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Lemma 1.5. Let \( \varphi : A \rightarrow B \) be a morphism in an additive category. Then \( \varphi \) is a monomorphism if and only if \( 0 \rightarrow A \) is its kernel, and \( \varphi \) is an epimorphism if and only if \( B \rightarrow 0 \) is its cokernel. | ## Proof. Let's do kernels this time.\n\nFirst assume \( \varphi : A \rightarrow B \) is a monomorphism. If \( \zeta : Z \rightarrow A \) is any morphism such that the composition \( Z \rightarrow A \rightarrow B \) is 0, then \( \zeta \) is 0 by Lemma 1.3, and in particular \( \zeta \) factors (uniquely) through \( 0 ... | No |
Lemma 1.8. In an abelian category \( \mathrm{A} \), every kernel is the kernel of its cokernel; every cokernel is the cokernel of its kernel. | Proof. I will prove the second half and leave the first half to the reader (Exercise 1.9).\n\nLet \( \varphi : A \rightarrow B \) be the cokernel of some morphism \( Z \rightarrow A \) ; since \( \mathrm{A} \) is abelian, \( \varphi \) has a kernel \( \iota : K \rightarrow A \) . The composition \( Z \rightarrow A \rig... | No |
Lemma 1.9. Let \( \varphi : A \rightarrow B \) be a morphism in an abelian category \( \mathrm{A} \), and assume that \( \varphi \) is both a monomorphism and an epimorphism. Then \( \varphi \) is an isomorphism. | Proof. By Lemma 1.5 the kernel of \( \varphi \) is \( 0 \rightarrow A \), since \( \varphi \) is a monomorphism. Similarly, \( B \rightarrow 0 \) is a cokernel of \( \varphi \) . Further, \( \varphi \) is the cokernel of \( 0 \rightarrow A \) and the kernel of \( B \rightarrow 0 \), by Lemma 1.8 .\n\nNow consider the i... | Yes |
For instance, fibered products (or 'pull-backs') exist in any abelian category, just as in \( R \) -Mod (cf. Exercise III 6.10). Consider a diagram\nin an abelian category. The fibered product of \( A \) and \( B \) over \( C \) is an object \( A{ \times }_{C}B \) with morphisms to \( A \) and \( B \), completing the c... | The fibered product may be constructed in this context, just as in the particular case of \( R \) -Mod, as the kernel of the difference of the two morphisms\nwhere \( {\rho }_{A},{\rho }_{B} \) are the morphisms making \( A \times B \) a product. | No |
Lemma 1.14. Let \( \varphi : A \rightarrow B \) be a morphism in an abelian category, and let \( \iota : K \rightarrow B \) be the kernel of the cokernel of \( \varphi \) . Then\n\n- \( \iota \) is a monomorphism;\n\n- \( \varphi \) factors through \( \iota \) ; and\n\n- \( \iota \) is initial with these properties. | By Lemma 1.4, \( \iota \) is a monomorphism. It is clear that \( \varphi \) factors through \( \iota \) : the composition \( A \rightarrow B \rightarrow \operatorname{coker}\varphi \) is the zero-morphism, so there is a naturally induced \( A \rightarrow K \) by the universal property of kernels. The more interesting p... | Yes |
Lemma 1.16. Let \( \varphi : A \rightarrow B \) be a morphism in an abelian category, and let \( \operatorname{im}\varphi : K \rightarrow B,\operatorname{coim}\varphi : A \rightarrow C \) be its image and coimage, respectively. Then the induced morphisms \( A \rightarrow K \) and \( C \rightarrow B \) are, respectively... | Proof. As usual, I will prove half of the statement and leave the other half to the reader (Exercise 1.20)\n\nTo verify that \( \bar{\varphi } : A \rightarrow K \) is an epimorphism, consider its image \( {K}^{\prime } \rightarrow K \) :\n\n\[ \n{K}^{\prime } \succ \xrightarrow[]{\text{ im }\bar{\varphi } = \ker \opera... | No |
For a slightly more interesting example, consider a diagram\n\n\n\nand the associated sequence\n\n\\[ \nD\\xrightarrow[]{\\left( {\\psi }^{\\prime },{\\varphi }^{\\prime }\\right) }A \\oplus B\\xrightarrow[]{\\left( ... | Indeed, the first assertion is trivial; | No |
Lemma 2.3. Let\n\n\n\nbe a fibered diagram in an abelian category, and assume \( \varphi \) is an epimorphism. Then \( {\varphi }^{\prime } \) is also an epimorphism. | Proof. First, observe that if \( \varphi : A \rightarrow C \) is an epimorphism, so is the map \( A \oplus B \rightarrow \) \( C \) considered in Example 2.2 Since epimorphisms are cokernels in an abelian category and cokernels are cokernels of their kernels (Lemma 1.8), we see that \( A \oplus B \rightarrow C \) is th... | Yes |
Lemma 2.5. \( z \sim 0 \Leftrightarrow z = 0 \). Further, a morphism \( \varphi : A \rightarrow B \) in \( \mathrm{A} \) is 0 if and only if \( \widehat{\varphi }\left( z\right) = 0 \) for all \( z \in \widehat{A} \). | Proof. According to the definition given above, \( z : Z \rightarrow A \) is equivalent to 0 if and only if there is an epimorphism \( W \rightarrow Z \) making the following diagram commute:\n\n\n\nSince \( W \right... | Yes |
Lemma 2.6. Let \( \varphi : A \rightarrow B \) be a morphism in A. Then\n\n- \( \varphi \) is a monomorphism if and only if \( \widehat{\varphi } \) is injective;\n\n- \( \varphi \) is an epimorphism if and only if \( \widehat{\varphi } \) is surjective. | Proof. As usual, I will propose a division of labor: the reader will prove the first statement (Exercise 2.6), and I will prove the second.\n\nAssume \( \varphi \) is an epimorphism, and let \( z : Z \rightarrow B \) represent an arbitrary ’element’ of \( \widehat{B} \) . Consider the fiber product:\n\n![23387543-548b-... | No |
Lemma 2.7. With notation as above, let \( \varphi : A \rightarrow B \) be a morphism in a small abelian category \( \mathrm{A} \), and let \( \widehat{\varphi } : \widehat{A} \rightarrow \widehat{B} \) be the corresponding function of pointed sets. Let \( \ker \varphi : K \rightarrow A \), resp., \( \operatorname{im}\v... | Proof. These statements are very close to the universal properties satisfied by kernel and image.\n\nThe reader will verify the statement about the kernel (Exercise 2.7). For the image, recall that we have a decomposition of \( \varphi \) ,\n\n\[ \varphi : A \rightarrow I\overset{\operatorname{im}\varphi }{ \mapsto }B ... | No |
Proposition 2.8. Let \( \mathrm{A} \) be a small abelian category. Then a sequence\n\n\[ A\overset{\varphi }{ \rightarrow }B\overset{\psi }{ \rightarrow }C \]\n\nin \( \mathrm{A} \) is exact if and only if the corresponding sequence\n\n\[ \widehat{A}\overset{\widehat{\varphi }}{ \rightarrow }\widehat{B}\overset{\wideha... | Proof. This now follows immediately from Lemma 2.7 exactness in A means that im \( \varphi = \ker \psi \), and exactness in \( {\operatorname{Set}}^{ * } \) means that the image of \( \widehat{\varphi } \) equals \( {\widehat{\psi }}^{-1}\left( 0\right) \) . By Lemma 2.7, these conditions are equivalent. | Yes |
Theorem 2.9 (Freyd-Mitchell theorem). Let \( \\mathrm{A} \) be a small abelian category. Then there is a fully faithful, exact functor \( \\mathrm{A} \\rightarrow R \) -Mod for a suitable ring \( R \) . | This functor is fully faithful: this means that one can in fact construct morphisms in an arbitrary (small) abelian category by working with elements. Indeed, this amounts to constructing the appropriate morphisms in the ambient category \( R \) -Mod, and fullness guarantees that these morphisms ’already’ exist in A. | No |
Lemma 3.3. \( \mathrm{C}\left( \mathrm{A}\right) \) is an abelian category. | I will leave to the reader the careful verification of this fact (Exercise 3.3). In broad terms, morphisms between two given complexes form an abelian group, essentially because if \( {\alpha }^{i} \) and \( {\beta }^{i} : {M}^{i} \rightarrow {N}^{i} \) are both collections of morphisms making the appropriate diagram c... | No |
For every integer \( i \), the assignment\n\n\[ \n{H}^{i} : {M}^{ \bullet } \mapsto {H}^{i}\left( {M}^{ \bullet }\right) \n\]\n\ndefines an additive covariant functor \( \mathrm{C}\left( \mathrm{A}\right) \rightarrow \mathrm{A} \) . | Of course, the statement means that each \( {H}^{i} \) induces in a natural (and functorial) way homomorphisms of abelian groups\n\n\[ \n{\operatorname{Hom}}_{\mathrm{C}\left( \mathrm{A}\right) }\left( {{M}^{ \bullet },{N}^{ \bullet }}\right) \rightarrow {\operatorname{Hom}}_{A}\left( {{H}^{i}\left( {M}^{ \bullet }\rig... | Yes |
Theorem 3.5 (Long exact cohomology sequence). The sequence determined as above by a short exact sequence of complexes is an exact sequence. | Proof. The proof is a diagram chase, which everyone should perform once by oneself in his or her lifetime. So it is mostly left as an exercise for the reader (Exercise 3.9). But I will stress the extent to which \( {H}^{i} \) is exact ’on the nose’: part of the claim in this theorem is that if\n\n\[ 0 \rightarrow {L}^{... | No |
Proposition 4.1. There is an exact triangle\n\n\n\nwhere the connecting morphism \( \delta \) is the morphism induced by \( {\alpha }^{ \bullet } \) in cohomology. | Proof. The existence of the triangle is a direct consequence of Theorem 3.5 all we have to check is that the connecting morphism indeed agrees with the morphism\n\ninduced by \( {\alpha }^{ \bullet } \) . Chasing the diagram\n\n of an object \( A \) of an abelian category A, as in Definition 3.2 and with \( {M}^{i} = 0 \) for \( i > 0 \), is the same as the datum of a quasi-isomorphism\n\n\[ \n{M}^{ \bullet }\xrightarrow[]{\text{ q-iso. }}\iota \left( A\right) \n\]\n\nwhere \( \iota \) places \( ... | Thus, quasi-isomorphisms may be viewed as generalizations of more simpleminded resolutions. Also note that the mapping cone of a resolution as in Example 4.4 is obtained (as the reader should check) by shifting the complex 'one step to the left’ and completing it with \( A \), obtaining the exact complex:\n\n\[ \n\cdot... | No |
Let \( {M}^{ \bullet } \) be an exact complex in \( \mathrm{C}\left( \mathrm{A}\right) \). Then the complex \( \mathcal{F}\left( {M}^{ \bullet }\right) \), obtained by applying \( \mathcal{F} \) to the objects and morphisms of \( {M}^{ \bullet } \), is a zero-object in \( \mathrm{D} \). | To verify the first claim, note that since \( {M}^{ \bullet } \) is exact, the zero-morphism: \( {M}^{ \bullet } \rightarrow \) \( {M}^{ \bullet } \) is a quasi-isomorphism; hence it is mapped to an invertible morphism by \( \mathcal{F} \) :\n\n\[ \mathcal{F}\left( {M}^{ \bullet }\right) \underset{{\operatorname{id}}_{... | No |
Proposition 4.10. If \( {\alpha }^{ \bullet },{\beta }^{ \bullet } : {L}^{ \bullet } \rightarrow {M}^{ \bullet } \) are homotopic morphisms of complexes, then \( {\alpha }^{ \bullet },{\beta }^{ \bullet } \) induce the same morphisms on cohomology: \( {H}^{ \bullet }\left( {L}^{ \bullet }\right) \rightarrow {H}^{ \bull... | Proof. Let \( \bar{\ell } \in {H}^{i}\left( {L}^{ \bullet }\right) \) . Then \( \bar{\ell } \) is represented by an element \( \ell \in \ker \left( {d}_{{L}^{ \bullet }}^{i}\right) \), and its images in \( {H}^{i}\left( {M}^{ \bullet }\right) \) under the morphisms induced by \( {\alpha }^{ \bullet },{\beta }^{ \bullet... | Yes |
Corollary 4.11. Homotopy equivalent complexes have isomorphic cohomology. | Proof. Indeed, morphisms \( {\alpha }^{ \bullet } : {L}^{ \bullet } \rightarrow {M}^{ \bullet },{\beta }^{ \bullet } : {M}^{ \bullet } \rightarrow {L}^{ \bullet } \) such that \( {\beta }^{ \bullet } \circ {\alpha }^{ \bullet } \) and \( {\alpha }^{ \bullet } \circ {\beta }^{ \bullet } \) are both homotopic to the iden... | Yes |
Lemma 4.13. With \( \mathcal{F} \) as above, if \( {\alpha }^{ \bullet } \sim {\beta }^{ \bullet } \) in \( \mathrm{C}\left( \mathrm{A}\right) \), then \( \mathrm{C}\left( \mathcal{F}\right) \left( {\alpha }^{ \bullet }\right) \sim \mathrm{C}\left( \mathcal{F}\right) \left( {\beta }^{ \bullet }\right) \) in \( \mathrm{... | Proof. The second assertion follows from the first. The first is an immediate consequence of the fact that \( \mathcal{F} \) is additive. Indeed, if \( h \) is a homotopy between \( {\alpha }^{ \bullet },{\beta }^{ \bullet } : {L}^{ \bullet } \rightarrow {M}^{ \bullet } \), then\n\n\[{\beta }^{i} - {\alpha }^{i} = {d}_... | Yes |
Theorem 4.14. Let \( \mathcal{F} : \mathrm{A} \rightarrow \mathrm{B} \) be an additive functor between two abelian categories. If \( {L}^{ \bullet },{M}^{ \bullet } \) are homotopy equivalent complexes in \( \mathrm{C}\left( \mathrm{A}\right) \), then the cohomology complexes\n\n\[ \n{H}^{ \bullet }\left( {\mathrm{C}\l... | The proof of this statement is essentially immediate after all our preparatory work, so it is left to the reader (Exercise 4.16). | No |
Lemma 5.3. Let \( \mathrm{A} \) be an abelian category. Then the homotopic category \( \mathrm{K}\left( \mathrm{A}\right) \) of complexes is an additive category. | ## Proof. Exercise 5.1. | No |
Proposition 5.4. Let \( \mathcal{F} : \mathrm{C}\left( \mathrm{A}\right) \rightarrow \mathrm{D} \) be an additive functor such that \( \mathcal{F}\left( {\rho }^{ \bullet }\right) \) is an isomorphism in \( \mathrm{D} \) for all quasi-isomorphisms \( {\rho }^{ \bullet } \) in \( \mathrm{C}\left( \mathrm{A}\right) \) . ... | ## Proof. Exercise 5.2 | No |
Lemma 5.11. Let \( {P}^{ \bullet } \) be a complex of projective objects of an abelian category A such that \( {P}^{i} = 0 \) for \( i > 0 \), and let \( {L}^{ \bullet } \) be a complex in \( \mathrm{C}\left( \mathrm{A}\right) \) such that \( {H}^{i}\left( {L}^{ \bullet }\right) = 0 \) for \( i < 0 \) .\n\nLet \( {\alp... | Proof. We have to construct morphisms \( {h}^{i} : {P}^{i} \rightarrow {L}^{i - 1} \) :\n\n\n\n\nsuch that\n\n(*)\n\n\[{\alpha }^{i} = {d}_{{L}^{ \bullet }}^{i - 1} \circ {h}^{i} + {h}^{i + 1} \circ {d}_{{P}^{ \bulle... | Yes |
Corollary 5.12. Let \( {P}^{ \bullet } \) be a bounded-above cochain complex of projectives of an abelian category \( \mathrm{A} \), and let \( {L}^{ \bullet } \) be an exact complex in \( \mathrm{C}\left( \mathrm{A}\right) \). Then every morphism of complexes \( {P}^{ \bullet } \rightarrow {L}^{ \bullet } \) is homoto... | This follows immediately from (a harmless shift of) Lemma 5.11 since every morphism to an exact complex has no choice but to induce the zero-morphism in cohomology. | Yes |
Corollary 5.13. Let \( {P}^{ \bullet } \) (resp., \( {Q}^{ \bullet } \) ) be a bounded-above exact complex of projec-tives (resp., a bounded-below exact complex of injectives). Then \( {P}^{ \bullet } \) (resp., \( {Q}^{ \bullet } \) ) is homotopy equivalent to the zero-complex. | ## Proof. Exercise 5.12. | No |
Lemma 5.14. Let \( \mathrm{A} \) be an abelian category, and let \( {\rho }^{ \bullet } : {L}^{ \bullet } \rightarrow {M}^{ \bullet } \) be a quasi-isomorphism in \( \mathrm{C}\left( \mathrm{A}\right) \) . Let \( {P}^{ \bullet } \) be a bounded-above complex of projectives, and let \( {\alpha }^{ \bullet } : {P}^{ \bul... | Proof. Let \( {h}^{i} : {P}^{i} \rightarrow {M}^{i - 1} \) define a homotopy between \( {\rho }^{ \bullet } \circ {\alpha }^{ \bullet } \) and 0, so that \( - {\rho }^{i} \circ {\alpha }^{i} = {d}_{{M}^{ \bullet }}^{i - 1} \circ {h}^{i} + {h}^{i + 1} \circ {d}_{{P}^{ \bullet }}^{i} \) . Consider the mapping cone \( {MC... | Yes |
To see that \( {\alpha }^{ \bullet } \) may not be zero on the nose even if \( {\rho }^{ \bullet } \circ {\alpha }^{ \bullet } = 0 \) , look back again at Example 4.6: | Here \( {\rho }^{ \bullet } \) is a quasi-isomorphism, and \( {\rho }^{ \bullet } \circ {\alpha }^{ \bullet } = 0 \) . According to Lemma 5.14, the (nonzero) morphism \( {\alpha }^{ \bullet } \) is homotopic to 0 . (Indeed, a homotopy is immediately visible. What is it?) | No |
Proposition 5.16. Let \( \mathrm{A} \) be an abelian category, and let \( {L}^{ \bullet } \) be a complex in \( \mathrm{C}\left( \mathrm{A}\right) \) . Let \( {P}^{ \bullet } \) in \( {\mathrm{C}}^{ - }\left( \mathrm{P}\right) \) be a bounded-above complex of projectives, and let \( {\alpha }^{ \bullet } : {L}^{ \bulle... | Proof. Since \( {\alpha }^{ \bullet } : {L}^{ \bullet } \rightarrow {P}^{ \bullet } \) is a quasi-isomorphism, the mapping cone \( {MC}{\left( \alpha \right) }^{ \bullet } \) of \( \alpha \) is an exact complex (Corollary 4.2). Let \( {\rho }^{ \bullet } \) be the morphism of complexes\n\n\[ \n{\rho }^{ \bullet } = \le... | Yes |
Lemma 6.3. Let \( A \) be an object of an abelian category A. Let \( {M}^{ \bullet } \) be a resolution of \( A \), and let \( {P}^{ \bullet } \) be any complex in \( {\mathrm{C}}^{ \leq 0}\left( \mathrm{P}\right) \) . Let\n\n\[ \varphi : {H}^{0}\left( {P}^{ \bullet }\right) \rightarrow {H}^{0}\left( {M}^{ \bullet }\ri... | Proof. We have to define \( {\alpha }^{i} : {P}^{i} \rightarrow {M}^{i} \) for all \( i \) . Since \( {\alpha }^{i} = 0 \) necessarily for \( i > 0 \) , we may as well replace \( {M}^{ \bullet } \) with its truncated version (cf. Exercise 3.1) and then extend both \( {P}^{ \bullet } \) and this complex as follows:\n\n!... | No |
Proposition 6.4. Any two projective (resp., injective) resolutions of an object \( A \) of an abelian category \( \mathrm{A} \) are homotopy equivalent. | ## (This is also a direct consequence of Lemma 6.3.) | No |
Proposition 6.5. Let \( {A}_{0},{A}_{1} \) be objects of an abelian category \( \mathrm{A} \), and let \( {P}_{i}^{ \bullet } \) be a projective resolution of \( {A}_{i}, i = 0,1 \) . Then every morphism \( \varphi : {A}_{0} \rightarrow {A}_{1} \) in \( \mathrm{A} \) is induced by a morphism \( {\alpha }^{ \bullet } : ... | Proof. By hypothesis, \( \varphi \) is a morphism \( {H}^{0}\left( {P}_{0}^{ \bullet }\right) \rightarrow {H}^{0}\left( {P}_{1}^{ \bullet }\right) \) . The complex \( {P}_{0}^{ \bullet } \) consists of projectives, and \( {P}_{1}^{ \bullet } \) is a resolution of \( {A}_{1} \) ; therefore a lift \( {\alpha }^{ \bullet ... | Yes |
Theorem 6.6. Assume the abelian category A has enough projectives, and let \( {L}^{ \bullet } \) be a complex in \( {\mathrm{C}}^{ - }\left( \mathrm{A}\right) \) . Then there exists a bounded-above complex of projectives \( {P}^{ \bullet } \) and a quasi-isomorphism \( {P}^{ \bullet } \rightarrow {L}^{ \bullet } \), an... | Proof. The proof of this result is admittedly rather technical, as it involves many of the tools that we have developed.\n\nConstruction of \( {P}^{ \bullet } \) . We may assume that \( {L}^{ \bullet } \) is in \( {\mathrm{C}}^{ \leq 0}\left( \mathrm{\;A}\right) \) ,\n\n\[ \cdots \rightarrow {L}^{-2}\overset{{d}_{{L}^{... | No |
Theorem 6.7. Let \( \\mathrm{A} \) be an abelian category with enough projectives. Then the functor \( \\mathrm{K}^{ - }\\left( \\mathrm{P}\\right) \\rightarrow \\mathrm{D}^{ - }\\left( \\mathrm{A}\\right) \) is an equivalence of categories. | Theorem 6.7 is proven in full detail in any more complete treatment of homological algebra. Since we have not actually constructed \( \\mathrm{D}^{ - }\\left( \\mathrm{A}\\right) \\), we cannot really prove this statement here; but we now know enough to appreciate why Theorem 6.7 should be true, in the sense that \( \\... | No |
If \( {\rho }^{ \bullet } \) is a quasi-isomorphism in \( {\mathrm{C}}^{ - }\left( \mathrm{A}\right) \), then \( \mathcal{P}\left( {\rho }^{ \bullet }\right) \) is an isomorphism in \( {\mathrm{K}}^{ - }\left( \mathrm{P}\right) \) . | For the first point, let \( {\rho }^{ \bullet } : {L}^{ \bullet } \rightarrow {M}^{ \bullet } \) be any morphism in \( {\mathrm{C}}^{ - }\left( \mathrm{A}\right) \) . By Theorem 6.6. \( {\rho }^{ \bullet } \) lifts to a morphism of resolutions: we have a diagram\n\n is (additive and) exact. Then I claim that \( \widehat{\mathrm{A}} \) is sent to \( \widehat{\mathrm{B}} \) by \( \mathrm{L}\mathcal{F} \) . | Indeed, let \( {P}^{ \bullet } \) be a complex in \( \widehat{\mathrm{A}} : {H}^{i}\left( {P}^{ \bullet }\right) = 0 \) for \( i \neq 0 \) . The image \( \mathrm{L}\mathcal{F}\left( {P}^{ \bullet }\right) \) is obtained by choosing a projective resolution \( {P}_{\mathcal{F}\left( {P}^{ \bullet }\right) }^{ \bullet } \... | Yes |
Proposition 7.3. The left-derived functor \( \mathrm{L}\mathcal{F} \) satisfies the following universal property:\n\n- There is a natural transformation\n\n\[ \mathrm{L}\mathcal{F} \circ {\mathcal{P}}_{\mathrm{A}} \sim {\mathcal{P}}_{\mathrm{B}} \circ \mathrm{K}\left( \mathcal{F}\right) \]\n\n- for every functor \( \ma... | Proof. If we have done our homework (and in particular Exercise 6.4), then we know that there is a natural transformation\n\n\[ {\mathcal{I}}_{\mathrm{A}} \circ {\mathcal{P}}_{\mathrm{A}} ⤳ {\operatorname{id}}_{{\mathrm{K}}^{ - }\left( \mathrm{A}\right) } \]\n\ncomposing on the left by \( {\mathcal{P}}_{\mathrm{B}} \ci... | No |
Every \( R \) -module \( N \) determines a functor \( {}_{ - }{ \otimes }_{R}N : M \mapsto M{ \otimes }_{R}N \) (see SVIII 2.2). The left-derived functor of \( {}_{ - }{ \otimes }_{R}N \) is denoted \( {}_{ - }{ \otimes }_{R}N \) and acts \( {\mathrm{D}}^{ - }\left( {R\text{-Mod}}\right) \rightarrow {\mathrm{D}}^{ - }\... | Indeed, the construction of \( {\operatorname{Tor}}_{i}^{R}\left( {M, N}\right) \) given in SVIII 2.4 matches precisely the ’concrete’ interpretation of the \( i \) -th left-derived functor given above. The reader may note that in SVIII 2.4 we used a free resolution of \( M \) ; free modules are projective, so this was... | No |
Lemma 7.8. Let\n\n(*)\n\n\[ 0 \rightarrow L \rightarrow M \rightarrow N \rightarrow 0 \]\n\nbe an exact sequence in an abelian category A with enough projectives. Assume \( {P}_{L}^{ \bullet },{P}_{N}^{ \bullet } \) are projective resolutions of \( L, N \), respectively. Then there exists an exact sequence\n\n\( \\left... | Proof. The hypotheses give us the solid part of the diagram\n\n\n\nand our task is to fill in the blanks with projective objects and morphisms so that all rows are exact, and the middle column is a resolution of \( M... | Yes |
Corollary 7.9. Let\n\n\\[ \n{M}^{ \bullet } : \;\cdots \rightarrow {M}^{-3} \rightarrow {M}^{-2} \rightarrow {M}^{-1} \rightarrow {M}^{0} \rightarrow 0 \n\\]\n\nbe a complex in an abelian category A with enough projectives. Then there is a complex of complexes:\n\n\\[ \n{P}_{{M}^{ \bullet }}^{ \bullet } : \;\cdots \rig... | Proof. Break up \\( {M}^{ \bullet } \\) into short exact sequences\n\n\\[ \n0 \rightarrow {K}^{i} \rightarrow {M}^{i} \rightarrow {I}^{i + 1} \rightarrow 0 \n\\]\n\ntogether with exact sequences\n\n\\[ \n0 \rightarrow {I}^{i} \rightarrow {K}^{i} \rightarrow {H}^{i} \rightarrow 0 \n\\]\n\nwhere \\( {K}^{i} \\) is the ke... | Yes |
Lemma 7.11. Let \( \mathrm{A} \) be an abelian category, and let\n\n\( \left( *\right) \)\n\n\[ 0 \rightarrow {L}^{ \bullet } \rightarrow {M}^{ \bullet } \rightarrow {P}^{ \bullet } \rightarrow 0 \]\n\nbe an exact sequence of complexes in \( \mathrm{A} \), where \( {P}^{i} \) is projective for all \( i \) . Let \( \mat... | Proof. Since \( {P}^{i} \) is projective, the sequence\n\n\[ 0 \rightarrow {L}^{i} \rightarrow {M}^{i} \rightarrow {P}^{i} \rightarrow 0 \]\n\nsplits (see the end of SVIII 6.1). It follows that\n\n\[ 0 \rightarrow \mathcal{F}\left( {L}^{i}\right) \rightarrow \mathcal{F}\left( {M}^{i}\right) \rightarrow \mathcal{F}\left... | No |
Theorem 7.12. Let \( \mathcal{F} : \mathrm{A} \rightarrow \mathrm{B} \) be an additive functor of abelian categories, and assume A has enough projectives. Every exact sequence\n\n\[ 0 \rightarrow L \rightarrow M \rightarrow N \rightarrow 0 \]\nin \( \mathrm{A} \) induces a long exact sequence ![23387543-548b-40c2-8595-... | Proof. By Lemma 7.8, the given exact sequence is induced by an exact sequence of projective resolutions\n\n\[ 0 \rightarrow {P}_{L}^{ \bullet } \rightarrow {P}_{M}^{ \bullet } \rightarrow {P}_{N}^{ \bullet } \rightarrow 0. \]\n\nBy Lemma 7.11, the corresponding sequence\n\n\[ 0 \rightarrow \mathcal{F}\left( {P}_{L}^{ \... | Yes |
Proposition 7.13. Let \( \mathcal{F} : \mathrm{A} \rightarrow \mathrm{B} \) be a right-exact additive functor. Then \( {\mathrm{L}}_{i}\mathcal{F} = \) 0 for \( i < 0 \), and \( {\mathrm{L}}_{0}\mathcal{F} \) is naturally isomorphic to \( \mathcal{F} \) . | Proof. Projective resolutions \( {P}^{ \bullet } \) of an object \( M \) of \( \mathrm{A} \) are in \( {\mathrm{C}}^{ \leq 0}\left( \mathrm{\;A}\right) \) : it follows that \( \mathrm{C}\left( \mathcal{F}\right) \left( {P}^{ \bullet }\right) \) is 0 in positive degree, hence so is its cohomology. Since \( {H}_{i} = {H}... | Yes |
Take \( G = \mathbb{Z} \) . Then \( \mathbb{Z}\left\lbrack G\right\rbrack \) is the ring \( \mathbb{Z}\left\lbrack {x,{x}^{-1}}\right\rbrack \) of Laurent polynomials. As \( \mathbb{Z}\left\lbrack {x,{x}^{-1}}\right\rbrack /\left( {1 - x}\right) \cong \mathbb{Z} \), with the trivial action (check this!), the complex | \[ \cdots \xrightarrow[]{\;}0\xrightarrow[]{\;}\mathbb{Z}\left\lbrack {x,{x}^{-1}}\right\rbrack \xrightarrow[]{\; \cdot \left( {1 - x}\right) \;}\mathbb{Z}\left\lbrack {x,{x}^{-1}}\right\rbrack \xrightarrow[]{\;}0\xrightarrow[]{\;}\cdots \] is a free, hence projective, resolution of \( \mathbb{Z} \), endowed with the t... | Yes |
Let \( G = {C}_{m} \) be a cyclic group of order \( m \) ; then \( \mathbb{Z}\left\lbrack G\right\rbrack \cong \mathbb{Z}\left\lbrack x\right\rbrack /\left( {{x}^{m} - 1}\right) \) . | Again it is not difficult to produce a projective resolution of \( \mathbb{Z} \) (with trivial action) in the category of \( \mathbb{Z}\left\lbrack {C}_{m}\right\rbrack \) -modules: letting \( N = 1 + x + \cdots + {x}^{m - 1} = \left( {1 - {x}^{m}}\right) /\left( {1 - x}\right) \), the reader will verify that the compl... | No |
Proposition 7.16. Let \( G \) be a finite group, and let \( M \) be a \( G \) -module. Then the group cohomology \( {H}^{i}\left( {G, M}\right) \) is the cohomology of the cochain complex | \[ 0 \rightarrow {C}^{0}\left( {G, M}\right) \overset{{d}_{G}^{0}}{ \rightarrow }{C}^{1}\left( {G, M}\right) \overset{{d}_{G}^{1}}{ \rightarrow }{C}^{2}\left( {G, M}\right) \overset{{d}_{G}^{2}}{ \rightarrow }\cdots \] induced by \( \left( \dagger \right) \). Tracing definitions, we see that for \( a \in {C}^{0}\left( ... | Yes |
Claim 7.18. \( {H}^{1}\left( {G,{F}^{ * }}\right) = 0 \) . | Indeed, with notation as above we have \( {H}^{1}\left( {G,{F}^{ * }}\right) \cong \ker {d}_{G}^{1}/\operatorname{im}{d}_{G}^{0} \), and we can compute this quotient explicitly. Let \( \alpha \in {C}^{1}\left( {G,{F}^{ * }}\right) \) ; denote by \( {\alpha }_{g} \) the image of \( g \) in \( {F}^{ * } \) by \( \alpha \... | Yes |
Let \( R \) be a commutative ring, and let\n\n\[ \n{F}_{ \bullet } : \;\cdots \rightarrow {F}_{2} \rightarrow {F}_{1} \rightarrow {F}_{0} \rightarrow 0 \n\] \n\nbe a resolution of an \( R \) -module \( M \) by flat \( R \) -modules. Then for every \( R \) -module \( N \) ,\n\n\[ \n{\operatorname{Tor}}_{i}^{R}\left( {M,... | Indeed, flat modules are acyclic with respect to \( \_ \otimes N \) (Example 8.2), so this is now a consequence of Theorem 8.3. | No |
Theorem 8.9. Let \( \mathrm{A} \) be an abelian category, and let\n\n(*) \n\n\[ \cdots \rightarrow {M}^{-3, \bullet } \rightarrow {M}^{-2, \bullet } \rightarrow {M}^{-1, \bullet } \rightarrow {M}^{0, \bullet } \rightarrow 0 \rightarrow \cdots \] \n\nbe a complex in \( {\mathrm{C}}^{ \leq 0}\left( {{\mathrm{C}}^{ \leq 0... | Proof. It is enough to prove the second statement: the first one follows by flipping the double complex corresponding to (*) (cf. Exercise 8.4).\n\n\( {}^{34}\mathrm{\;A} \) clever way out of the sign quagmire in this computation is to choose another way to get a double complex out of \( {\operatorname{Hom}}_{\mathbf{A... | Yes |
Example 8.10. Here is a taste of how convenient Theorem 8.9 is. Let \( {P}^{ \bullet } \) be a complex in \( {\mathrm{C}}^{ \leq 0}\left( \mathrm{\;A}\right) \), where each \( {P}^{i} \) is projective, and let \( {L}^{ \bullet } \) be an exact complex in \( {\mathrm{C}}^{ \geq 0}\left( \mathrm{\;A}\right) \). Since \( ... | This recovers Corollary 5.12, up to easy adjustments, taking care of the boundedness hypothesis on \( {L}^{ \bullet } \) . | Yes |
Let \( \mathrm{A} \) be an abelian category, and denote by \( {\mathrm{A}}^{\prime } \) the category \( {\mathrm{C}}^{ \leq 0}\left( \mathrm{\;A}\right) \) . Let \( {N}^{ \bullet } \) be an object of \( {\mathrm{A}}^{\prime } \), and let\n\n(*) \n\n\[ \cdots \rightarrow {M}^{-3, \bullet }\rightarrow {M}^{-2, \bullet }\... | The second statement follows from the first, by flipping the corresponding double complex about the main diagonal.\n\nThe first statement follows from Theorem 8.9 and Claim 8.11 Indeed, let \( {M}_{N}^{\bullet , \bullet } \) be the exact complex\n\n\[ \cdots \rightarrow {M}^{-2, \bullet } \rightarrow {M}^{-1, \bullet }... | Yes |
Theorem 8.13. Let \( M, N \) be modules over a commutative ring \( R \), and let \( {P}_{M}^{ \bullet } \) , resp., \( {P}_{N}^{ \bullet } \), be projective resolutions of \( M \), resp., \( N \) . Then\n\n\[ \n{H}^{i}\left( {{P}_{M}^{ \bullet }{ \otimes }_{R}N}\right) \cong {H}^{i}\left( {M{ \otimes }_{R}{P}_{N}^{ \bu... | Proof. Apply Theorem 8.12 to the complex\n\n\( \left( *\right) \)\n\n\[ \n\cdots \rightarrow {P}_{M}^{-2}{ \otimes }_{R}{P}_{N}^{ \bullet } \rightarrow {P}_{M}^{-1}{ \otimes }_{R}{P}_{N}^{ \bullet } \rightarrow {P}_{M}^{-0}{ \otimes }_{R}{P}_{N}^{ \bullet } \rightarrow 0 \rightarrow \cdots .\n\]\n\nSince each \( {P}_{N... | Yes |
Theorem 8.14. Let \( M, N \) be modules over a commutative ring \( R \), and let \( {P}_{M}^{ \bullet } \) , resp., \( {Q}_{N}^{ \bullet } \), be a projective resolution of \( M \), resp., an injective resolution of \( N \) . Then\n\n\[ \n{H}^{i}\left( {{\operatorname{Hom}}_{R}\left( {{P}_{M}^{ \bullet }, N}\right) }\r... | The proof of Theorem 8.14 is left to the reader (Exercise 8.9): Example 8.8 and the strategy extensively discussed above will hopefully make this a very easy task. | No |
Proposition 1.1. \( \Lambda \) is of finite-representation type if and only if it satisfies\n\n(a) every simple \( F \) in \( \left( {{\;\operatorname{mod}\;\Lambda },\mathrm{{Ab}}}\right) \) is finitely presented, and\n\n(b) every nonzero \( F \) in \( \left( {{\;\operatorname{mod}\;\Lambda },\mathrm{{Ab}}}\right) \) ... | The following description of the simple functors \( F \) : \( {\;\operatorname{mod}\;\Lambda } \rightarrow \mathrm{{Ab}} \) is given in [2]. The starting point is the observation that because an indecomposable object \( M \) of \( {\;\operatorname{mod}\;\Lambda } \) has a local ring, \( \left( {M,\;}\right) \) has a un... | Yes |
Proposition 1.2. An artin algebra \( \Lambda \) is of finite representation type if and only if every nonzero \( F \) in (mod \( \Lambda \), Ab) has a simple subobject. | Returning to the case of an arbitrary left artin ring \( \Lambda \), we recall (see [2]) that an \( F \) in (mod \( \Lambda \), Ab) has a simple subobject if and only if there is an \( M \) in mod \( \Lambda \) such that there is an \( x \) in \( F\left( M\right) \) that is universally minimal, i.e., \( x \neq 0 \) and... | No |
Theorem 1.5. Let \( F : {\;\operatorname{mod}\;\Lambda } \rightarrow \mathrm{{Ab}} \) be a nonzero functor that has no simple subfunctors. Suppose \( M \) is in \( {\;\operatorname{mod}\;\Lambda } \) and \( x \) is a nonzero element in \( F\left( M\right) \) . Then there exists a sequence of morphisms\n\n\[ M\overset{{... | Proof. Suppose \( x \in F\left( M\right) \) is not zero. Then by Proposition 1.4 we know there is an epimorphism \( {f}_{0} : M \rightarrow {M}_{0} \) such that \( F\left( {f}_{0}\right) \left( x\right) = {x}_{0} \) is minimal in \( F\left( {M}_{0}\right) \) . Suppose we have defined, for \( k \geq 0 \), the sequence o... | Yes |
Proposition 2.2. Let \( F \) be a finitely presented object of (mod \( \Lambda \), Ab). Then its extension \( \widetilde{F} \) : Mod \( \Lambda \rightarrow \mathrm{{Ab}} \) commutes with arbitrary products. | Proof. Since \( F \) is finitely presented, we know there is an exact sequence \( \left( {N,\;}\right) \rightarrow \left( {M,\;}\right) \rightarrow \widetilde{F} \rightarrow 0 \) with \( N, M \) in mod \( \Lambda \) . Then we have the commutative exact diagram\n\n be an indecomposable module in mod \( \Lambda \) such that the simple functor \( F = \left( {M,\;}\right) /\mathrm{r}\left( {M,\;}\right) \) is finitely presented. Suppose \( {\left\{ {M}_{i}\right\} }_{i \in I} \) is a family of indecomposable modules in \( {\;\operatorname{mod}\;\Lambda } \... | Proof. Since \( F = \left( {M,}\right) /r\left( {M,}\right) \) we know that if \( X \) is an indecomposable module in mod \( \Lambda \), then \( F\left( X\right) \neq 0 \) if and only if \( X \cong M \) (see Section 1). Suppose \( M \) is a summand of \( \mathop{\prod }\limits_{{i \in I}}{X}_{i} \) with the \( {X}_{i} ... | Yes |
Corollary 3.2. Let \( M \) be an indecomposable \( \Lambda \) -module in mod \( \Lambda \) and \( {\left\{ {X}_{i}\right\} }_{i \in I} \) a family of arbitrary \( \Lambda \) -modules. If \( M \) is isomorphic to a summand of the product \( \mathop{\prod }\limits_{{i \in I}}{X}_{i} \), then \( M \) is isomorphic to a su... | Proof. Let \( F \) be the simple functor \( \left( {M,\;}\right) /\mathrm{r}\left( {M,\;}\right) \) . Since \( M \) is a summand of \( \mathop{\prod }\limits_{{i \in I}}{X}_{i} \), we have that \( F\left( {\mathop{\prod }\limits_{{i \in I}}{X}_{i}}\right) \neq 0 \) . But \( F\left( {\mathop{\prod }\limits_{{i \in I}}{X... | Yes |
Corollary 3.3. Suppose \( X \) in Mod \( \Lambda \) has no finitely generated indecomposable summands. Then \( \left( {\left( {X,\;}\right), F}\right) = 0 \) for all \( F \) in \( \left( {{\;\operatorname{mod}\;\Lambda },\mathrm{{Ab}}}\right) \) of finite length. | Proof. Since \( F \) is of finite length, there is a finite filtration \( \left( 0\right) = \) \( {F}_{0} \subset {F}_{1} \subset \cdots \subset {F}_{n} = F \) of \( F \) with the property \( {F}_{i + 1}/{F}_{i} \) is a simple object of \( \left( {{\;\operatorname{mod}\;\Lambda },\mathrm{{Ab}}}\right) \) for \( i = 0,\... | Yes |
Proposition 3.4. Let \( F \) be a finitely generated functor in (mod \( \Lambda \), Ab). Then \( F \) has finite length if and only if there are only a finite number of nonisomorphic indecomposable \( M \) in mod \( \Lambda \) such that \( F\left( M\right) \neq 0 \) . | Proof. In [2], it was shown that an \( F \) in \( \left( {{\;\operatorname{mod}\;\Lambda },\mathrm{{Ab}}}\right) \) is of finite length if and only if (a) there is only a finite number of nonisomorphic indecomposable \( M \) in mod \( \Lambda \) such that \( F\left( M\right) \neq 0 \), and (b) \( F\left( M\right) \) ha... | Yes |
Theorem 3.5. The following statements are equivalent for an indecomposable \( \Lambda \) -module \( M \) in mod \( \Lambda \) .\n\n(a) \( \left( {M, N}\right) \neq 0 \) for an infinite number of nonisomorphic indecomposable \( N \) in \( {\;\operatorname{mod}\;\Lambda } \).\n\n(b) \( \left( {M, X}\right) \neq \left( 0\... | Proof. (a) implies (b). In view of Proposition 3.4, the hypothesis that \( \left( {M, N}\right) \neq 0 \) for an infinite number of nonisomorphic indecomposable \( N \) in \( {\;\operatorname{mod}\;\Lambda } \) is the same as the hypothesis that \( \left( {M,}\right) \) does not have finite length. Consequently the sub... | Yes |
Proposition 3.7. Let \( X \) in Mod \( \Lambda \) be a filtered inverse limit of finitely generated \( \Lambda \) -modules. Then\n\n(a) If \( Z = \underline{\lim }{Z}_{i} \) in \( \operatorname{Mod}\Lambda \), then \( {\operatorname{Ext}}_{\Lambda }{}^{i}\left( {Z, X}\right) \cong \underline{\lim }{\operatorname{Ext}}^... | Proof. (a) Since \( X \) is a filtered inverse limit of finitely generated \( \Lambda \) - modules, we know that \( X = D\left( Y\right) \) for some \( Y \) in \( \operatorname{Mod}{\Lambda }^{\text{op }} \) . Now it is well known (see [5]) that there are functorial isomorphisms\n\n\[{\operatorname{Ext}}_{\Lambda }^{i}... | Yes |
Proposition 3.8. Suppose \( f : X \rightarrow Y \) is a morphism in \( \operatorname{Mod}\Lambda \) and \( X \) is a filtered inverse limit of finitely generated \( \Lambda \) -modules. Then the following statements are equivalent:\n\n(a) \( f : X \rightarrow Y \) is a splittable monomorphism.\n\n(b) The morphism of fu... | Proof. (a) implies (b). This is trivial.\n\n(b) implies (c). Since \( \otimes f : \otimes X \rightarrow \otimes Y \) is a monomorphism, it follows that \( f : X \rightarrow Y \) is a monomorphism. Suppose \( 0 \rightarrow X\overset{f}{ \rightarrow }Y\overset{g}{ \rightarrow }Z \rightarrow 0 \) is exact. Then for each f... | Yes |
Corollary 2. If a morphism of commutative DGA \( k \) -algebras \( \widehat{A} \rightarrow A \) induces an isomorphism \( H\left( \widehat{A}\right) \approx H\left( A\right) \) with \( {H}^{0}\left( A\right) \approx k \), then there are induced isomorphisms of the Eilenberg-Moore spectral sequences | \[ {E}_{r}\left( {\bar{B}\left( \widehat{A}\right) }\right) \approx {E}_{r}\left( {\bar{B}\left( A\right) }\right) ,\;r \geq 1. \] | Yes |
Let \( w \) be the normalized volume element of \( {S}^{n}, n > 1 \) , and let \( A \) be the DG subalgebra of \( \Lambda \left( {S}^{n}\right) \) having \( \{ 1, w\} \) as a basis. Then \( {\bar{B}}_{Z}\left( {H\left( {A;Z}\right) }\right) \) has as a basis\n\n\[ \left\lbrack \begin{matrix} \end{matrix}\right\rbrack ,... | Let \( {A}^{\prime } \) be DG subalgebra of \( \Lambda \left( {\Omega {S}^{n}}\right) \) having a basis consisting of iterated integrals\n\n\[ \text{1,}\int w,\;\int {ww},\ldots \]\n\n(5.1)\n\nThese iterated integrals represent a basis for the integral cohomology classes of \( \Omega {S}^{n} \) . We also obtain the kno... | No |
Let \( X = \Omega {S}^{n}, n \geq 3 \) being odd. Let \( {A}^{\prime } \) be the DG subalgebra of \( \Lambda \left( {\Omega {S}^{n}}\right) \) having as a basis the iterated integrals\n\n\[ 1,\;{u}_{1} = \int w,\;{u}_{2} = \int {ww},\ldots \]\n\nas given in (5.1). Then the deRham theorem holds for \( {A}^{\prime } \) o... | According to [3],(4.1.1),\n\n\[ {u}_{r} \land {u}_{s} = \left( \begin{matrix} r + s \\ r \end{matrix}\right) {u}_{r + s} \]\n\nwhere \( \left( \begin{matrix} r + s \\ r \end{matrix}\right) \) is the binomial coefficient. Thus \( {H}_{ * }\left( {{\Omega }^{2}{S}^{n};k}\right) \) is isomorphic to the homology of the DG ... | Yes |
Example 3. Let \( X \) be an \( \left( {n - 1}\right) \) -connected \( {2n} \) -dimensional orientable compact \( {C}^{\infty } \) manifold, \( n > 1 \) . Let \( {w}_{1},\ldots ,{w}_{m} \) be closed \( n \) -forms such that \( {\bar{w}}_{1},\ldots ,{\bar{w}}_{m} \) is an integral basis for \( {H}^{n}\left( X\right) \) ... | \[ J{w}_{j} \land {w}_{k} + d{w}_{jk} = {c}_{jk}w \] and \( d{w}_{jk} \in N \) . Let \( A \) be the DG subalgebra of \( \Lambda \left( X\right) \) spanned by \( 1, w,{w}_{i} \) , \( {w}_{ij} \), where \( i, j = 1,\ldots, m \) . Then the map \[ A \rightarrow H\left( A\right) \] given by \( {w}_{i} \mapsto {\bar{w}}_{i},... | Yes |
Proposition 1.1. \( M \times c \) is left exact if and only if \( \left( {Y, M}\right) \) is filtered. | Proof. If \( \left( {Y, M}\right) \) is filtered, then \( M \) is a filtered colimit of representables. Since tensoring with representables preserves finite limits by (1), and since filtered colimits in Sets preserve finite limits, it follows that \( M \times {}_{\mathbb{C}} \) is left exact.\n\nNow suppose that \( M \... | Yes |
Theorem 2.2. If \( M \) is a \( {\mathbb{C}}^{\text{op }} \) -set such that \( {M\alpha } \) is an injection for all \( \alpha \in \mathbb{C} \), then \( \mathbb{Z}M \) is flat if and only if \( \operatorname{aff}\left( {Y, M}\right) \) has filtered components. | Proof. We have seen the \ | No |
Lemma 2.3. The components of aff \( \mathbb{C} \) are filtered if and only if every diagram in \( \mathbb{C} \) of the form\n\n\n\ncan be completed to a (not necessarily commutative) square in \( \mathbb{C} \), and for... | Proof. If \( \beta = \sum {r}_{i}{\beta }_{i} \) and \( {\beta \alpha } = \beta {\alpha }^{\prime } \), then using the first assumption we can choose, for each object in \( \mathbb{C} \) that appears as the codomain of a \( {\beta }_{i} \), a morphism \( \gamma \) with common codomain. Composing each \( {\beta }_{i} \)... | Yes |
Corollary 2.6. If \( U : \mathbb{C} \rightarrow \mathbb{D} \) takes all morphisms to epimorphisms, then the left adjoint of \( {\mathrm{{Ab}}}^{\mathbb{D}} \rightarrow {\mathrm{{Ab}}}^{\mathbb{C}} \) is exact if and only if \( \operatorname{aff}\left( {U, D}\right) \) has filtered components for all \( D \in \left| \ma... | Proof. The left adjoint composed with the evaluation functor at \( D \) is given by tensoring with \( \mathbb{{ZD}}\left( {{U}_{ - }, D}\right) \) . Hence the left adjoint is exact if and only if \( \mathbb{Z}\mathbb{D}\left( {{U}_{ - }, D}\right) \) is flat for all \( D \) . By Corollary 2.5 this is true if and only i... | Yes |
Proposition 3.1. If \( M \) is a \( \mathbb{C} \) -set, then \( {\operatorname{Hom}}_{\mathbb{C}}\left( {M,}\right) \) preserves epimorphisms if and only if the components of \( {\left( Y, M\right) }^{\text{op }} \) have right zeros. Moreover \( {\operatorname{Hom}}_{\mathbb{C}}\left( {M,\;}\right) \) preserves finite ... | Proof. If \( {\operatorname{Hom}}_{\mathbb{C}}\left( {M,}\right) \) preserves epimorphisms, then the natural map\n\n\[ \n{\bigoplus }_{x \in M}\mathbb{C}\left( {\left| x\right| , \rightarrow }\right) \rightarrow M \n\]\n\nsplits. The splitting must take each indecomposable subset of \( M \) into a single term of the co... | No |
Lemma 4.1 (Laudal). If \( \mathbb{M} \) is a monoid and \( \mathbb{Z} \) is projective, then \( \mathbb{M} \) has a right zero. | Proof. The splitting of the augmentation \( \mathbb{{ZM}} \rightarrow \mathbb{Z} \) gives rise to a family of elements \( {e}_{1},\ldots ,{e}_{k} \) of \( \mathbb{M} \) and nonzero coefficients \( {r}_{1},\ldots ,{r}_{k} \) in \( \mathbb{Z} \) such that \( \sum {r}_{i} = 1 \) (showing \( k \geq 1 \) ), and such that\n\... | Yes |
Corollary 4.3. Let \( \mathbb{C} \) be any small category. Then lim: \( {\mathrm{{Ab}}}^{\mathbb{C}} \rightarrow \mathrm{{Ab}} \) is exact if and only if the components of \( \mathbb{C} \) have right zeros. | Proof. This follows since lim is given by homing with \( \Delta \mathbb{Z} \in {\mathrm{{Ab}}}^{\mathbb{C}} \). | No |
Corollary 4.4. If all morphisms of \( \mathbb{C} \) are monomorphisms, then \( \mathbb{Z}M \) is projective if and only if it is free. | Proof. If all morphisms of \( \mathbb{C} \) are monomorphisms and \( \mathbb{Z}M \) is projective, then as we saw in Section \( 2,{M\alpha } \) is an injection for all \( \alpha \) . Therefore by the theorem, \( M \) is a disjoint union of retracts of representables. But again since all morphisms of \( \mathbb{C} \) ar... | Yes |
Corollary 4.5. If \( U : \mathbb{C} \rightarrow \mathbb{D} \) takes all morphisms to monomorphisms, then the right adjoint to \( {\mathrm{{Ab}}}^{\mathbb{D}} \rightarrow {\mathrm{{Ab}}}^{\mathbb{C}} \) is exact if and only if the components of \( {\left( D, U\right) }^{\text{op }} \) have right zeros for all \( D \in \... | Proof. The right adjoint composed with the evaluation functor at \( D \) is given by homing with \( \mathbb{Z}\mathbb{D}\left( {D,{U}_{ - }}\right) \) . Hence the right adjoint is exact if and only if the components of \( {\left( Y,\mathbb{D}\left( D,{U}_{ - }\right) \right) }^{\text{op }} \) have right zeros for all \... | Yes |
Theorem 4.6. Let \( \mathbb{C} \) be a category in which the only isomorphisms are identities, and such that an equation \( {\alpha \beta } = \alpha \) implies \( \beta = 1 \) . If \( R \) is any nonzero ring and \( M \) is a \( \mathbb{C} \) -set with \( {RM} \) a projective \( R\mathbb{C} \) -module, then \( {RM} \) ... | Proof. The hypothesis on \( \mathbb{C} \) is easily seen to be equivalent to the single property \( {\alpha \beta \gamma } = \alpha \) implies \( \beta = 1 \) . Let \( \mu \) be a splitting for the natural epimorphism\n\n\( x \in M \) Then if \( y \) is an element of \( M,\mu \left( y\right) \) has the form\n\n\[ \mu \... | No |
Lemma 4.7. If \( \left( {\alpha, e}\right) \) appears in the first sum in \( \mu \left( y\right) \), then \( e \) is minimal. Moreover, no term of the form \( \left( {1, e}\right) \) with \( e \) minimal can appear in the second sum. | Proof. The second statement is clear since there would be nothing to cancel \( e \) after application of the augmentation \( \varepsilon \) . To prove the first assertion, suppose \( {\beta z} = e \) . Then \( {\alpha \beta z} = {\alpha e} = y \), so \( {\alpha \beta \mu }\left( z\right) = \mu \left( y\right) \) . It f... | Yes |
Lemma 4.8. Given a diagram in \( {\left( Y, M\right) }^{\text{op }} \)\n\n\n\nwith \( e \) and \( f \) minimal, we must have \( e = f \) and \( \beta = \gamma \) . | Proof. Since \( e \) and \( f \) are minimal, \( \mu \left( e\right) \) and \( \mu \left( f\right) \) have the form\n\n\[ \mu \left( e\right) = \left( {1, e}\right) + \mathop{\sum }\limits_{{{\alpha x} \neq e}}{r}_{\alpha, x, e}\left( {\alpha, x}\right) \]\n\n\[ \mu \left( f\right) = \left( {1, f}\right) + \mathop{\sum... | Yes |
Lemma 1. Let \( G \) be an orientable, \( n \) -dimensional Poincaré duality group and \( E \) a subgroup.\n\n(a) If \( \left\lbrack {G : E}\right\rbrack = \infty \), then \( {H}_{n}\left( {E;\mathbb{Z}}\right) = 0 \) .\n\n(b) If \( \left\lbrack {G : E}\right\rbrack = s < \infty \), then the homomorphism \( {H}_{n}\lef... | Proof. By a Shapiro lemma and Poincaré duality we have\n\n\[ \n{H}_{n}\left( {E;\mathbb{Z}}\right) \cong {H}_{n}\left( {G;\mathbb{Z}\left\lbrack G\right\rbrack { \otimes }_{\mathbb{Z}\left\lbrack E\right\rbrack }\mathbb{Z}}\right) \cong {H}^{0}\left( {G;\mathbb{Z}\left\lbrack G\right\rbrack { \otimes }_{\mathbb{Z}\left... | Yes |
Lemma 2. Let \( G \) be a surface group and \( \rho : G \rightarrow \mathbb{Z} \) be an onto homomorphism. Then \( K = \ker \rho \) is a free group. | Proof. Let \( A \) be a \( K \) -module. By another Shapiro lemma and Poincaré duality\n\n\[ \n{H}^{2}\left( {K;A}\right) \cong {H}^{2}\left( {G;{\operatorname{Hom}}_{\mathbb{Z}\left\lbrack K\right\rbrack }\left( {\mathbb{Z}\left\lbrack G\right\rbrack, A}\right) }\right) \cong {H}_{0}\left( {G;{\operatorname{Hom}}_{\ma... | No |
Proposition 3. The functor \( {S}^{\prime }I : {\left( \mathrm{{ca}}\right) }_{0} \rightarrow {\mathrm{{ab}}}_{0}\left( \mathrm{{cc}}\right) \) is the right adjoint (or just adjoint) of \( F : {\mathrm{{ab}}}_{0}\left( \mathrm{{cc}}\right) \rightarrow {\left( \mathrm{{ca}}\right) }_{0} \) . | We apply the functor \( {S}^{\prime }I \) to the case \( A = k \oplus M \) where the product is zero on \( M \), i.e., \( M \cdot M = 0 \) . Thus \( {S}^{\prime }\left( {I\left( {k \oplus M}\right) }\right) = {S}^{\prime }\left( M\right) \) has a bicommutative Hopf algebra structure. The underlying algebra \( F{S}^{\pr... | No |
The morphism \( {H}_{ * }\left( {{P}^{\infty }\left( \mathbf{C}\right), k}\right) \rightarrow {G}_{m}{H}_{ * }\left( {\mathbf{Z} \times {BU}, k}\right) \) considered at the end of Section 2 is just the adjunction morphism \( \beta \left\lbrack {x,2}\right\rbrack : {S}^{\prime }\left\lbrack {x,2}\right\rbrack \rightarro... | \[ {R}_{g}{H}_{ * }\left( {{P}^{\infty }\left( \mathbf{C}\right), k}\right) \rightarrow {H}_{ * }\left( {\mathbf{Z} \times {BU}, k}\right) \] which is an isomorphism. To see this, recall that \( {H}_{ * }\left( {{P}^{\infty }\left( \mathbf{C}\right), k}\right) \rightarrow Q{H}_{ * }\left( {{BU}, k}\right) \) is an isom... | Yes |
Proposition 4. For a morphism \( u : G \rightarrow {G}_{m}\left( R\right) \) in \( {\operatorname{gr}}_{0}\left( \mathrm{{cc}}\right) \), the adjoint morphism \( f : {R}_{g}\left( G\right) \rightarrow R \) is an isomorphism if and only if \( \bar{u} : I\left( G\right) \rightarrow Q\left( {{G}_{a}\left( R\right) }\right... | Proof. The direct implication follows from the construction. For the converse observe that we have only to show that \( {G}_{a}\left( f\right) = \bar{f} : {G}_{a}{R}_{g}\left( G\right) \rightarrow {G}_{a}\left( R\right) \) is an isomorphism since \( R = k\left( \varepsilon \right) \otimes {G}_{a}\left( R\right) \) as a... | Yes |
Let \( k \) be an \( {\mathbf{F}}_{2} \) -algebra. The morphism \( {H}_{ * }\left( {{P}^{\infty }\left( \mathbf{R}\right), k}\right) \rightarrow \) \( {G}_{m}{H}_{ * }\left( {\mathbf{Z} \times \mathrm{{BO}}, k}\right) \) defined by the natural inclusion \( {P}^{\infty }\left( \mathbf{R}\right) \rightarrow 1 \times {BO}... | By Proposition 4 the adjunction morphism\n\n\[ \n{R}_{g}{H}_{ * }\left( {{P}^{\infty }\left( \mathbf{R}\right), k}\right) \rightarrow {H}_{ * }\left( {\mathbf{Z} \times {BO}, k}\right) \n\]\n\nis an isomorphism. | No |
Let \( k \) be a \( \mathbf{Z}\left\lbrack \frac{1}{2}\right\rbrack \) -algebra. The composite map \( {P}^{\infty }\left( \mathbf{C}\right) \rightarrow \) \( {BU} \rightarrow {BO} \) factors by \( {P}^{\infty }\left( \mathbf{C}\right) /\left( {\pm 1}\right) \rightarrow {BO} \), where the action of \( \pm 1 \) on \( {P}... | By Proposition 4 the adjunction morphism\n\n\[ \n{R}_{g}{H}_{ * }\left( {{P}^{\infty }\left( \mathbf{C}\right) /\left( {\pm 1}\right), k}\right) \rightarrow H\left( {\mathbf{Z} \times {BO}, k}\right) \n\] \n\nis an isomorphism. | No |
Theorem 3. (1) The map \( {BG}{L}_{1}\left( \bar{F}\right) \rightarrow 1 \times {BGL}\left( \bar{F}\right) \subset \mathbf{Z} \times {BGL}\left( \bar{F}\right) \) induces a morphism \( {H}_{ * }\left( {{BG}{L}_{1}\left( \bar{F}\right) }\right) \rightarrow {G}_{m}{H}_{ * }\left( {\mathbf{Z} \times {BGL}\left( \bar{F}\ri... | \[ {R}_{g}{H}_{ * }\left( {{BG}{L}_{1}\left( \bar{F}\right) }\right) \rightarrow {H}_{ * }\left( {\mathbf{Z} \times {BGL}\left( \bar{F}\right) }\right) \] is an isomorphism. | Yes |
Theorem 2. If \( C \) is a coalgebra and either \( {H}_{ * }{\Omega C} \) is of finite type or \( R \) is a field, then there is a reduced coalgebra \( D \), and a morphism of coalgebras \( f : C \rightarrow {B\Omega }\left( D\right) \) such that \( \Omega \left( f\right) \) is a chain equivalence. | Proof. The hypotheses imply that \( {I\Omega }\left( C\right) \) is isomorphic with \( {X}^{\prime } \oplus {X}^{\prime \prime } \) where \( {X}^{\prime } \) is a coproduct of elementary complexes and \( {X}^{\prime \prime } \) is contractible. Applying the technical theorem, the result follows easily. | No |
Proposition 2.2. If \( {S}^{\prime } \in {S}_{\mathbf{v}} \) and \( \varphi : S \rightarrow T \) is a morphism, then there exists \( {T}^{\prime } \in {T}_{\mathbf{v}} \) such that\n\n\[ \n{S}^{\prime } \subset {T}^{\prime }{\varphi }^{-1} \n\]\n\nFurthermore, if \( {S}^{\prime } \) is a type \( \mathbf{V} \) semigroup... | Proof. Let \( \psi : T \rightarrow V \) be a type \( \mathbf{V} \) morphism of \( T \) . Then\n\n\[ \n{\varphi \psi } : S \rightarrow V,\;V \in \mathbf{V} \n\]\nso there exists \( W \in {V}_{\mathrm{E}} \) such that\n\n\[ \n{S}^{\prime } \subset W{\left( \varphi \psi \right) }^{-1} = W{\psi }^{-1}\varphi \n\]\n\nBy The... | Yes |
If \( \varphi : S \rightarrow T \) is a functional morphism and \( {S}^{\prime } \in {S}_{\mathbf{v}} \), then \( {S}^{\prime }\varphi \in {T}_{\mathbf{v}} \). Furthermore, if \( {S}^{\prime } \) is a type \( \mathbf{V} \) semigroup, then so is \( {S}^{\prime }\varphi \). | Proof. If \( {S}^{\prime } \in {S}_{\mathbf{v}} \), there exists \( {T}^{\prime } \in {T}_{\mathbf{v}} \) such that\n\n\[ \n{S}^{\prime } \subset {T}^{\prime }{\varphi }^{-1} \n\]\n\nBut since \( \varphi \) is a function, we have\n\n\[ \n{S}^{\prime }\varphi \subset {T}^{\prime }{\varphi }^{-1}\varphi \subset {T}^{\pri... | Yes |
Proposition 2.4. (a) \( {S}^{\prime } \) is a type I subsemigroup of \( S \) iff (2.11) holds. | Proof. (a) Since elementary subsemigroups belong to \( {\mathbf{R}}_{\mathbf{s}} \), they are clearly \( {R}_{1} \), so \( {S}^{\prime } \) being type I in \( S \) implies (2.11). Assume (2.11) holds and let \( \varphi : S \rightarrow A \) be a morphism with \( A \) aperiodic. We must show the existence of an elementar... | Yes |
Proposition 2.5. If \( {S}^{\prime } \in {S}_{\mathbf{v}} \) and \( S \in \mathbf{V} \), then \( {S}^{\prime } \in {S}_{\mathbf{E}} \) . | Proof. Consider the identity morphism \( I : S \rightarrow S \) . There must be a \( W \in {S}_{\mathbf{E}} \) such that \( {S}^{\prime } \subset W{I}^{-1} = W \) | Yes |
Proposition 3.1. Let \( \varphi : S \rightarrow T \) be a morphism. Then\n\n\[ \n{Sl} \leq {\varphi l} + {Tl} \n\] | Proof. Let \( {Sl} = n > 0 \) and let (3.1) be a series for \( S \) of length \( n \) . By Proposition 2.2, there exists a type I semigroup \( {T}_{1}{}^{\prime } \subset T \) such that\n\n\[ \n{T}_{1} \subset {T}_{1}^{\prime }{\varphi }^{-1} \n\]\n\nLet \( {\varphi }_{1} : {T}_{1} \rightarrow {T}_{1}{}^{\prime } \) be... | Yes |
Corollary 3.2. Let \( \varphi : S \rightarrow T \) be an aperiodic morphism. Then\n\n\[ \n{Sl} \leq {Tl} \n\] | Proof. Since all elementary subsemigroups of \( T \) are aperiodic, we have \( {\varphi l} = 0 \) | No |
Corollary 3.3. If \( X \) and \( Y \) are ts’s and \( X \prec Y \), then\n\n\[ {Xl} \leq {Yl} \] | Proof. \( \left( {Q, S}\right) \prec \left( {P, T}\right) \) implies the existence of an aperiodic morphism \( \varphi : S \rightarrow T \) . See Exercise XII,4.6 of [2]\n\nTherefore \( l \) satisfies (A1). We next show that \( l \) satisfies (A3); it suffices to prove | No |
Proposition 3.4. \( \left( {{S}_{1} \times {S}_{2}}\right) l \leq \sup \left\{ {{S}_{1}l,{S}_{2}l}\right\} \) | Proof. We may assume \( \left( {{S}_{1} \times {S}_{2}}\right) l = n > 0 \) . Let\n\n\[ \n{S}_{1} \times {S}_{2} \supset {T}_{1} \supset {U}_{1} \supset \cdots \supset {T}_{n} \supset {U}_{n} \n\] \n\nbe a series for \( {S}_{1} \times {S}_{2} \) of length \( n \) . Let\n\n\[ \n{\pi }_{i} : {S}_{1} \times {S}_{2} \right... | Yes |
Proposition 3.5. Let \( X \) be a ts. Then \[ \bar{X}l = {Xl} \] | Proof. Let \( X = \left( {Q, S}\right) \) . Then \( \bar{X} = \left( {Q,{S}^{\prime }}\right) \) where \( {S}^{\prime } \) is the subsemigroup of \( {PF}\left( Q\right) \) generated by \( S \) and \( \{ \widetilde{q} : q \in Q\} \) . Let \[ I = \left\{ {t \in {S}^{\prime } \mid t = {t}_{1}\dot{q}{t}_{2}\text{ for some ... | Yes |
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