form a sequence MATH, and let MATH be the colimit of all the MATH. Then MATH, by REF . Note that MATH is the colimit of the MATH, by commuting colimits. REF then shows that MATH, as required. To complete the proof, we must construct the complex MATH. The construction we give is fairly complicated; we do not know if there is a simpler one. Let us denote the MATH-module homology of a complex MATH by MATH and let us denote the torsion submodule of a MATH-module MATH by MATH. Then the class MATH is represented by a homomorphism MATH, for some left ideal MATH of MATH such that MATH is in the torsion theory. The class MATH must map to MATH in MATH, since MATH is a quasi-isomorphism. This means that there is a subideal MATH of MATH with MATH also in the torsion theory, such that the composite MATH is the zero map. We need to construct MATH so that this map is already the zero map when MATH replaces MATH. For each MATH, choose an element MATH in MATH with MATH whose homology class MATH is a representative for MATH. Since the above composite is MATH, there is an ideal MATH such that MATH is in the torsion theory, and MATH in MATH. This means that, for every MATH, there is an element MATH such that MATH. We define MATH to be the smallest subcomplex of MATH containing MATH and all the MATH. It is clear from the construction that MATH goes to MATH in MATH. Since there are MATH choices for MATH and MATH, the smallest subcomplex of MATH-modules containing MATH and the MATH has size MATH. REF then shows MATH, as required. |
math/9909024 | The second statement is an immediate corollary of the first. By REF , the maps of MATH are injective quasi-isomorphisms. Now suppose MATH is an injective quasi-isomorphism. To show that MATH, we show that MATH has the left lifting property with respect to MATH. So suppose MATH is in MATH, and we have a commutative diagram as follows. MATH . Let MATH denote the set of partial lifts MATH, where MATH is a subcomplex of MATH containing MATH such that the map MATH is a quasi-isomorphism, and MATH is a partial lift in our diagram. Then MATH is obviously partially ordered and nonempty. REF and the argument used in the proof of REF imply that a chain in MATH has an upper bound. NAME 's lemma then gives us a maximal element MATH of MATH. Suppose MATH is not all of MATH, and choose an element MATH in MATH but not in MATH. Let MATH denote the subcomplex of MATH generated by MATH, so MATH by REF . Since MATH is a quasi-isomorphism, REF implies that there is a complex MATH containing MATH such that MATH and the map MATH is a quasi-isomorphism. Let MATH denote the subcomplex of MATH generated by MATH and MATH. Then the map MATH is in MATH, since it is a pushout of MATH. Since MATH, there is an extension of MATH to MATH, contradicting the maximality of MATH. Therefore we must have had MATH, and so MATH, as required. |
math/9909024 | If MATH is exact, then clearly MATH preserves injections and all quasi-isomorphisms, so preserves cofibrations and trivial cofibrations. Conversely, suppose MATH preserves cofibrations and trivial cofibrations. We can think of an exact sequence as a complex MATH with no homology, so MATH is a trivial cofibration. Then MATH must also be a trivial cofibration, so MATH must be exact. |
math/9909024 | NAME implies that MATH has the right lifting property with respect to MATH if and only if the map MATH is surjective. Applying this when MATH is the map MATH for MATH, we find that, if MATH is a MATH-fibration, then MATH is surjective. Furthermore, if MATH is a MATH-fibration, then MATH is in MATH. Applying the above criterion, we find that MATH is dimensionwise MATH-flasque. Conversely, suppose MATH is a surjection for all MATH and MATH is dimensionwise MATH-flasque. Suppose MATH is in MATH. We have an exact sequence MATH and a similar exact sequence that is in fact short exact when MATH is replaced by MATH. By pulling back the exact sequence for MATH through the map MATH, we obtain the following commutative diagram whose top row is short exact and whose bottom row only misses being short exact because the right map is not necessarily surjective. MATH . Since MATH is dimensionwise MATH-flasque, the left-hand vertical map is surjective. A standard diagram chase, as in the snake lemma, then show that the middle vertical map is surjective, so MATH is a MATH-fibration. |
math/9909024 | Recall that the functor MATH is left adjoint to the functor that takes MATH to MATH, the cycles in MATH. This implies that MATH is in MATH if and only if it is a MATH-fibration and the map MATH is surjective for all MATH and MATH. Let MATH. If MATH, then the map MATH is as well. Hence the map MATH is surjective for all MATH and all MATH. Since MATH is a set of generators for MATH, this implies that the map MATH is surjective, and hence that MATH has no homology. A similar argument shows that MATH is surjective, and so the long exact sequence implies that MATH is a quasi-isomorphism. |
math/9909024 | The only if implication is clear. Suppose MATH is surjective for all MATH, and let MATH. Suppose MATH is in MATH. In particular, this means that MATH is dimensionwise MATH-flasque, so MATH. In order to show that MATH is in MATH, we must show that, given MATH, a map MATH, and a map MATH such that MATH, there is a map MATH such that MATH and MATH. First choose MATH such that MATH, using the fact that MATH is surjective. Then MATH, so MATH. Since MATH is in MATH, there is a map MATH such that MATH. Now let MATH. |
math/9909024 | By REF , it suffices to show that MATH is in MATH. But MATH is an acyclic dimensionwise flasque complex, and so MATH is also acyclic. Hence the map MATH is surjective, and so MATH is in MATH. |
math/9909024 | Since MATH is acyclic, we have a short exact sequence MATH . Since MATH for all MATH, this gives us an exact sequence MATH and isomorphisms MATH for MATH. Thus MATH for all MATH. Since MATH has finite projective dimension, this implies MATH for all MATH. It follows that MATH is acyclic. |
math/9909024 | Note first that the MATH-flasque objects of MATH coincide with the injective objects, by CITE. REF implies that if MATH is an acyclic, dimensionwise injective, complex, then MATH is acyclic for all MATH. Hence REF gives us a model structure. Any MATH-fibration is a surjection with dimensionwise injective kernel, by REF , and therefore must be a dimensionwise split surjection. Conversely, a dimensionwise split surjection with dimensionwise injective kernel certainly satisfies the conditions of REF , so is a MATH-fibration. |
math/9909024 | We first show that the locally frees generate MATH. CITE shows that every quasi-coherent sheaf is a colimit of finitely presented sheaves. On a noetherian scheme, finitely presented sheaves are coherent, and thus, since MATH has enough locally frees, are quotients of locally free sheaves of finite rank. Now let MATH be a locally free sheaf of finite rank, and MATH a quasi-coherent sheaf of MATH-modules on MATH. By the corollary to CITE, we have MATH where MATH denotes sheaf Hom and the cohomology groups are sheaf cohomology. If MATH is finite-dimensional, we can apply NAME 's vanishing theorem CITE to conclude that these cohomology groups are MATH for large enough MATH. If MATH is separated, then we can apply CITE to reach the same conclusion, using the fact that MATH is quasi-coherent. This does not complete the proof, because these are MATH groups in MATH rather than in MATH. However, these two possibly different MATH groups in fact coincide, because the exact inclusion functor MATH has a right adjoint and left inverse MATH CITE whenever MATH is quasi-compact and quasi-separated, as any noetherian scheme is. In detail, given a quasi-coherent sheaf MATH, we can first take an injective resolution MATH of MATH in MATH and apply MATH to get a complex of injectives MATH in MATH. We claim that MATH is still exact. To see this, consider the short exact sequence MATH . After we apply MATH, we get a long exact sequence involving the derived functors MATH of MATH. However, MATH for MATH, by the last paragraph of CITE. Furthermore, MATH for MATH because MATH is injective. It follows that MATH for MATH as well. Repeating this argument on the short exact sequence MATH we find that MATH for MATH, and, by induction, that MATH for all MATH and MATH. Hence MATH is still exact, and so is an injective resolution of MATH in MATH. Applying MATH to MATH and using adjointness, we find that, if MATH and MATH are both quasi-coherent, then MATH, completing the proof. |
math/9909024 | The fact that the locally free sheaves form a set of weak generators follows from CITE. To see that they are small, in the triangulated sense, we use the result of CITE. We must then show that, if MATH is a locally free sheaf of finite rank, the functor MATH preserves all transfinite compositions. Since we are on a noetherian scheme, we can take the transfinite composition in the category of presheaves CITE. It is then easy to check the desired result. |
math/9909024 | We apply REF , taking the set MATH to be the canonical inclusions MATH. We use REF . One can easily check that MATH; this is essentially the definition of sheaf cohomology. In particular, MATH has finite hereditary global dimension if and only if each MATH has finite projective dimension. Also, since the restriction of a flasque sheaf is still flasque and flasque sheaves have no cohomology, MATH if MATH and MATH is a complex of flasque sheaves. So REF applies, and REF gives us the desired model structure. The characterization of fibrations in REF translates into surjections of presheaves with dimensionwise flasque kernel. However, sheaf surjections with flasque kernel are also presheaf surjections, so we get the claimed characterization of fibrations. |
math/9909024 | REF follows from the vanishing theorem CITE of NAME, since an open subspace of a finite-dimensional noetherian space is still a finite-dimensional noetherian space. REF is an immediate consequence of CITE. |
math/9909024 | The maps of MATH are all degreewise split monomorphisms on each stalk. Every cofibration is a retract of a transfinite composition of pushouts of maps of MATH, by the small object argument CITE. Since retracts, transfinite compositions, and pushouts all commute with the operation of taking stalks and preserve split monomorphisms, every cofibration will be a degreewise split monomorphism on each stalk. The cokernel of a cofibration will of course be cofibrant, so to complete the proof it suffices to show that every cofibrant object is NAME. Every cofibrant object MATH is a retract of the colimit of a transfinite sequence MATH, where each map MATH is a pushout of a map of MATH and MATH. Since colimits commute with tensor products and homology, it suffices to show that, if MATH is NAME, so is MATH. On each stalk, the maps of MATH are degreewise split monomorphisms with degreewise flat cokernel, so the same will be true of MATH. Thus, if MATH is a complex of flat sheaves, so is MATH. Now suppose MATH is an acyclic complex and MATH is a map of MATH. Again, since the maps of MATH are degreewise split monomorphisms on each stalk, the map MATH will still be injective. Thus the map MATH will be injective. It therefore suffices to show that MATH is a quasi-isomorphism, by REF . In case MATH is of the form MATH, both the domain and codomain of MATH are contractible. The same will be true of MATH, so MATH will be a quasi-isomorphism. In case MATH is of the form MATH, the codomain of MATH is contractible, so it suffices to show that MATH is acyclic. But, since MATH is flat, we have MATH, so we are done. |
math/9909024 | As explained in CITE, since monomorphisms and injective quasi-isomorphisms are closed under retracts, transfinite compositions, and pushouts, it suffices to check this theorem when the flat cofibration is in fact a map of MATH, and the flat trivial cofibration is a map of MATH. We begin with REF . Suppose that MATH is a monomorphism, and suppose MATH is the map MATH. Let MATH denote the domain of MATH, and suppose MATH. If MATH, then the stalk of MATH at MATH is MATH; if MATH, then the stalk of MATH at MATH is MATH; and if MATH is not in MATH, then the stalk of MATH at MATH is MATH. The stalk of the codomain of MATH at MATH is MATH if MATH is in MATH, and MATH otherwise, and the map MATH does the obvious thing on the stalks. Hence MATH is a monomorphism. Furthermore, the domain and codomain of MATH are contractible, so the same will be true for MATH. Thus MATH will be a quasi-isomorphism, completing the proof of REF . To complete the proof of REF , we must show that MATH is a monomorphism, where now MATH is the map MATH. In this case, the stalk of the domain MATH of MATH at a point MATH is MATH if MATH, and otherwise is MATH. The stalk of the codomain of MATH at MATH is MATH if MATH, and otherwise is MATH. The map MATH does the obvious thing, and so is a monomorphism. For REF , we can assume MATH is the map MATH. Then the codomain of MATH is contractible, so it suffices to show that the domain MATH of MATH has no homology. Since MATH is an injective quasi-isomorphism, and MATH is flat, MATH is also an injective quasi-isomorphism. Hence its pushout MATH is also an injective quasi-isomorphism. Since MATH is contractible, it follows that MATH has no homology. Finally, for REF , we can assume that both MATH and MATH are maps of MATH. To calculate MATH in this case, use the easily checked (on stalks) fact that MATH. It follows that MATH and MATH and that MATH is an amalgamation of MATH and MATH. With these identities in hand, the proof is a calculation we leave to the reader. |
math/9909024 | The fact that MATH is a symmetric monoidal model category is immediate from REF . Suppose MATH is cofibrant. Then MATH preserves trivial cofibrations in any symmetric monoidal model category. So, in order to see that MATH preserves quasi-isomorphisms, it suffices to show that, if MATH is a trivial fibration, then MATH is a quasi-isomorphism. Let MATH denote the kernel of MATH, so that MATH is an acyclic complex. Since cofibrant objects are degreewise flat, MATH is still surjective, and its kernel is MATH. Since cofibrant objects are NAME, MATH is still acyclic, so the long exact sequence completes the proof. In general, the total derived functor of the tensor product is defined by MATH, where MATH (respectively, MATH) is a functorial cofibrant replacement for MATH (respectively, MATH). But, since the map MATH is a quasi-isomorphism, MATH is isomorphic in the derived category to MATH. |
math/9909024 | It is well-known that the derived category of any abelian category is triangulated, but this also follows, in a stronger sense of the word triangulated, from the results of CITE. We have already seen that the flat model caetgory is a symmetric monoidal model category, so the derived category is also closed symmetric monoidal in a way that is compatible with the triangulation (see CITE, with one technical point dealt with by CITE). Since the flat model structure is cofibrantly generated, the cofibers of the generating cofibrations form a set of weak generators CITE. In our case, these are the objects MATH (the cofibers of the maps of MATH are trivial in the derived category). Because MATH is noetherian, the presheaf colimit of a direct system of sheaves coincides with the sheaf colimit CITE. It follows from this that MATH commutes with direct colimits. The results of CITE then show that MATH is small (in the triangulated sense) in the derived category. |
math/9909024 | Suppose MATH is a flat trivial cofibration, and MATH is an arbitrary object. Then MATH is a monomorphism, so applying REF shows that MATH is an injective quasi-isomorphism. REF completes the proof. |
math/9909024 | It suffices to show that, if MATH is a cofibrant MATH-module, then MATH preserves weak equivalences, by CITE. The proof of this is very similar to the proof of the corresponding fact in REF , so we leave it to the reader. |
math/9909028 | In CITE we proved this statement for the following composition: MATH where MATH is a closed manifold and MATH is an excisive triad. If MATH has the boundary the proof is the same except for the last step: for any MATH . This formula follows from definitions of the NAME class and the fundamental class, see CITE. Now we obtain the above statement by putting MATH . |
math/9909028 | Let MATH be given by MATH . Let's consider the diagram on the chain level and start in the left upper corner with MATH where MATH is a MATH-chain and MATH is a MATH-chain. Then, in terms of the NAME approximation, we get MATH in the right lower corner. |
math/9909028 | By REF we have MATH . |
math/9909028 | The first square trivially commutes. For the second, going MATH we get MATH . Now going MATH, from REF we get-MATH . The third square commutes as MATH . |
math/9909028 | (compare CITE) We start in the left upper corner of the above diagram with MATH, where MATH . Let MATH . Then going MATH we get MATH by REF . Let MATH with MATH . Then going MATH we get: MATH . Thus MATH . Therefore by definition of the NAME homomorphism we have MATH . Now the statement follows from REF . |
math/9909028 | In CITE the formula is proven for MATH but the proof is valid for any MATH . |
math/9909028 | Consider the following commutative diagram: MATH . Then going from the left to the right we get MATH . Hence MATH therefore MATH. Thus the next diagram is commutative: MATH . Then MATH so MATH . Now the statement follows from the definition of MATH. |
math/9909028 | Let MATH. Then we have MATH . |
math/9909028 | Let MATH and suppose MATH for some MATH . Then we have MATH . Therefore MATH and the statement follows. |
math/9909028 | The fact that MATH is a homeomorphism implies two things. First, MATH is well defined. Hence from the naturality of the NAME isomorphism we have MATH where MATH is the NAME isomorphism for MATH . Second, since MATH it follows that MATH and, therefore, MATH is well defined REF . In the computation below we also use REF (a similar statement can be proven independently for MATH), REF and the trivial fact that MATH . We have MATH . Observe that the NAME homomorphism MATH is well defined without the restriction MATH . |
math/9909028 | It follows from REF . |
math/9909028 | If MATH then MATH by REF . The rest follows from REF . |
math/9909028 | It follows from REF . |
math/9909028 | Suppose MATH, a NAME class, contain only the zero group, see CITE. Then MATH is an ideal of abelian groups. Observe that MATH for MATH and MATH for MATH . Then MATH is a MATH-epimorphism for MATH, so REF is satisfied. |
math/9909028 | Consider the following commutative diagram: MATH where MATH is the inclusion. Since MATH is an isomorphism, for any MATH we have the following MATH . Therefore MATH . |
math/9909029 | First, let me quickly point out that in order to verify unfoldability in the extension it suffices merely to lift the ground model embeddings. This is because if MATH is any structure of size MATH in MATH, then it has a name MATH of size MATH in MATH, and this name can be placed into the domain of a ground model canonical unfoldability embedding MATH. If this ground model embedding lifts to MATH, then MATH is included in MATH, the domain of an unfoldability embedding in MATH, and this is sufficient, since as I remarked earlier the restriction MATH then provides an unfoldability embedding for MATH. So fix any MATH and any canonical MATH-unfoldability embedding MATH in MATH, and choose any MATH. I will show how to lift the embedding to the extension. For simplicity, let me initially assume that MATH; I will consider the general case subsequently. Below the condition MATH, a partial function with only one element in its domain, we may split the conditions of MATH into their small part, with domain below MATH, and their tail part, with domain above MATH. Since the collection of such small parts is precisely MATH again, this splitting procedure shows that MATH is isomorphic to MATH. In order to lift the embedding, therefore, my strategy will first be to find a generic for MATH. To do so, I will employ a factor technique of NAME 's from the strong cardinal context. Specifically, let MATH, the seed hull of MATH in MATH. This is an elementary substructure of MATH. Let MATH be the NAME collapse, and factor the embedding to produce MATH and MATH, where MATH, MATH and MATH. MATH . Because there are only MATH many functions MATH in MATH, it follows that MATH, and hence also MATH, have size MATH. Let MATH, where MATH. Let MATH so that MATH. Below MATH, the forcing MATH factors as MATH, where MATH is MATH-closed in MATH. I claim that MATH in MATH. To see why, suppose that MATH is in MATH. Each MATH, being in MATH, must have the form MATH for some function MATH. Since MATH, the sequence MATH is also in MATH, so the sequence MATH is in MATH. Thus, by simply evaluating these functions at MATH, the sequence MATH is also in MATH, and I have established my claim. Continuing now with the main line of argument, since MATH, we may line up the MATH-many dense subsets of MATH in MATH into a MATH-sequence in MATH and diagonalize against these dense sets to produce in MATH a MATH-generic MATH. Specifically, we recursively construct a descending sequence of conditions which gradually meets more and more of the dense sets; at limit stages, the sequence constructed so far is in MATH, since MATH, and there is a condition below it by the closure of the forcing MATH in MATH. So the constuction continues up to MATH. If MATH is the filter generated by this descending sequence, then, it is constructed in MATH but it meets all the required dense sets to be MATH-generic for MATH. Since MATH is MATH-generic it is also MATH-generic, and so the function MATH is MATH-generic for MATH. Thus, MATH lifts to MATH with MATH. Moving over to MATH, now, I will argue that the filter generated by MATH is MATH-generic for MATH. Every element of MATH has the form MATH for some function MATH in MATH and some MATH in MATH. Thus, since MATH, every element of MATH has the form MATH for some function MATH in MATH. And we may assume that MATH, since this is enough to ensure that MATH is in MATH. Thus, if MATH is a dense open subset of MATH in MATH, then MATH must be MATH for some such MATH. We may assume without loss of generality that MATH is a dense open subset of MATH for every MATH in MATH. But MATH is actually MATH-closed in MATH, so we may intersect all these MATH to obtain a single dense set MATH in MATH which is contained in MATH for every MATH. It follows that MATH is contained in MATH. Also, since MATH meets MATH by genericity, it follows that MATH meets MATH. Thus, MATH meets MATH, as desired. Since MATH is MATH-generic, it is also MATH-generic, and so MATH is MATH-generic for MATH. Thus, we may lift the embedding MATH to MATH where MATH, resulting in the following diagram. MATH . In particular, MATH, as desired. As I promised, let me now in conclusion treat the general case, in which MATH is possibly less than MATH. In this case, I simply use instead the condition MATH in the place of the earlier one, where MATH is the next inaccessible beyond MATH and MATH in MATH. This condition ensures MATH while still ensuring the closure of MATH. And since this was all that was required of MATH in the previous argument, the theorem is proved. |
math/9909029 | NAME has pointed out REF that if an unfoldable cardinal MATH is indestructible by MATH then it is indestructible by MATH for any MATH. So in order to prove the theorem it suffices merely to obtain a model in which the unfoldability of MATH is indestructible by MATH. Suppose that MATH is unfoldable in MATH. Let MATH be a MATH-generic fast function. The results of the previous section establish that MATH is unfoldable in MATH, that every canonical MATH-unfoldability embedding in MATH lifts to one in MATH, and furthermore that we have complete freedom to specify the value of MATH for these lifted embeddings. Define now in MATH the reverse NAME MATH-iteration MATH which adds a NAME subset to MATH at stage MATH for every MATH, with trivial forcing at all other stages, and suppose that MATH is MATH-generic for MATH. Let MATH be a NAME subset of MATH added generically over MATH. The desired model will be MATH. Fix any MATH and any canonical MATH-unfoldability embedding MATH in the ground model MATH. I aim to lift this embedding to an embedding MATH in the forcing extension. As an initial step, REF allows us to lift this embedding to MATH with MATH. The forcing MATH factors as MATH and so we may use MATH as the generic up to stage MATH. The tail forcing MATH is MATH-closed in MATH, since the next nontrivial stage of forcing is the next element of MATH, and this is neccessarily beyond MATH. As in the proof of REF , we factor: let MATH and MATH, leading to the following diagram, where MATH and MATH. MATH . Let me argue, in some detail, that MATH in MATH. Since every set in MATH, a model of MATH, can be enumerated, a sequence of sets is equivalent in the model to a sequence of ordinals. So it suffices for me to show that MATH, where MATH. Indeed, I will show that MATH. Suppose that MATH. Since MATH, it follows that actually MATH is in MATH, since the forcing to add the tail MATH is MATH-closed. Thus, MATH has a MATH-name MATH in MATH. This name gives rise to a sequence MATH of MATH-names, where MATH is a nice MATH-name for the MATH ordinal in the sequence named by MATH. Since the forcing MATH has size less than MATH, a nice MATH-name for a MATH-sequence or ordinals can itself be coded with a sequence of ordinals of length less than MATH. Thus, since MATH in MATH, each name MATH is in MATH. Thus also the sequence MATH of names is also in MATH. And so, interpreting those names by the generic MATH, we conclude that MATH itself is in MATH, as I had claimed. A similar argument establishes that MATH in MATH and MATH in MATH. For these claims, it once again suffices to consider only MATH in the appropriate model; and once again, a MATH sequence of ordinals in MATH is actually in MATH, and so the earlier argument shows that a name for a sequence of ordinals transforms to a sequence of names for ordinals in MATH. One then evaluates these names at the relevent generic object to obtain the original sequence in the desired model. Returning to the main line of argument, now, I may factor MATH as MATH. Using MATH as the generic up to stage MATH, it remains to construct a tail generic MATH. This I will do by the same diagonalization technique as in REF . The first step is to observe that since MATH has size MATH, so also does MATH. Thus, in MATH I may line up the dense subsets of MATH in MATH into a MATH-sequence. I now construct recursively a descending MATH-sequence of conditions which meet these dense sets one-by-one. At limit stages, the claim of the previous paragraph ensures that the construction up to that stage exists in MATH, and since MATH is MATH-closed in that model, there is a condition below it and the construction proceeds up to MATH. The filter MATH generated by this descending sequence of conditions exists in MATH, but because it meets the relevent dense sets, it is MATH-generic for MATH. The embedding MATH therefore lifts to MATH with MATH. Since the critical point of MATH is bigger than MATH, the embedding MATH lifts trivially to MATH. To lift MATH through the rest of the MATH forcing, I will employ an argument similar the previous case to show that the filter MATH generated by MATH is MATH-generic for MATH. Namely, any dense set MATH has the form MATH for some MATH with MATH and some MATH. I may assume without loss of generality that MATH is dense in MATH for every MATH. Let MATH be the intersection of all MATH. Thus, MATH is contained in MATH. But by the genericity of MATH, we know MATH meets MATH, and so MATH meets MATH. Thus, MATH meets MATH, as desired. Putting everything together, I conclude that MATH is MATH-generic for MATH. So we can lift MATH to MATH with MATH, resulting in the following diagram: MATH . It remains to lift through the MATH forcing. Here, we observe that MATH, being a bounded subset of MATH in MATH, is a condition in MATH. The forcing MATH is MATH-closed, so we may use the same factor argument to diagonalize to produce a MATH-generic MATH below the condition MATH. Again, MATH will generate a MATH-generic filter MATH for MATH. So we may lift the embedding to MATH, as desired. MATH . At this point, I have established that any canonical MATH-unfoldability embedding MATH in the ground model MATH can be lifted to a MATH-unfoldability embedding MATH in the forcing extension. It follows that MATH is unfoldable in MATH, because if MATH is a structure of size MATH in MATH, then it has a name MATH of size MATH in MATH, and for any MATH this name can be placed into such a MATH for which I will have lifted the embedding. The structure MATH therefore exists in MATH, and the restriction MATH provides an unfoldability witness embedding for MATH. So MATH is unfoldable in MATH. The final step is to realize that since adding two NAME subsets to MATH is the same as adding one, if we force over MATH to add another NAME subset MATH, then we may regard MATH as a single NAME subset of MATH and employ the previous argument to conclude that MATH is unfoldable in MATH. So the unfoldability of MATH is indestructible in MATH by MATH, as desired. |
math/9909029 | This is a simple, general observation. If MATH is strongly unfoldable in MATH, then for every MATH there is a canonical MATH-unfoldability embedding MATH with MATH. In the case of REF , this embedding lifts to MATH. Since MATH, it follows that MATH, and so the lifted embeddings witness the strong unfoldability of MATH in MATH. Similarly, in the case of the Main REF , the embedding lifts to MATH, and since, once again, both MATH and the generics MATH are in MATH, the lifted embedding remains sufficiently strong. The argument involving MATH then shows that further forcing with MATH must preserve the strong unfoldability of MATH. |
math/9909029 | Use the model MATH of the Main Theorem. I have proved already that if MATH is strongly unfoldable in MATH, then it remains so in MATH. It follows by the argument involving MATH at the conclusion of the proof of the Main Theorem, that the strong unfoldability of MATH is indestructible by further forcing with MATH. Now suppose that MATH is MATH-generic for further forcing by MATH for some cardinal MATH, and, towards a contradiction, that MATH remains strongly unfoldable in MATH. Suppose that MATH is a nice model of MATH in MATH of size MATH. It follows that MATH is a nice model of MATH in MATH. Suppose that MATH is a strong MATH-unfoldability embedding. NAME, MATH has the form MATH. I claim that MATH. First, I observe that MATH, and so since the tail forcing is closed, MATH. Thus, if MATH and MATH, then every initial segment of MATH is in MATH, and hence in MATH. Since the iteration MATH admits a gap below MATH - that is, it factors as MATH where MATH and MATH is MATH-closed for some MATH - it follows by the NAME Lemma of CITE that it cannot add any new subsets of MATH all of whose initial segments are in the ground model. So MATH must be in MATH. It follows from this, in particular, that none of the sets in MATH are in MATH. Furthermore, since the forcing MATH has size MATH, it can add at most MATH many of the sets in MATH (and these are added at stage MATH if at all). Since the rest of the forcing MATH does not add any additional subsets to MATH, most of the sets in MATH are not in MATH. This contradicts the fact that the embedding was supposed to be MATH-strongly unfoldable. |
math/9909029 | Simply force to MATH, and then force with MATH. The previous arguments establish that this forcing will preserve the unfoldability but not the strong unfoldability of MATH. |
math/9909029 | Suppose that MATH is MATH-generic for the (possibly proper class) reverse NAME iteration MATH that forces with MATH at every inaccessible stage MATH, and trivial forcing at all other stages. It has been argued elsewhere that MATH is a model of ZFC. I will show that every strongly unfoldable cardinal MATH from MATH is preserved to MATH. In order to do so, I will lift the canonical ground model strong unfoldability embeddings MATH to strong unfoldability embeddings MATH in the extension, where MATH is any NAME subset of MATH in MATH. This suffices because any set MATH in MATH is actually in MATH for some such MATH, and therefore in MATH for a suitably large choice of MATH. Choose any non-cardinal MATH and any canonical strong unfoldability embedding MATH in the ground model, where MATH is a nice model of MATH, MATH, MATH and MATH where MATH. Consider the forcing MATH in MATH. Since MATH, the iteration up to stage MATH in MATH and MATH is the same, and I may factor MATH as MATH, where MATH is the forcing beginning from stage MATH. And since MATH is MATH-generic for MATH, it is also MATH-generic. Thus, in order to lift the embedding, I must merely construct in MATH a MATH-generic for MATH. Since MATH is not a cardinal, the first nontrivial stage of forcing in MATH occurs at the next innaccessible of MATH beyond MATH, and this is larger than MATH. Thus, MATH is MATH-closed in MATH. Let MATH. My earlier arguments establish that MATH and MATH in MATH. Furthermore, since MATH has size MATH, there are only MATH many functions MATH to represent elements of MATH, and so MATH has size MATH. Now let MATH be the set of interpretations of MATH-names in MATH by the generic object MATH. (Note: since MATH may not be MATH-generic, it may be that MATH has more ordinals than MATH.) The set MATH also has size MATH. I claim that MATH. To see this, simply verify the NAME criterion: if MATH for some MATH, then the boolean value MATH, which is in MATH, is non-zero (and met by MATH). Thus, there is some name MATH with MATH. Since MATH meets MATH, it follows that MATH, and the criterion is verified. Next, I claim that MATH in MATH. Since any set can be enumerated, it suffices to show that MATH in MATH. So suppose that MATH is a MATH-sequence of ordinals in MATH, with MATH. By the closure of the tail forcing, we know that actually MATH, and so MATH has a MATH-name MATH. This name gives rise in MATH to a sequence MATH of nice MATH-names for ordinals. Since such names can be coded with a MATH-sequence of ordinals, they are all in MATH. Consequently, the whole MATH-sequence of these names must also be in MATH. Thus, the sequence of interpretations of these names by the generic MATH, namely the sequence MATH, must be in MATH, as desired. Enumerate in MATH the MATH many dense subsets of MATH in MATH, and construct a descending MATH-sequence of conditions that meet these dense sets one-by-one. At limit stages, the sequence constructed so far lies in MATH because MATH, and since the forcing MATH is MATH-closed, the construction may continue up to MATH. Let MATH be the filter generated by the resulting descending sequence. By construction, MATH is MATH-generic. I claim that MATH is neccessarily MATH-generic. To see this, suppose MATH is an open dense subset of MATH in MATH. The set MATH must have the form MATH for some sequence MATH of names and some other parameter MATH. Let MATH be the intersection of all MATH which are open dense subsets of MATH for any MATH. Thus MATH and since MATH is MATH-closed in MATH, the set MATH remains dense. Furthermore, since by intersecting over all MATH we have eliminated the need for the parameter MATH, it follows that MATH. Consequently, since MATH is MATH-generic, it meets MATH, and hence also MATH, as desired. I may therefore lift the embedding to MATH where MATH. Suppose now that MATH is one of the generic NAME sets added at stage MATH by MATH. By rearranging the sets in MATH if neccessary, I may assume that MATH is the first set added at stage MATH in MATH, and hence also in MATH. In particular, MATH is a bounded subset of MATH in MATH, and therefore a condition in MATH in that model. By the same technique just used to construct MATH, I may construct a filter below the condition MATH in MATH that is generic over MATH. And once again, the argument shows that this filter will be fully MATH-generic. Thus, I may lift the embedding to MATH, where MATH is the (union of) the generic filter I just mentioned. Since MATH and MATH, it follows that MATH, so this lifted embedding is a MATH-strong unfoldability embedding. Since MATH can be taken to be arbitrarily large, it follows that MATH is strongly unfoldable in MATH. And since MATH clearly forces the failure of the GCH at every inaccessible cardinal, the theorem is proved. |
math/9909030 | Assume MATH is faithful; if MATH is a monomorphism of MATH which is not an isomorphism, and let MATH is the cokernel of MATH, then, if MATH is an isomorphism, it follows that MATH since MATH is left exact. Therefore MATH, which is a contradiction. Conversely, assume MATH is a family of generators of MATH and let MATH be a morphism in MATH such that MATH. If MATH is the canonical decomposition of MATH with MATH monomorphism and MATH epimorphism, then MATH and if MATH is the kernel of MATH, we get that MATH is an isomorphism. Since MATH is a family of generators, MATH is an isomorphism and therefore MATH, hence MATH. |
math/9909030 | CASE: The inclusion functor MATH preserves injectivity since it is right adjoint to the exact functor MATH. If MATH is MATH-injective, then any MATH-monomorphism MATH is also a MATH-monomorphism, and so splits. CASE: It follows from REF . |
math/9909030 | CASE: It suffices to notice that the inclusion functor MATH, being right adjoint to the localizing functor MATH, is left exact. CASE: Localizing the exact sequence MATH with MATH, we get that MATH that implies MATH. |
math/9909030 | Suppose the pair MATH defines MATH as a quotient category of MATH. Also, suppose MATH defines MATH as a quotient category of MATH. Denote by MATH . Then MATH, being a composition of exact fuctors, is an exact functor. Similarly, MATH, being a composition of fully faithful fuctors, is a fully faithful functor. Furthermore, given MATH and MATH, we have MATH . Hence MATH is a left adjoint functor to MATH. Thus the pair MATH defines MATH as a quotient category of MATH with respect to the localizing subcategory MATH. By construction of MATH it is easily seen that MATH and MATH. |
math/9909030 | Suppose MATH is a NAME topology on MATH; then by MATH denote the following subcategory in MATH: MATH if and only if for any MATH the kernel MATH. We claim that MATH is a localizing subcategory. Indeed, let MATH and MATH be a monomorphism, and MATH; then MATH. Suppose now MATH is an epimorphism, MATH. Since MATH is projective, there is MATH such that MATH whence MATH, and hence MATH. Let us show now that MATH is closed under extensions. To see this, consider a short exact sequence MATH with MATH. Let MATH; then MATH. Consider MATH such that MATH. As MATH, there exists MATH such that MATH. Therefore MATH. From MATH it follows that MATH. Thus MATH is a NAME subcategory. If MATH, then there is MATH such that MATH (since MATH is finitely generated). Because MATH is a NAME subcategory, it follows that any finite direct sum of objects from MATH belongs to MATH, and so MATH, hence MATH is a localizing subcategory. Conversely, assume that MATH is a localizing subcategory in MATH. Let MATH. Obviously that MATH. If MATH and MATH is a morphism, then MATH with MATH the canonical epimorphism. Since MATH, it follows that MATH. It remains to check MATH. Assume that MATH and let MATH be such that for any MATH with MATH the ideal MATH. Let us consider an exact sequence MATH . Since MATH, one has MATH. Let MATH be the canonical epimorphism, MATH for MATH. Because MATH, it follows that MATH. In particular, if MATH, then MATH, and hence we obtain then that MATH belongs to MATH. Since MATH is closed under extensions, we conclude that MATH. |
math/9909030 | By definition the funcor MATH is exact, faithful by REF , and by REF it preserves direct limits. Clearly that MATH also preserves finitely presented modules. From NAME Lemma it easily follows that MATH . Let us show that for any finitely presented MATH-modules MATH there is an isomorphism MATH . To see this, we consider the following commutative diagram with exact rows: MATH . Since MATH is projective, there exists MATH such that MATH. In turn, because MATH, it follows that MATH. Since MATH is projective, there exists MATH such that MATH. Thus there results a commutative diagram MATH where MATH, MATH. We get then that there exists MATH such that MATH. Consider now right MATH-modules MATH and MATH. Write them as direct limits MATH and MATH of finitely presented MATH-modules MATH, MATH. One has MATH that is MATH is a fully faithful functor. It thus remains to check that any functor MATH is isomorphic to MATH for some MATH. To see this, choose for MATH a projective presentation MATH with MATH, MATH some sets of indices. Because MATH is fully faithful, there exists MATH such that MATH. Define MATH by the exact sequence MATH . We result in the following commutative diagram MATH . Therefore we obtain that MATH, that is, MATH is an equivalence. |
math/9909030 | By the slight modification the proof of the first part repeats the proof of REF . Furthermore, if MATH, then MATH is a finitely generated projective generator for MATH whence it easily follows that MATH, in view of the NAME Theorem, is equivalent to the category of right MATH-modules MATH with MATH. |
math/9909030 | As above, we can construct a morphism MATH, MATH with MATH. One analogously varifies that MATH is a MATH-homomorphism. Then from definitions of the morphisms MATH and MATH it easily follows that the diagram with exact rows MATH is commutative whence one gets MATH. Let us show that MATH is a monomorphism or equivalently, MATH is a monomorphism for any MATH-torsionfree object MATH. Indeed, assume that MATH is MATH-torsionfree and let MATH be such that MATH. Then there exists an element MATH such that the restriction MATH. But this implies that MATH is contained in MATH whence MATH (see remark on p. REF), and so MATH. Because MATH is MATH-torsionfree, it follows that MATH, that is MATH. Thus MATH. Therefore, if MATH, then MATH. Indeed, we obtain then that MATH, hence MATH. Therefore MATH if and only if MATH. It remains to check that MATH. Let MATH. We suffice to show that the ideal MATH is an element of MATH. Indeed, if MATH, then there exists MATH such that MATH since MATH is projective. Further, because the sequence MATH is exact and MATH, by assumtion, belongs to MATH, it will follow then that MATH belongs to MATH. Without loss a generality we can assume that MATH is a monomorphism and identify MATH with MATH. We put MATH. Let MATH be a MATH-homomorphism representing MATH in the direct limit MATH and MATH be such that MATH. Let us consider the following commutative diagram MATH . Recall that the element MATH is represented by the composed morphism MATH (see sequence REF). In view of that MATH, we have MATH, and therefore MATH. Since both squares of the diagram are pullback, it follows that the outer square is pullback, and so there exists a morphism MATH such that MATH whence MATH. By MATH we deduce that MATH. |
math/9909030 | To begin with, we shall show that MATH is MATH-torsionfree. Let MATH be a subobject of MATH belonging to MATH, MATH and let MATH be represented by MATH in the direct limit MATH. Suppose also MATH is such that MATH; then the equality MATH implies that the image of the composed map MATH in MATH equals to zero, that is (using the properties of direct limits) there is an ideal MATH such that the restriction MATH representing MATH to MATH equals to zero, so that MATH. Then MATH, that is, MATH. But MATH is MATH-torsionfree whence MATH, hence MATH. Because this holds for every MATH such that MATH, we infer that MATH and since MATH, it follows that MATH is also an element of MATH. But in that case, MATH, being an image of zero homomorphism MATH from MATH, equals to zero. Since it holds for any MATH, we deduce that MATH. Let us prove that a module MATH is MATH-closed if and only if MATH is an isomorphism. Indeed, if MATH is MATH-closed, then, in view of the preceding Proposition, MATH, that is MATH is a monomorphism. Since MATH, there exists a morphism MATH such that MATH, that is, MATH is a direct summand of MATH and since MATH is MATH-torsionfree, we conclude that MATH that imply MATH is an isomorphism. On the other hand, if for MATH the morphism MATH is an isomorphism, then MATH, that is MATH is MATH-torsionfree. Now if we showed that every short exact sequence MATH with MATH splits, it would follow then that MATH for any MATH that implies MATH would be MATH-closed. To see this, consider a commutative diagram MATH where the bottom row is exact and MATH. We deduce that MATH is an isomorphism. Hence MATH that implies MATH is a split monomorphism. Thus to see that MATH is MATH-closed for any module MATH, it suffices to show that MATH is an isomorphism. To begin, let us prove that MATH is an isomorphism. From construction of MATH it follows that MATH. Let MATH be the canonical epimorphism. If we apply the functor MATH to the commutative diagram MATH we obtain MATH. Since MATH, it follows that MATH is an isomorphism, and so MATH is an isomorphism. Further, since MATH is a functorial morphism, the following relations MATH and MATH hold. Applying the functor MATH to REF, one gets MATH and since MATH is an isomorphism, one has MATH. Then from equalities REF it follows that MATH . Because MATH is MATH-torsionfree, according to the first part of the proof of the preceding Proposition MATH is a monomorphism, and so from REF it follows that MATH, that is MATH is an isomorphism, hence MATH is MATH-closed. In particular, if we consider an exact sequence MATH and apply the exact localizing functor MATH, one obtains MATH whence it immediately follows that MATH is a MATH-envelope of MATH. |
math/9909030 | CASE: Let us consider a short exact sequence in MATH which induces an exact sequence MATH . If MATH and MATH, then evidently MATH, so MATH is closed under extensions. Let MATH, MATH, MATH, MATH be an epimorphism. If we consider an exact sequence MATH one obtains MATH. Conversely, consider the identity morphism MATH of the object MATH. Then MATH. It remains to check that MATH. It suffices to show that any short exact sequence MATH splits. By assumtion for MATH there is MATH such that MATH. So MATH splits. In turn, if MATH is cocomplete, then the fact that MATH is closed under under taking coproducts follows from the functor MATH commutes with direct sums. CASE: MATH is trivial. MATH: Suppose that MATH, MATH is an arbitrary morphism; then MATH, by hypothesis, belongs to MATH. Substituting MATH for MATH in REF, we get that the canonical morphism MATH is an isomorphism for all MATH. The converse is proved similar to REF . MATH: From the first assertion it follows that MATH is closed under extensions. Suppose now that in exact sequence REF the object MATH and MATH, MATH. Then MATH whence it easily follows that MATH. Further, if we consider exact sequence REF with MATH, we obtain that also MATH. If MATH is a NAME category, then MATH is closed under taking coproducts, and so it is a localizing subcategory CITE. |
math/9909030 | Indeed, if MATH is a localizing subcategory such that any object MATH is MATH-closed, then for any MATH, MATH, the object MATH, and hence, if we consider exact sequence REF with MATH, we get that the homomorphism MATH is an isomorphism. Therefore, in view of REF , MATH, that is MATH. |
math/9909030 | The necessary condition is straightforward. Assume the converse. Let us consider the commutative diagram MATH where the bottom row is exact, the couple MATH is the kernel of MATH, MATH is an epimorphism, MATH for any MATH (here MATH), MATH are the canonical monomorphisms, and MATH is a unique morphism that makes the diagram commute. Let MATH and MATH a morphism such that MATH. Then MATH for any MATH. By assumption it follows that MATH for any MATH, and therefore MATH that implies MATH since MATH is an epimorphism. Thus the canonical homomorphism induced by MATH is a monomorphism. Consider now a morphism MATH. Then MATH. Let MATH be such that MATH. The morphisms MATH induce the morphism MATH such that MATH. Suppose now that MATH is an arbitrary finite subset of MATH, MATH is the canonical monomorphism MATH, the couple MATH is the kernel of MATH, MATH is a unique morphism such that MATH, MATH is a unique morphism such that MATH, and MATH is the canonical map MATH, so that MATH. As the diagram MATH is pullback and MATH, there exists a morphism MATH such that MATH and MATH. Then we have MATH. In turn, MATH whence, using the hypothesis, we conclude that MATH for all MATH. But also MATH for all MATH. Consequently, MATH. On the other hand, since MATH and the square from the diagram MATH is pullback, there exists MATH such that MATH and MATH. Then MATH, that is, (if we identify MATH with his image in MATH) MATH coincides with MATH on MATH for all finite MATH. Because MATH, we conclude that MATH. From REF we have then MATH. But since MATH and MATH is a monomorphism, it follows that MATH. Hence MATH, and therefore there is a morphism MATH such that MATH. But then MATH and since MATH is an epimorphism, it follows that MATH as was to be proved. |
math/9909030 | Denote by MATH the localizing subcategory cogenerated by MATH. We need to show that MATH. Let MATH. Consider a short exact sequence MATH . It induces an exact sequence MATH for any MATH. If MATH, then MATH and also MATH since MATH is a MATH-torsionfree object. Therefore MATH, and hence MATH that implies MATH. On the other hand, suppose MATH; then MATH since MATH is a MATH-torsionfree object. Hence we obtain that MATH and MATH that means MATH is MATH-closed. But MATH, by hypothesis, is the largest localizing subcategory such that any MATH is MATH-closed, hence MATH. |
math/9909030 | Easy. |
math/9909030 | The necessary condition is straightforward. Let MATH be an arbitrary cyclic subobject of MATH; then MATH. Indeed, there is an epimorphism MATH for some MATH and if we consider a short exact sequence MATH it will follow then MATH since the homomorphism MATH, by hypothesis, is an isomorphism. Therefore, in view of REF , MATH. Consider now a short exact sequence MATH . It induces an exact sequence MATH . But MATH (since MATH), and hence MATH. Our proof will be finished if we show that any short exact sequence MATH splits and then MATH. To see this, let us consider the following commutative diagram MATH where the lower right square is pullback, so that the morphisms MATH and MATH are epimorphisms since MATH and MATH are epimorphisms. As MATH is a projective MATH-module, it follows that MATH is a split monomorphism. Let MATH be the canonical projection onto MATH, that is MATH. By assumtion there exists a morphism MATH such that MATH. But MATH, and hence MATH. Then there exists MATH such that MATH whence MATH, that is MATH is a split monomorphism that finishes the proof. |
math/9909030 | Consider the commutative diagram MATH where the rows are exact, the couple MATH is the kernel of MATH. Because MATH, there exists a unique morphism MATH such that MATH. We claim that MATH is an epimorphism. Indeed, let us consider the folowing diagram MATH where MATH is the kernel of MATH, MATH, and the middle arrow means a unique morphism that makes the diagram commute. As the outer and right squares are pullback, it follows that also the left square is pullback. But then the epimorphism MATH implies MATH is an epimorphism. Suppose that MATH is MATH-covering and MATH a morphism with MATH. Then MATH, and so MATH, hence MATH since MATH is MATH-negligible. Thus the homomorphism MATH induced by MATH is a monomorphism. Suppose now that a morphism MATH satisfies MATH. Since MATH is the cokernel of MATH, we deduce that there exists a morphism MATH such that MATH. Since MATH is MATH-negligible, there exists MATH such that MATH. But then MATH. Conversely, suppose the sequence of Lemma is exact and MATH is an arbitrary morphism MATH. Since MATH is projective, there exists a morphism MATH such that MATH, so that MATH is the kernel of MATH. If a morphism MATH satisfies MATH, then MATH whence MATH since MATH is a monomorphism. Let now MATH. Because MATH and MATH, it follows that there exists MATH such that MATH, that is MATH, and therefore MATH. Thus the canonical homomorphism MATH is an isomorphism. From REF it follows that MATH is MATH-negligible. |
math/9909030 | Our assertion is proved by the slight modification of the proof of the preceding Lemma. Namely it is necessary to substitute MATH for MATH in the respective places and then apply REF . |
math/9909030 | CASE: From REF it follows that MATH is a faithful functor. To see that it is full, we must show that if MATH and MATH are objects in MATH and MATH is a functor, then MATH is of the form MATH for some MATH. Denote by MATH the set of all morphisms MATH and put MATH. For each MATH we let MATH denote the corresponding injection. There exists a unique morphism MATH such that MATH for each MATH, and MATH is an epimorphism since MATH is a family of generators. Similarly there exists a unique morphism MATH such that MATH for each MATH. Let MATH be the kernel of MATH. We can show that MATH, then MATH factors as MATH for some MATH and for each MATH, MATH we get MATH, and our assertion would be proved. So we need to check that MATH. For each finite subset MATH of MATH and each MATH there are canonical morphisms MATH, MATH and MATH. Let MATH be the kernel of the composed morphism MATH. Since MATH is the direct limit of the kernels MATH for all finite subsets MATH of MATH, it suffices to show that MATH. Now for each MATH, MATH we have, using the fact that MATH is a functor, that MATH since MATH. Since this holds for arbitrary MATH, it follows that MATH. CASE: Let MATH be the largest localizing subcategory in MATH for which all modules of the form MATH are MATH-closed. This subcategory exists by REF and it is cogenerated by the class of injective modules of the form MATH. Let MATH; then the respective MATH-negligible objects and MATH-covering morphisms will be referred to as negligible and covering respectively, omitting the prefix MATH. Since every module MATH is MATH-closed, there is a functor MATH such that MATH with MATH an inclusion functor. We must show that MATH is an equivalence. Since MATH is full and faithful by REF , also MATH is full and faithful, and thus it suffices to show that every MATH-closed module is isomorphic to a module of the form MATH. To see this, for each MATH-module MATH we choose an exact sequence MATH with MATH, MATH some sets of indices. Then MATH induces in a natural way a morphism MATH in MATH. To be precise, for each MATH we have a functorial isomorphism MATH and analogously, one has a functorial isomorphism MATH . The morphism MATH induces the functorial morphism MATH . According to CITE, there exists a unique morphism MATH such that MATH. Now let us define MATH by the exact sequence MATH . We now apply the functor MATH to REF. If we knew that MATH preserves colimits, we would have then the following isomorphisms MATH and obtain the following commutative diagram in MATH with exact in MATH rows, and that MATH is MATH-closed would imply MATH. To conclude the proof it thus remains to show: MATH . First we prove the exactness of MATH. The functor MATH, where MATH is the respective localizing functor, is obviously left exact, and so it suffices to prove that MATH preserves epimorphisms. This means that if MATH is an epimorphism in MATH, then the morphism MATH of MATH is covering, that is, in view of REF , that for any object MATH of MATH and MATH we have the exact sequence MATH induced by any MATH-homomorphism MATH. Since MATH is full and faithful and MATH, we deduce that there exists a morphism MATH such that MATH. Therefore we have the commutative diagram MATH with exact rows. The short exact sequence MATH induces the exact sequence MATH and so, since MATH is fully faithful, we have the exact sequence MATH . But since MATH is left exact, MATH is isomorphic to MATH, MATH is obviously isomorphic to MATH and these isomorphisms are functorial. Thus sequences REF are exact for all MATH, MATH and now it suffices to apply REF . It remains to prove that MATH preserves direct sums. Actually we prove a little more, namely that it preserves direct unions. Suppose MATH is a directed family of subobjects of MATH such that MATH with MATH some set of indices. We need to show that the canonical monomorphism MATH is covering. Let MATH be an arbitrary morphism MATH. Since MATH is a projective object of MATH, there exists a morphism MATH such that MATH where MATH. Because MATH is fully faithful, there exists a morphism MATH such that MATH. Let MATH be the fibered product associated to the scheme MATH with MATH the canonical monomorphism. Since MATH is left exact, it follows that MATH is the fibered product associated to the scheme MATH . We obtain thus that MATH is the fibered product associated to the scheme MATH that implies MATH. Since MATH and MATH is fully faithful, one obtains MATH for all MATH. Hence we get, in view of REF , that MATH is negligible. Thus the functor MATH is an equivalence. This concludes the proof of the NAME Theorem. |
math/9909030 | It suffices to note that MATH is again a family of generators for MATH (that directly follows from definition). |
math/9909030 | By REF the category MATH is equivalent to the quotient category of MATH with MATH the family of generators. Since the representable functor MATH is a finitely generated projective generator for MATH, it follows that MATH, in view of the NAME Theorem, is equivalent to the category of right MATH-modules MATH whence our assertion easily follows. |
math/9909030 | MATH: Straightforward. MATH: Because the left square of the commutative diagram MATH is pullback, it follows that MATH is a monomorphism. Hence MATH since MATH is closed under subobjects. MATH: Suppose MATH and MATH is a MATH-homomorphism such that MATH; then MATH. Because MATH, it follows that MATH since MATH. Therefore MATH and by MATH we deduce that MATH. Since MATH is MATH-closed, we get that MATH. |
math/9909030 | To begin, we shall show that MATH is localizing. Indeed, let MATH be a short exact sequence in MATH. If we apply the exact functor MATH, we shall get a short exact sequence of abelian groups MATH whence it easily follows that MATH if and only if MATH and MATH that implies MATH is a NAME subcategory. Furthermore, if we consider the map MATH with MATH and MATH a monomorpism induced by the canonical monomorphism MATH, it will be follow that MATH is closed under taking coproducts, and therefore MATH is a localizing subcategory. In the rest of the proof we shall show that MATH is a projective generator of MATH. First let us consider a short exact sequence MATH in MATH. It induces an exact sequence in MATH with MATH. If we apply the exact functor MATH, one gets the following commutative diagram of abelian groups: MATH where vertical arrows are isomorphisms. Therefore MATH is MATH-projective. It remains to check that MATH is a generator of MATH. Let MATH be an arbitrary object of MATH, MATH; then there exists a morphism MATH with MATH, MATH for all MATH. We have an exact sequence in MATH . Let MATH be an arbitrary morphism. By construction of MATH we have that MATH, and so there exists a morphism MATH such that MATH (since MATH is projective) that evidently implies MATH, that is, MATH. But then, MATH is a MATH-epimorphism. Thus any object MATH is a quotient object of MATH whence it is easily follows that MATH is a generator of MATH. Now let us show that the restriction of MATH to MATH defines an equivalence of MATH and MATH. For MATH there is an exact sequence of the form REF MATH with MATH, MATH. Hence MATH induces an isomorphism: MATH . But then we have that MATH with MATH for any MATH. Now, if we consider projective presentation REF for MATH, we obtain a commutative diagram MATH with exact rows and vertical arrows isomorphisms. Thus, MATH that implies MATH is a fully faithful functor. Finally let MATH. Consider a MATH-projective presentation of MATH where MATH. Then there exists MATH such that MATH, hence MATH, as was to be proved. |
math/9909030 | It suffices to note that isomorphisms REF induce a ring isomorphism MATH . Now our assertion follows from REF since MATH is a generator for MATH. |
math/9909030 | Denote by MATH; then MATH. From the preceding Theorem it follows that MATH is localizing and that MATH is a MATH-projective generator that implies MATH is a family of projective generators of MATH. In view of isomorphisms REF, there is an equivalence of MATH and MATH. Now our assertion follows from REF . |
math/9909030 | By REF MATH is a family of projective generators for MATH. It thus remains to show that every MATH is MATH-finitely generated. To see this, we consider an object MATH of MATH. Let MATH be a directed family of MATH-subobjects of MATH such that MATH. Because MATH are also MATH-subobjects, it follows that the MATH-direct union MATH is a subobject of MATH and the quotient object MATH belongs to MATH. Indeed, if we apply the exact localizing functor MATH commuted with direct limits to the short exact sequence MATH we shall obtain then the short exact sequence MATH . Whence MATH, that is MATH. For any MATH one has then MATH . Thus, MATH . By REF MATH. The fact that the localizing functor MATH induces an equivalence between MATH and MATH is proved similar to REF . |
math/9909030 | According to the preceding Theorem the localized object MATH is a finitely generated projective generator, and so, in view of the NAME Theorem and REF , the quotient category MATH is equivalent to the category of modules over the ring MATH. |
math/9909030 | If MATH is coexact, then any short exact sequence MATH in MATH is also exact in MATH. Suppose MATH is projective; then we have the following commutative diagram MATH with exact rows and vertical arrows being isomorphisms. Hence MATH is MATH-projective. Conversely, we need to show that any MATH-epimorphism is MATH-epimorphism. Consider exact sequence REF. By REF MATH. By assumption the morphism MATH is an epimorphism, where MATH, hence MATH is an epimorphism. Therefore MATH for every MATH that implies MATH since MATH is a family of generators. |
math/9909030 | MATH: Straightforward. MATH: Let MATH be a MATH-direct limit of MATH, MATH. Denote by MATH the subset of MATH consisting of pairs MATH with MATH and for any MATH we put MATH, MATH. By CITE MATH with MATH induced by MATH, MATH and MATH the canonical morphism. By hypothesis the inclusion functor MATH is exact and commutes with coproducts, and so we obtain MATH as was to be proved. |
math/9909030 | Suppose that MATH commutes with direct limits. Let us consider an exact sequence MATH with MATH, MATH a MATH-envelope of MATH. Applying the left exact functor MATH, we have MATH since MATH. So MATH is MATH-torsionfree. Conversely, let MATH be a direct system of objects from MATH. As above, there is an exact sequence MATH . Because the direct limit functor is exact, one gets an exact sequence MATH . Since MATH is closed under taking direct limits, it follows that MATH. Now, if we apply MATH to sequence REF, one obtains MATH since MATH. |
math/9909030 | Equivalence MATH follows from REF . MATH: By REF. Now our assertion follows from REF . MATH: Let MATH be a MATH-envelope of MATH. Then the composed map MATH with MATH the canonical morphism is, by hypothesis, an isomorphism. Hence, MATH . By REF MATH. MATH: If MATH there is an epimorphism MATH for some MATH. Since MATH, it follows that MATH. MATH: Suppose MATH, that is MATH for some MATH. Write MATH as a directed sum of MATH-finitely generated subobjects MATH. Then MATH. By assumtion there is MATH such that MATH, and hence MATH. MATH: First let us show that MATH where MATH. Indeed, suppose MATH; then MATH with MATH a MATH-envelope of MATH. By assumtion there is a finitely generated subobject MATH of MATH such that MATH. Then there is MATH such that MATH. One has MATH . So MATH that implies MATH. Further, the isomorphism MATH is functorial in MATH, and so MATH. MATH: The direct limit MATH may be described as a quotient object of a coproduct MATH. To be precise, let MATH be the subset of MATH consisting of pairs MATH with MATH and for each MATH we put MATH where MATH is the canonical monomorphism for MATH, MATH is the canonical morphism for MATH. Then MATH, where MATH runs over all finite subsets of MATH. By assumtion both MATH and MATH are MATH-closed, and therefore MATH, as a quotient object of MATH-closed objects, is MATH-torsionfree. MATH: Suppose that MATH is MATH-closed. Write MATH as a direct union of MATH-closed subobjects. Then MATH since MATH. Thus MATH. In turn, let MATH. Write MATH as a directed sum of MATH. Then MATH whence MATH for some MATH. |
math/9909030 | It suffices to note that for each MATH the representable functor MATH commutes with direct limits (see NAME REF) and there is a presentation MATH of MATH by objects from MATH; and then our proof literally repeats REF . In turn, if MATH there is an epimorphism MATH with MATH; then MATH. According to CITE we can choose MATH such that MATH and MATH. Hence MATH. |
math/9909030 | Suppose MATH is an equivalence. Then every MATH finitely generated and projective in MATH. By REF MATH is coexact and by REF it is of finite type. Conversely, let MATH be of finite type and coexact. By REF every MATH is projective in MATH and by REF it is finitely generated in MATH. Now our assertion follows from REF . |
math/9909030 | Let MATH and MATH; then there is a morphism MATH such that MATH, where MATH is the canonical epimorphism and MATH. Consider in MATH the commutative diagram MATH in which MATH is an epimorphism. Because MATH, there is a finitely generated subobject MATH of MATH such that MATH. There exists an epimorphism MATH with MATH. So MATH is an epimorphism. It is easily seen that MATH, hence MATH. |
math/9909030 | Let MATH be a contravariant left exact functor, MATH and MATH such that MATH. Here MATH denotes a unique morphism such that MATH. Then there exists an epimorphism MATH in MATH such that MATH. But MATH is a monomorphism of abelian groups and thus MATH. So MATH is MATH-torsionfree. Now let MATH and MATH a morphism in MATH. There exists an epimorphism MATH such that MATH. By REF MATH. One has a morphism MATH. Since MATH, there exists an epimorphism MATH and thus one gets an exact sequence MATH . We have the following commutative diagram of abelian groups: MATH . So, MATH . Then there exists an element MATH such that MATH. Thus MATH. Since MATH and MATH is MATH-torsionfree, it follows that MATH. Thus MATH is a morphism prolonging MATH. The uniqueness of MATH follows from the fact that MATH is MATH-torsionfree. So MATH is MATH-closed. |
math/9909030 | Our assertion would be proved, if we showed that MATH. Because for every MATH the functor MATH is left exact, in view of REF , it follows that MATH is MATH-closed, and so MATH. Conversely, let MATH. We can choose a projective presentation of MATH . By REF MATH is of finite type, and so any coproduct of MATH-closed objects is a MATH-closed object. Therefore there exists a morphism MATH such that MATH. Furthermore, MATH is an epimorphism in MATH since MATH. Thus, without loss of generality, we can assume that for every MATH there is an exact sequence MATH where MATH is a MATH-epimorphism. According to CITE MATH is a direct limit MATH of epimorphisms MATH in MATH, and so MATH. By REF every MATH, hence MATH. |
math/9909030 | MATH: It follows from REF . MATH: By REF MATH. Let us show that for any MATH the object MATH is coherent. Let MATH be MATH-finitely generated subobject of MATH. From CITE it follows that there is a MATH-finitely generated subobject MATH of MATH such that MATH. Since MATH is coherent, the object MATH. By REF MATH. MATH: Easy. MATH: MATH is of finite type by REF . MATH: It follows from REF . MATH: Since any finitely generated subobject of a coherent object is coherent, our assertion follows from REF . MATH: Straightforward. MATH: It follows from CITE. |
math/9909030 | CITE has shown that MATH is MATH-injective if and only if it is right exact. Therefore the functor MATH is coh-injective. Let MATH be coh-injective. Define MATH by putting MATH. Now our assertion is proved similar to CITE. |
math/9909030 | Write MATH as a direct limit of finitely presented objects MATH. For every MATH consider the following commutative diagram MATH in which the right square is pullback and MATH the canonical morphism. Since for MATH the relations MATH hold, there exists a unique MATH such that MATH. Clearly that the system MATH is direct and MATH. By hypothesis there exists a morphism MATH such that MATH, and hence there exists MATH such that MATH, that is .each sequence MATH splits. Conversely, if each MATH splits, then the sequence MATH is exact. The fact that MATH, MATH, commutes with direct limits implies REF . |
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