Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
values |
|---|---|---|
Proposition 10.5. (i) Every retract of a projective object is projective. | Proof. (i) Given \( P\overset{\varrho }{ \rightleftarrows }Q,\varrho \sigma = 1, P \) projective, and\n\n\n\nchoose \( {\psi }^{\prime } : P \rightarrow A \) so that \( \varepsilon {\psi }^{\prime } = \varphi \varrho \... | Yes |
Proposition 10.6. (i) A coproduct of free objects is free. | Proof. (i) Since Fr has a right adjoint, it maps coproducts to co-products. (Coproducts in \( \mathfrak{S} \) are disjoint unions.) | No |
Lemma 1.1. The square\n\n\n\n(1.2)\n\nis a pull-back diagram if and only if the sequence\n\n\\[ 0 \rightarrow Y\\xrightarrow[]{\\{ \\alpha ,\\beta \\} }A \\oplus B\\xrightarrow[]{\\langle \\varphi , - \\psi \\rangle }X... | Proof. We have to show that the universal property of the pull-back of \\( \\left( {\\varphi ,\\psi }\\right) \\) is the same as the universal property of the kernel of \\( \\langle \\varphi , - \\psi \\rangle \\) . But it is plain that two maps \\( \\gamma : Z \\rightarrow A \\) and \\( \\delta : Z \\rightarrow B \\) ... | Yes |
Lemma 1.2. If the square (1.2) is a pull-back diagram, then\n\n(i) \( \beta \) induces \( \ker \alpha \overset{ \sim }{ \rightarrow }\ker \psi \) ;\n\n(ii) if \( \psi \) is an epimorphism, then so is \( \alpha \) . | Proof. Part (i) has been proved in complete generality in Theorem II.6.2. For part (ii) we consider the sequence \( 0 \rightarrow {Y}^{\{ \alpha ,\beta \} } \rightarrow A \oplus {B}^{\langle \varphi , - \psi \rangle } \rightarrow X \) , which is exact by Lemma 1.1. Suppose \( a \in A \) . Since \( \psi \) is epimorphic... | Yes |
Lemma 1.3. Let\n\n\n\nbe a commutative diagram with exact rows. Then the right-hand square is a pull-back diagram. | Proof. Let\n\n\n\nbe a pull-back diagram. By Lemma \( {1.2\varepsilon } \) is epimorphic and \( \varphi \) induces an isomorphism \( \ker \varepsilon \cong B \) . Hence we obtain an extension\n\n\( B\overset{\mu }{ \ri... | No |
Lemma 2.1. \( {\pi }^{ * } \) does not depend on the chosen \( \pi : {P}^{\prime } \rightarrow P \) but only on \( \alpha : {A}^{\prime } \rightarrow A \) . | Proof. Let \( {\pi }_{i} : {P}^{\prime } \rightarrow P, i = 1,2 \), be two homomorphisms lifting \( \alpha \) and inducing \( {\sigma }_{i} : {R}^{\prime } \rightarrow R \), so that the following diagram is commutative for \( i = 1,2 \n\n and \( {R}^{\prime } \rightarrowtail {P}^{\prime }\overset{{\varepsilon }^{\prime }}{ \rightsquigarrow }A \) be two projective presentations of \( A \) . Then\n\n\[ \left( {{1}_{A};{P}^{\prime }, P}\right) : {\operatorname{Ext}}_{A}... | Proof. Let \( \pi : P \rightarrow {P}^{\prime } \) and \( {\pi }^{\prime } : {P}^{\prime } \rightarrow P \) both lift \( {1}_{A} : A \rightarrow A \) . By formulas (2.2) and (2.3) we obtain \( \left( {{1}_{A};P,{P}^{\prime }}\right) \circ \left( {{1}_{A};{P}^{\prime }, P}\right) = \left( {{1}_{A};P, P}\right) = 1 : {\o... | Yes |
Corollary 2.5. The set \( E\left( {A, B}\right) \) of equivalence classes of extensions has a natural abelian group structure. | Proof. This is obvious, since \( {\operatorname{Ext}}_{A}\left( {A, B}\right) \) carries a natural abelian group structure and since \( \eta : E\left( {-, - }\right) \rightarrow {\operatorname{Ext}}_{A}\left( {-, - }\right) \) is a natural equivalence. | No |
Proposition 2.6. If \( P \) is projective and \( I \) injective, then \( {\operatorname{Ext}}_{A}\left( {P, B}\right) = 0 \) \( = {\operatorname{Ext}}_{\Lambda }\left( {A, I}\right) \) for all \( \Lambda \) -modules \( A, B \) . | Proof. By Theorem 2.4 \( {\operatorname{Ext}}_{A}\left( {P, B}\right) \) is in one-to-one correspondence with the set \( E\left( {P, B}\right) \), consisting of classes of extensions of the form \( B \gg E \gg P \) . By Theorem I.4.7 short exact sequences of this form split. Hence \( E\left( {P, B}\right) \) contains o... | Yes |
Lemma 3.1. Let\n\n\n\n(3.1)\n\nbe a commutative diagram with exact rows. Then \( \varphi \) induces an isomorphism\n\n\[ \Phi : \ker \theta {\alpha }_{2}/\left( {\ker {\alpha }_{2} + \ker \varphi }\right) \overset{ \... | Proof. First we show that \( \varphi \) induces a homomorphism of this kind. Let \( x \in \ker \theta {\alpha }_{2} \) ; plainly \( {\varphi x} \in \operatorname{im}\varphi \) . Since \( 0 = \theta {\alpha }_{2}x = {\beta }_{2}{\varphi x},{\varphi x} \in \operatorname{im}{\beta }_{1} \) . If \( x \in \ker {\alpha }_{2}... | Yes |
For any projective presentation \( R\overset{\mu }{ \mapsto }P\overset{\varepsilon }{ \rightarrow }A \) of \( A \) and any injective presentation \( B\overset{ \cdot }{ \rightarrowtail }{I}^{n} \rightarrowtail S \) of \( B \), there is an isomorphism \[ \sigma : {\operatorname{Ext}}_{\Lambda }^{\varepsilon }\left( {A, ... | Proof. Consider the following commutative diagram with exact rows and columns  (3.2) The reader easily checks that \( \operatorname{Ker}{\sum }_{1} = {\overline{\operatorname{Ext}}}_{A}^{v}\left( {A, B}\right) \) and... | Yes |
Lemma 4.1. (i) \( {\operatorname{Ext}}_{A}\left( {{\bigoplus }_{i}{A}_{i}, B}\right) \cong \mathop{\prod }\limits_{i}{\operatorname{Ext}}_{A}\left( {{A}_{i}, B}\right) \) . | Proof. We only prove assertion (i), leaving the other to the reader. For each \( i \) in the index set we choose a projective presentation \( {R}_{i} \mapsto {P}_{i} \rightarrow {A}_{i} \) of \( {A}_{i} \) . Then \( {\bigoplus }_{i}{R}_{i} \mapsto {\bigoplus }_{i}{P}_{i} \rightarrow {\bigoplus }_{i}{A}_{i} \) is a proj... | No |
Lemma 5.4. To a short exact sequence \( {A}^{\prime }\overset{\varphi }{ \rightarrow }A\overset{\psi }{ \rightarrow }{A}^{\prime \prime } \) and to projective presentations \( {\varepsilon }^{\prime } : {P}^{\prime } \rightarrow {A}^{\prime } \) and \( {\varepsilon }^{\prime \prime } : {P}^{\prime \prime } \rightarrow ... | Proof. Let \( P = {P}^{\prime } \oplus {P}^{\prime \prime } \), let \( \iota : {P}^{\prime } \rightarrow {P}^{\prime } \oplus {P}^{\prime \prime } \) be the canonical injection. \( \pi : {P}^{\prime } \oplus {P}^{\prime \prime } \rightarrow {P}^{\prime \prime } \) the canonical projection. We define \( \varepsilon \) b... | Yes |
Corollary 5.5. The \( A \) -module \( A \) is projective if and only if \( {\operatorname{Ext}}_{A}\left( {A, B}\right) = 0 \) for all \( A \) -modules B. | Proof. Suppose \( A \) is projective. Then \( 1 : A\widetilde{ \rightarrow }A \) is a projective presentation, whence \( {\operatorname{Ext}}_{A}\left( {A, B}\right) = 0 \) for all \( \Lambda \) -modules \( B \) . Conversely, suppose \( {\operatorname{Ext}}_{A}\left( {A, B}\right) = 0 \) for all \( \Lambda \) -modules ... | Yes |
Corollary 5.6. The \( \Lambda \) -module \( B \) is injective if and only if \( {\operatorname{Ext}}_{\Lambda }\left( {A, B}\right) = 0 \) for all \( \Lambda \) -modules \( A \) . | [] | No |
Corollary 5.7. Let \( \Lambda \) be a principal ideal domain. Then the homomorphisms \( {\psi }_{ * } : {\operatorname{Ext}}_{A}\left( {A, B}\right) \rightarrow {\operatorname{Ext}}_{A}\left( {A,{B}^{\prime \prime }}\right) \) in sequence (5.3) and \[ {\varphi }^{ * } : {\operatorname{Ext}}_{A}\left( {A, B}\right) \rig... | Proof. Over a principal ideal domain \( \Lambda \) submodules of projective modules are projective. Hence in diagram (5.4) \( R \) is projective; thus \[ {\psi }_{ * } : {\operatorname{Hom}}_{\Lambda }\left( {R, B}\right) \rightarrow {\operatorname{Hom}}_{\Lambda }\left( {R,{B}^{\prime \prime }}\right) \] is epimorphic... | Yes |
Lemma 6.2. Let \( A \) be an abelian group of countable rank. If every subgroup of \( A \) of finite rank is free, then \( A \) is free. | Proof. By hypothesis there is a maximal countable linearly independent set \( T = \left( {{a}_{1},{a}_{2},\ldots ,{a}_{n},\ldots }\right) \) of elements of \( A \) . Let \( {A}_{n} \) be the subgroup of \( A \) consisting of all elements \( a \in A \) linearly dependent on \( \left( {{a}_{1},{a}_{2},\ldots ,{a}_{n}}\ri... | Yes |
Proposition 7.1. For any left \( \Lambda \) -module \( B, - { \otimes }_{A}B : {\mathfrak{M}}_{A}^{r} \rightarrow \mathfrak{{Ab}} \) is a covariant functor. For any right \( \Lambda \) -module \( A, A{ \otimes }_{A} - : {\mathfrak{M}}_{A}^{l} \rightarrow \mathfrak{{Ab}} \) is a covariant functor. Moreover, \( - { \otim... | The proof is left to the reader. | No |
For any right \( A \) -module \( A \), the functor \( A{ \otimes }_{A} - : {\mathfrak{M}}_{A}^{l} \rightarrow \mathfrak{{Ab}} \) is left adjoint to the functor \( {\operatorname{Hom}}_{\mathbb{Z}}\left( {A, - }\right) : \mathfrak{{Ab}} \rightarrow {\mathfrak{M}}_{A}^{l} \) . | The left-module structure of \( {\operatorname{Hom}}_{\mathbb{Z}}\left( {A, - }\right) \) is induced by the right-module structure of \( A \) (see Section I.8). We have to show that there is a natural transformation \( \eta \) such that for any abelian group \( G \) and any left \( \Lambda \) -module \( B \)\n\n\[ \eta... | No |
Proposition 7.3. (i) Let \( \left\{ {B}_{j}\right\}, j \in J \), be a family of left \( \Lambda \) -modules and let \( A \) be a right \( \Lambda \) -module. Then there is a natural isomorphism\n\n\[ A{ \otimes }_{\Lambda }\left( {{\bigoplus }_{j \in J}{B}_{j}}\right) \overset{ \sim }{ \rightarrow }{\bigoplus }_{j \in ... | Proof. By the dual of Theorem II.7.7 a functor possessing a right adjoint preserves coproducts and cokernels. | No |
Proposition 7.4. Every projective module is flat. | Proof. A projective module \( P \) is a direct summand in a free module. Hence, since \( A{ \otimes }_{A} \) - preserves sums, it suffices to show that free modules are flat. By the same argument it suffices to show that \( \Lambda \) as a left module is flat. But this is trivial since \( A{ \otimes }_{A}A \cong A \) .... | Yes |
Proposition 8.1. If \( A \) (or \( B \) ) is projective, then\n\n\[{\operatorname{Tor}}_{\varepsilon }^{A}\left( {A, B}\right) = 0 = {\overline{\operatorname{Tor}}}_{\eta }^{A}\left( {A, B}\right) . | Proof. Since \( A \) is projective, the short exact sequence \( R\overset{\mu }{ \rightarrow }P\overset{\varepsilon }{ \rightarrow }A \) splits, i.e. there is \( \kappa : P \rightarrow R \) with \( {\kappa \mu } = {1}_{R} \) . Hence\n\n\[{\kappa \mu } \otimes 1 = \left( {\kappa \otimes 1}\right) \left( {\mu \otimes 1}\... | Yes |
Theorem 8.2. Let \( A \) be a right \( A \) -module and \( {B}^{\prime }\overset{\kappa }{ \mapsto }B\overset{v}{ \rightarrow }{B}^{\prime \prime } \) an exact sequence of left \( \Lambda \) -modules, then there exists a connecting homomorphism \( \omega : {\operatorname{Tor}}^{A}\left( {A,{B}^{\prime \prime }}\right) ... | \[ {\operatorname{Tor}}^{A}\left( {A,{B}^{\prime }}\right) \overset{{\kappa }_{ \star }}{ \rightarrow }{\operatorname{Tor}}^{A}\left( {A, B}\right) \overset{{v}_{ \star }}{ \rightarrow }{\operatorname{Tor}}^{A}\left( {A,{B}^{\prime \prime }}\right) \overset{\omega }{ \rightarrow }A{ \otimes }_{A}{B}^{\prime } \] \[ {\x... | Yes |
Theorem 8.3. Let \( B \) be a left \( \Lambda \) -module and let \( {A}^{\prime }\overset{\kappa }{ \rightarrowtail }A\overset{v}{ \rightsquigarrow }{A}^{\prime \prime } \) be an exact sequence of right \( \Lambda \) -modules. Then there exists a connecting homomorphism \( \omega : {\operatorname{Tor}}^{A}\left( {{A}^{... | Proof. We only prove Theorem 8.2; the proof of Theorem 8.3 may be obtained by replacing Tor by Tor. Consider the projective presentation\n\n\( R\overset{\mu }{ \rightarrow }P\overset{\varepsilon }{ \rightarrow }A \) and construct the diagram:\n\n be a principal ideal domain. Then the homomorphisms \( {\kappa }_{ * } : {\operatorname{Tor}}^{A}\left( {A,{B}^{\prime }}\right) \rightarrow {\operatorname{Tor}}^{A}\left( {A, B}\right) \) in sequence (8.3) and \[ {\kappa }_{ * } : {\operatorname{Tor}}^{A}\left( {{A}^{\prime }, B}\right... | Proof. By Corollary I.5.3 \( R \) is a projective right \( \Lambda \) -module, hence the map \( {\kappa }_{ * } : R{ \otimes }_{A}{B}^{\prime } \rightarrow R{ \otimes }_{A}B \) in diagram (8.5) is monomorphic, whence the first assertion. Analogously one obtains the second assertion. | Yes |
Lemma 2.2. \( {\partial }_{n} : {C}_{n} \rightarrow {C}_{n - 1} \) induces \( {\widetilde{\partial }}_{n} : \operatorname{coker}{\partial }_{n + 1} \rightarrow \ker {\partial }_{n - 1} \) with \( \ker {\widetilde{\partial }}_{n} = {H}_{n}\left( \mathbf{C}\right) \) and \( \operatorname{coker}{\widetilde{\partial }}_{n}... | Proof. Since im \( {\partial }_{n + 1} \subseteq \ker {\partial }_{n} \) and \( \operatorname{im}{\partial }_{n} \subseteq \ker {\partial }_{n - 1} \) the differential \( {\partial }_{n} \) induces a map \( {\widetilde{\partial }}_{n} \) as follows:\n\n\[ \operatorname{coker}{\partial }_{n + 1} = {C}_{n}/\operatorname{... | Yes |
Proposition 3.1. If the two chain maps \( \varphi ,\psi : \mathbf{C} \rightarrow \mathbf{D} \) are homotopic, then \( H\left( \varphi \right) = H\left( \psi \right) : H\left( \mathbf{C}\right) \rightarrow H\left( \mathbf{D}\right) \) . | Proof. Let \( z \in \ker {\partial }_{n} \) be a cycle in \( {C}_{n} \) . If \( \sum : \varphi \rightarrow \psi \), then\n\n\[ \left( {\psi - \varphi }\right) z = \partial {\sum z} + \sum \partial z = \partial {\sum z} \]\n\nsince \( \partial z = 0 \) . Hence \( \psi \left( z\right) - \varphi \left( z\right) \) is a bo... | Yes |
Lemma 3.3. Let \( \varphi \simeq \psi : C \rightarrow D \) and \( {\varphi }^{\prime } \simeq {\psi }^{\prime } : D \rightarrow E \), then\n\n\[{\varphi }^{\prime }\varphi \simeq {\psi }^{\prime }\psi : C \rightarrow E.\] | Proof. Let \( \psi - \varphi = \partial \sum + \sum \partial \) ; then\n\n\[{\varphi }^{\prime }\psi - {\varphi }^{\prime }\varphi = {\varphi }^{\prime }\partial \sum + {\varphi }^{\prime }\sum \partial = \partial \left( {{\varphi }^{\prime }\sum }\right) + \left( {{\varphi }^{\prime }\sum }\right) \partial .\]\n\nAlso... | Yes |
Lemma 3.4. Let \( F : {\mathfrak{M}}_{A} \rightarrow {\mathfrak{M}}_{{A}^{\prime }} \) be an additive functor. If \( \mathbf{C} \) and \( \mathbf{D} \) are chain complexes of \( \Lambda \) -modules and \( \varphi \simeq \psi : \mathbf{C} \rightarrow \mathbf{D} \), then \( {F\varphi } \simeq {F\psi } : F\mathbf{C} \righ... | Proof. Let \( \sum : \mathbf{\varphi } \rightarrow \mathbf{\psi } \), then\n\n\[ \n{F\psi } - {F\varphi } = F\left( {\psi - \varphi }\right) = F\left( {\partial \sum + \sum \partial }\right) = F\partial {F\sum } + {F\sum F}\partial .\n\]\n\nHence \( F\mathbf{\sum } : F\mathbf{\varphi } \rightarrow F\mathbf{\psi } \) . | Yes |
Corollary 3.5. If \( \varphi \simeq \psi : C \rightarrow D \) and if \( F \) is an additive functor, then \( H\left( {F\varphi }\right) = H\left( {F\psi }\right) : H\left( {FC}\right) \rightarrow H\left( {FD}\right) \) . | We remark that Lemma 3.3 enables one to associate with the category of chain complexes and chain maps the category of chain complexes and homotopy classes of chain maps. The passage is achieved simply by identifying two chain maps if and only if they are homotopic. The category so obtained is called the homotopy catego... | No |
Lemma 4.2. To every \( \Lambda \) -module \( A \) there exists a projective resolution. | Proof. Choose a projective presentation \( {R}_{1} \rightarrowtail {P}_{0} \rightarrow A \) of \( A \) ; then a projective presentation \( {R}_{2} \rightarrowtail {P}_{1} \rightarrow {R}_{1} \) of \( {R}_{1} \), etc. Plainly the complex\n\n\[ \mathbf{P} : \cdots \rightarrow {P}_{n}\overset{{\partial }_{n}}{ \rightarrow... | Yes |
Proposition 4.3. Two projective resolutions of \( A \) are canonically of the same homotopy type. | Proof. Let \( \mathbf{C} \) and \( \mathbf{D} \) be two projective resolutions of \( A \) . By Theorem 4.1 there exist chain maps \( \varphi : \mathbf{C} \rightarrow \mathbf{D} \) and \( \psi : \mathbf{D} \rightarrow \mathbf{C} \) inducing the identity in \( {H}_{0}\left( \mathbf{C}\right) = A = {H}_{0}\left( \mathbf{D... | Yes |
Proposition 5.1. Let \( \mathbf{P},\mathbf{Q} \) be two projective resolutions of \( A \) . Then there is a canonical isomorphism\n\n\[ \n\eta = {\eta }_{\mathbf{P},\mathbf{Q}} : {L}_{n}^{\mathbf{P}}{TA} \rightarrow {L}_{n}^{\mathbf{Q}}{TA},\;n = 0,1,\ldots \n\] | Proof. Let \( \mathbf{\eta } : \mathbf{P} \rightarrow \mathbf{Q} \) be a chain map inducing \( {1}_{A} \) . Its homotopy class is uniquely determined; moreover it is clear from Proposition 4.3 that \( \eta \) is a homotopy equivalence. Hence we obtain a canonical isomorphism\n\n\[ \n\eta = {1}_{A}\left( {\mathbf{P},\ma... | Yes |
Proposition 5.2. Let \( T : {\mathfrak{M}}_{A} \rightarrow \mathfrak{A}\mathfrak{b} \) be right exact, then \( {L}_{0}T \) and \( T \) are naturally equivalent. | Proof. Let \( \mathbf{P} \) be a projective resolution of \( A \) . Then \( {P}_{1} \rightarrow {P}_{0} \rightarrow A \rightarrow 0 \) is exact. Hence \( T{P}_{1} \rightarrow T{P}_{0} \rightarrow {TA} \rightarrow 0 \) is exact. It follows that \( {H}_{0}\left( {T\mathbf{P}}\right) \cong {TA} \) . Plainly the isomorphis... | No |
Proposition 5.3. For \( P \) a projective \( \Lambda \) -module \( {L}_{n}{TP} = 0 \) for \( n = 1,2,\ldots \) and \( {L}_{0}{TP} = {TP} \) . | Proof. Clearly \( \mathbf{P} : \cdots \rightarrow 0 \rightarrow {P}_{0} \rightarrow 0 \) with \( {P}_{0} = P \) is a projective resolution of \( P \) . | No |
Proposition 5.4. The functors \( {L}_{n}T : {\mathfrak{M}}_{A} \rightarrow \mathfrak{{Ab}}, n = 0,1,\ldots \) are additive. | Proof. Let \( \mathbf{P} \) be a projective resolution of \( A \) and \( \mathbf{Q} \) a projective resolution of \( B \), then\n\n\[ \mathbf{P} \oplus \mathbf{Q} : \cdots \rightarrow {P}_{n} \oplus {Q}_{n} \rightarrow {P}_{n - 1} \oplus {Q}_{n - 1} \rightarrow \cdots \rightarrow {P}_{0} \oplus {Q}_{0} \rightarrow 0 \]... | Yes |
Proposition 5.5. Let \( {K}_{q}\overset{\mu }{ \rightarrow }{P}_{q - 1} \rightarrow {P}_{q - 2} \rightarrow \cdots \rightarrow {P}_{0} \rightarrow A \) be an exact sequence with \( {P}_{0},{P}_{1},\ldots ,{P}_{q - 1} \) projective. Then if \( T \) is right exact, and \( q \geqq 1 \), the sequence \[ 0 \rightarrow {L}_{... | Proof. Let \( \cdots \rightarrow {P}_{q + 1} \rightarrow {P}_{q} \rightarrow {K}_{q} \rightarrow 0 \) be an exact sequence with \( {P}_{q},{P}_{q + 1},\ldots \), projective. Then the complex  is a projective resoluti... | Yes |
Theorem 6.1. Let \( T : {\mathfrak{M}}_{A} \rightarrow \mathfrak{{Ab}} \) be an additive functor and let \( {A}^{\prime }\overset{{\alpha }^{\prime }}{ \rightarrow }A\overset{{\alpha }^{\prime \prime }}{ \rightarrow }{A}^{\prime \prime } \) be a short exact sequence. Then there exist connecting homomorphisms\n\n\[ \n{\... | Proof. By Lemma III. 5.4 we can construct a diagram with exact rows\n\n\n\nwith \( {P}_{0}^{\prime },{P}_{0},{P}_{0}^{\prime \prime } \) projective. Clearly, \( {P}_{0} = {P}_{0}^{\prime } \oplus {P}_{0}^{\prime \pri... | Yes |
Proposition 6.2. Let \( \tau : T \rightarrow {T}^{\prime } \) be a natural transformation between additive covariant functors \( T,{T}^{\prime } : {\mathfrak{M}}_{A} \rightarrow \mathfrak{{Ab}} \) and let the diagram be commutative with short exact rows. Then the following diagrams are commutative: ![d528cc54-d89e-46eb... | The proof is left to the reader. | No |
Theorem 6.3. Let the sequence \( {T}^{\prime }\overset{{\tau }^{\prime }}{ \rightarrow }T\overset{{\tau }^{\prime \prime }}{ \rightarrow }{T}^{\prime \prime } \) of additive functors \( {T}^{\prime }, T,{T}^{\prime \prime } : {\mathfrak{M}}_{A} \rightarrow \mathfrak{{Ab}} \) be exact on projectives. Then, for every \( ... | Proof. Choose a projective resolution \( \mathbf{P} \) of \( A \) and consider the sequence of complexes \[ 0 \rightarrow {T}^{\prime }P\overset{{\tau }^{\prime }}{ \rightarrow }{TP}\overset{{\tau }^{\prime \prime }}{ \rightarrow }{T}^{''}P \rightarrow 0 \] which is short exact since \( {T}^{\prime }\overset{{\tau }^{\... | Yes |
Proposition 6.4. Let \( \alpha : A \rightarrow {A}^{\prime } \) be a homomorphism of \( A \) -modules and let\n\n\n\nbe a commutative diagram of additive functors and natural transformations such that the rows are ex... | The proof is left to the reader. | No |
Proposition 7.1. \( {\operatorname{Ext}}_{A}^{1}\left( {A, B}\right) \cong {\operatorname{Ext}}_{A}\left( {A, B}\right) \) . | Proof. We consider the projective presentation \( {R}_{1} \rightarrowtail {P}_{0}\overset{\varepsilon }{ \rightsquigarrow }A \) of \( A \) and apply Proposition 5.8. We obtain the exact sequence\n\n\[ \n\cdots \rightarrow {\operatorname{Hom}}_{A}\left( {{P}_{0}, B}\right) \rightarrow {\operatorname{Hom}}_{A}\left( {{R}... | Yes |
If \( P \) is projective and if \( I \) is injective, then \[ {\operatorname{Ext}}_{A}^{n}\left( {P, B}\right) = 0 = {\operatorname{Ext}}_{A}^{n}\left( {A, I}\right) \;\text{ for }\;n = 1,2,\ldots . \] | The first assertion is immediate by Proposition 5.3. To prove the second assertion, we merely remark that \( {\operatorname{Hom}}_{A}\left( {-, I}\right) \) is an exact functor, so that its \( {n}^{\text{th }} \) derived functor is zero for \( n \geqq 1 \) . | Yes |
Proposition 7.4. Let \( {B}^{\prime }\overset{{\beta }^{\prime }}{ \rightarrow }B\overset{{\beta }^{\prime \prime }}{ \rightarrow }{B}^{\prime \prime } \) be a short exact sequence, then the sequence \( {\operatorname{Hom}}_{A}\left( {-,{B}^{\prime }}\right) \overset{{\beta }_{ * }}{ \rightarrow }{\operatorname{Hom}}_{... | This is trivial. | No |
Proposition 8.2. For any \( A \) and any short exact sequence \( {B}^{\prime } \rightarrowtail B \rightarrow {B}^{\prime \prime } \) the following square is commutative | Proof. Choose an injective presentation \( {B}^{\prime } \rightarrowtail I \rightarrow S \) of \( {B}^{\prime } \) and construct \( \varphi ,\psi \) such that the diagram\n\n\n\nis commutative. We then embed (8.7) as... | Yes |
Proposition 10.3. Every natural transformation\n\n\[ \Phi : {\operatorname{Ext}}_{A}^{1}\left( {A, - }\right) \rightarrow {\operatorname{Ext}}_{A}^{1}\left( {{A}^{\prime }, - }\right) \]\n\nis induced by a homomorphism \( \alpha : {A}^{\prime } \rightarrow A \) . | Proof. Since \( {\operatorname{Ext}}_{A}^{1}\left( {{A}^{\prime }, - }\right) \) is additive we may apply Theorem 10.1. We have that \( \left\lbrack {{\operatorname{Ext}}_{A}^{1}\left( {A, - }\right) ,{\operatorname{Ext}}_{A}^{1}\left( {{A}^{\prime }, - }\right) }\right\rbrack \overset{ \sim }{ \rightarrow }{\widetilde... | Yes |
Theorem 11.2. There are natural isomorphisms\n\n\[ \n\Gamma : \left\lbrack {{\operatorname{Ext}}_{A}^{n}\left( {B, - }\right), A{ \otimes }_{A} - }\right\rbrack \widetilde{ \rightarrow }{\operatorname{Tor}}_{n}^{A}\left( {A, B}\right) \;\text{ for }\;n = 0,1,\ldots \n\] | Proof. We only have to observe that \( A{ \otimes }_{A} - \) is right exact and that \( {\operatorname{Tor}}_{n}^{A}\left( {A, B}\right) = {L}_{n}\left( {A{ \otimes }_{A} - }\right) \left( B\right) \) . The assertion then follows from Corollary 10.2. \( ▱ \) | No |
Theorem 12.1. (i) If \( U \) has a left adjoint \( F : {\mathfrak{M}}_{\Lambda } \rightarrow {\mathfrak{M}}_{{\Lambda }^{\prime }} \) and if \( U \) preserves surjections (i.e., if \( U \) is exact), then \( F \) sends projectives to projectives. | Now let \( U \) satisfy the hypotheses of Theorem 12.1 (i), let \( A \) be a \( \Lambda \) -module and let \( {B}^{\prime } \) be a \( {\Lambda }^{\prime } \) -module. Choose a projective resolution\n\n\[ P : \cdots \rightarrow {P}_{n} \rightarrow {P}_{n - 1} \rightarrow \cdots \rightarrow {P}_{0} \]\n\nof \( A \), and... | Yes |
Proposition 12.2. If \( {\Lambda }^{\prime } \) is flat as a right \( \Lambda \) -module via \( \gamma \), then\n\n\[ \Phi : {\operatorname{Ext}}_{{A}^{\prime }}^{n}\left( {{A}^{\prime }{ \otimes }_{A}A,{B}^{\prime }}\right) \overset{ \sim }{ \rightarrow }{\operatorname{Ext}}_{A}^{n}\left( {A,{U}^{\gamma }{B}^{\prime }... | Proof. This is clear since the functor \( F \) given by (12.4) is then exact. | No |
Theorem 12.5. Let \( {U}^{\gamma } : {\mathfrak{M}}_{{A}^{\prime }} \rightarrow {\mathfrak{M}}_{A} \) be the change-of-rings functor induced by \( \gamma : \Lambda \rightarrow {\Lambda }^{\prime } \) . Then (i) if \( {\Lambda }^{\prime } \) is a projective (left) \( \Lambda \) -module via \( \gamma \) , \( {U}^{\gamma ... | Proof. (i) The hypothesis implies that \( \bar{F} \) preserves epimorphisms, so \( {U}^{\gamma } \) sends projectives to projectives. (ii) The hypothesis implies that \( F \) preserves monomorphisms, so \( {U}^{\gamma } \) sends injectives to injectives. | Yes |
Theorem 1.1. Let \( \\mathbf{C} \) be a chain complex of right \( \\Lambda \) -modules, \( \\mathbf{D} \) a chain complex of left \( \\Lambda \) -modules and \( \\mathbf{E} \) a chain complex of abelian groups. Then there is a natural isomorphism of chain complexes of abelian groups\n\n\[ \n\\operatorname{Hom}\\left( {... | Proof. We have already observed the basic adjointness relation (Theorem III.7.2)\n\n\[ \n\\operatorname{Hom}\\left( {{C}_{-p}{ \\otimes }_{\\Lambda }{D}_{-q},{E}_{r}}\\right) \\cong {\\operatorname{Hom}}_{\\Lambda }\\left( {{C}_{-p},\\operatorname{Hom}\\left( {{D}_{-q},{E}_{r}}\\right) }\\right) .\n\]\n\n(1.7)\n\nThis ... | Yes |
Lemma 2.2. Let \( \mathbf{H} \) be a graded module over the p.i.d. A. Then there exists a free chain complex \( \mathbf{C} \) over \( \Lambda \) such that \( \mathbf{H}\left( \mathbf{C}\right) \cong \mathbf{H} \) . | Proof. Let \( 0 \rightarrow {R}_{p} \rightarrow {F}_{p} \rightarrow {H}_{p} \rightarrow 0 \) be a free presentation of \( {H}_{p} \) . Set\n\n\[ \n{C}_{p} = {F}_{p} \oplus {R}_{p - 1} \]\n\n\[ \n\partial \left( {x, y}\right) = \left( {y,0}\right) ,\;x \in {F}_{p},\;y \in {R}_{p - 1}. \]\n\nThen \( \partial \partial = 0... | Yes |
Lemma 2.3. Let \( \\mathbf{C},\\mathbf{D} \) be chain complexes over the p.i.d. \( \\Lambda \) and let \( \\mathbf{C} \) be free. Let \( \\psi : H\\left( \\mathbf{C}\\right) \\rightarrow H\\left( \\mathbf{D}\\right) \) be a homomorphism. Then there exists a chain map \( \\varphi : \\mathbf{C} \\rightarrow \\mathbf{D} \... | Proof. Consider \( 0 \\rightarrow {B}_{p} \\rightarrow {Z}_{p} \\rightarrow {H}_{p} \\rightarrow 0,{C}_{p}\\overset{{\\partial }_{p}}{ \\rightarrow }{B}_{p - 1} \), where everything relates to the chain complex \( \\mathbf{C} \). Since \( {B}_{p - 1} \) is free, it follows that \( {C}_{p} = {Z}_{p} \\oplus {Y}_{p} \), ... | Yes |
Theorem 2.5. (Universal coefficient theorem in homology.) Let \( A \) be a p.i.d., let \( \mathbf{C} \) be a flat chain complex over \( \Lambda \) and let \( A \) be a \( \Lambda \) -module. Then there is a natural short exact sequence \[ 0 \rightarrow {H}_{n}\left( \mathbf{C}\right) { \otimes }_{A}A\overset{\zeta }{ \... | Proof. The only part of the assertion requiring proof is the final phrase. That the splitting is unnatural in \( \mathbf{C} \) is attested by the example given to prove the unnaturality of the splitting of (2.1). Thus it remains to prove the naturality of the splitting of (2.13) in the variable \( A \) . If \( \mathbf{... | Yes |
Lemma 3.2. If \( \mathbf{D} \) is free, the construction of \( \mathbf{\varphi } \) from \( \mathbf{\psi } \) in Lemma 2.3 induces a homomorphism\n\n\[ \theta : \mathop{\prod }\limits_{{q - p = n}}{\operatorname{Hom}}_{\Lambda }\left( {{H}_{p}\left( \mathbf{C}\right) ,{H}_{q}\left( \mathbf{D}\right) }\right) \rightarro... | Proof. It is plain that the only assertion to be established is that the homology class of \( \varphi \) in \( {H}_{n}\left( {{\operatorname{Hom}}_{A}\left( {\mathbf{C},\mathbf{D}}\right) }\right) \) is independent of the choice of \( {\varphi }^{1} \) . Consider therefore a family of morphisms\n\n\[ {\alpha }_{-p, q} ... | Yes |
Theorem 3.3. (Universal coefficient theorem in cohomology.) Let \( \Lambda \) be a p.i.d., let \( \mathbf{C} \) be a free chain complex over \( \Lambda \), and let \( B \) be a \( \Lambda \) -module. Then there is a natural short exact sequence\n\n\[ 0 \rightarrow {\operatorname{Ext}}_{A}\left( {{H}_{n - 1}\left( C\rig... | Proof. Only a few remarks are required. First, the notation\n\n\[ {H}^{n}\left( {{\operatorname{Hom}}_{A}\left( {C, B}\right) }\right) \]\n\nis unambiguous, since the cohomology modules of the cochain complex \( \left( {{\operatorname{Hom}}_{A}\left( {C, B}\right) ,\operatorname{Hom}\left( {\partial ,1}\right) }\right)... | No |
Proposition 4.1. Let \( {C}^{\prime }, C,{C}^{\prime \prime } \) be chain complexes of abelian groups. Then there is a natural isomorphism\n\n\[ \n\left( {{\mathbf{C}}^{\prime } \otimes \mathbf{C}}\right) \otimes {\mathbf{C}}^{\prime \prime } \cong {\mathbf{C}}^{\prime } \otimes \left( {\mathbf{C} \otimes {\mathbf{C}}^... | First, we consider (4.1). We take \( {\mathbf{C}}^{\prime },\mathbf{C},{\mathbf{C}}^{\prime \prime } \) to be resolutions of abelian groups \( {A}^{\prime }, A,{A}^{\prime \prime } \) . Thus, for example \( {C}_{1} = R,{C}_{0} = F,{C}_{p} = 0, p \neq 0,1 \), and \( {\partial }_{1} \) is the inclusion \( R \subseteqq F ... | Yes |
Theorem 4.2. Let \( {A}^{\prime }, A,{A}^{\prime \prime } \) be abelian groups. There is then an unnatural isomorphism\n\n\[ \n\operatorname{Tor}\left( {{A}^{\prime }, A}\right) \otimes {A}^{\prime \prime } \oplus \operatorname{Tor}\left( {{A}^{\prime } \otimes A,{A}^{\prime \prime }}\right) \cong {A}^{\prime } \otimes... | Proof. We simply show why (4.4) is natural. A homomorphism \( \varphi : A \rightarrow B \) induces a unique homotopy class of chain maps \( \varphi : C\left( A\right) \rightarrow C\left( B\right) \) , where \( \mathbf{C}\left( A\right) ,\mathbf{C}\left( B\right) \) are resolutions of \( A, B \) . Thus from \( {\varphi ... | Yes |
Corollary 4.4. If \( A \) is torsion-free, then \( \operatorname{Ext}\left( {A, B}\right) \) is divisible, for all \( B \) . | Proof. It follows from (4.6) that, if \( A \) is torsion-free, then\n\n\[ \operatorname{Ext}\left( {{A}^{\prime },\operatorname{Ext}\left( {A, B}\right) }\right) = 0 \]\n\nfor all \( {A}^{\prime }, B \) . This means that \( \operatorname{Ext}\left( {A, B}\right) \) is injective, that is, divisible, for all \( B \) . | Yes |
Corollary 4.7. If \( \operatorname{Ext}\left( {A,\mathbb{Z}}\right) = 0,\operatorname{Hom}\left( {A,\mathbb{Z}}\right) = 0 \), then \( A = 0 \) . | Proof. By (4.5) we infer \( \operatorname{Ext}\left( {{A}^{\prime } \otimes A,\mathbb{Z}}\right) = 0 \) for all \( {A}^{\prime } \) . Now, since \( \operatorname{Ext}\left( {A,\mathbb{Z}}\right) = 0, A \) is torsion-free. Thus if \( A \neq 0 \), take \( {A}^{\prime } = \mathbb{Q} \) . Then \( \mathbb{Q} \otimes A \) is... | Yes |
Corollary 4.8. There is no abelian group \( A \) such that \( \operatorname{Ext}\left( {A,\mathbb{Z}}\right) = \mathbb{Q} \) , \( \operatorname{Hom}\left( {A,\mathbb{Z}}\right) = 0 \) . | Proof. Since \( \operatorname{Ext}\left( {A,\mathbb{Z}}\right) = \mathbb{Q}, A \) is a non-zero torsion-free group. Again by (4.5) we infer\n\n\[ \operatorname{Ext}\left( {\mathbb{Q} \otimes A,\mathbb{Z}}\right) \cong \operatorname{Hom}\left( {\mathbb{Q},\mathbb{Q}}\right) \]\n\nBut Hom \( \left( {\mathbb{Q},\mathbb{Q}... | Yes |
Proposition 1.1. Let \( R \) be a ring. To any function \( f : G \rightarrow R \) with \( f\left( {xy}\right) = f\left( x\right) \cdot f\left( y\right) \) and \( f\left( 1\right) = {1}_{R} \) there exists a unique ring homomorphism \( {f}^{\prime } : \mathbb{Z}G \rightarrow R \) such that \( {f}^{\prime }i = f \) . | Proof. We define \( {f}^{\prime }\left( {\mathop{\sum }\limits_{{x \in G}}m\left( x\right) x}\right) = \mathop{\sum }\limits_{{x \in G}}m\left( x\right) f\left( x\right) \) which obviously is the only ring homomorphism for which \( {f}^{\prime }i = f \) . | Yes |
Lemma 1.2. (i) As an abelian group IG is free on the set\n\n\[ \nW = \{ x - 1 \mid 1 \neq x \in G\} \n\] | Proof. (i) Clearly, the set \( W \) is linearly independent. We have to show that it generates \( {IG} \) . Let \( \mathop{\sum }\limits_{{x \in G}}m\left( x\right) x \in {IG} \), then \( \mathop{\sum }\limits_{{x \in G}}m\left( x\right) = 0 \) . Hence \( \mathop{\sum }\limits_{{x \in G}}m\left( x\right) x = \mathop{\s... | Yes |
Lemma 1.3. Let \( U \) be a subgroup of \( G \) . Then \( \mathbb{Z}G \) is free as left (or right) U-module. | Proof. Choose \( \left\{ {x}_{i}\right\} ,{x}_{i} \in G \), a system of representatives of the left cosets of \( U \) in \( G \) . The underlying set of \( G \) may be regarded as the disjoint union of the sets \( {x}_{i}U \) . Clearly, the part of \( \mathbb{Z}G \) linearly spanned by \( {x}_{i}U \) for fixed \( i \) ... | Yes |
Proposition 3.1. Let \( A, B \) be \( G \) -modules. Then\n\n\[ \n{H}^{0}\left( {G, A}\right) = {A}^{G},\;{H}_{0}\left( {G, B}\right) = {B}_{G}.\n\]\n\nIf \( A, B \) are trivial \( G \) -modules, then\n\n\[ \n{H}^{0}\left( {G, A}\right) = A,\;{H}_{0}\left( {G, B}\right) = B.\n\] | Proof. It is immediate that, in case the \( G \) -action is trivial, \( {A}^{G} = A \) and \( {B}_{G} = B \) . | No |
Lemma 4.1. \( \mathbb{Z}{ \otimes }_{G}{IG} = {IG}/{\left( IG\right) }^{2} \cong {G}_{ab} \). | Proof. The first equality is already proved, so we have only to show that \( {IG}/{\left( IG\right) }^{2} \cong {G}_{ab} \). By Lemma 1.2 the abelian group \( {IG} \) is free on \( W = \{ x - 1 \mid 1 \neq x \in G\} \). The function \( \psi : W \rightarrow G/{G}^{\prime } \) defined by\n\n\[ \psi \left( {x - 1}\right) ... | Yes |
Theorem 5.1. The homomorphism \( \eta : \operatorname{Der}\left( {G, A}\right) \rightarrow {\operatorname{Hom}}_{G}\left( {{IG}, A}\right) \) defined by\n\n\[ \left( {\eta \left( d\right) }\right) \left( {y - 1}\right) = d\left( y\right) ,\;y \in G \]\n\nis a natural isomorphism. | Proof. Given a derivation \( d : G \rightarrow A \), we claim that the group homomorphism \( \eta \left( d\right) = {\varphi }_{d} : {IG} \rightarrow A \) defined by \( {\varphi }_{d}\left( {y - 1}\right) = {dy}, y \in G \), is a \( G \) - module homomorphism. Indeed\n\n\[ {\varphi }_{d}\left( {x\left( {y - 1}\right) }... | Yes |
Proposition 5.3. Suppose given a group \( G \) and a \( G \) -module \( A \) . To every group homomorphism \( f : X \rightarrow G \) and to every \( f \) -derivation \( d : X \rightarrow A \) (i.e. \( d \) is a derivation if \( A \) is regarded as an \( X \) -module via \( f \) ), there exists a unique group homomorphi... | The proof is obvious; \( h \) is defined by \( {hx} = \left( {{dx},{fx}}\right), x \in X \), and it is straightforward to check that \( h \) is a homomorphism. | No |
Theorem 5.5. The augmentation ideal IF of a group \( F \) which is free on the set \( S \) is the free \( \mathbb{Z}F \) -module on the set \( S - 1 = \{ s - 1 \mid s \in S\} \) . | Proof. We show that any function \( f \) from the set \( \{ s - 1 \mid s \in S\} \) into an \( F \) -module \( M \) may be uniquely extended to an \( F \) -module homomorphism \( {f}^{\prime } : {IF} \rightarrow M \) . First note that uniqueness is clear, since \( \{ s - 1 \mid s \in S\} \) generates \( {IF} \) as \( F... | Yes |
Corollary 5.6. For a free group \( F \), we have\n\n\[ \n{H}^{n}\left( {F, A}\right) = 0 = {H}_{n}\left( {F, B}\right)\n\]\n\nfor all \( F \) -modules \( A, B \) and all \( n \geqq 2 \) . | Proof. I \( F \rightarrowtail \mathbb{Z}F \rightarrow \mathbb{Z} \) is an \( F \) -free resolution of \( \mathbb{Z} \) . | Yes |
Lemma 6.1. If \( N\overset{\iota }{ \rightarrow }G\overset{\nu }{ \rightarrow }Q \) is an exact sequence of groups, then \( \mathbb{Z}{ \otimes }_{N}\mathbb{Z}G \cong \mathbb{Z}Q \) as right \( G \) -modules. | Proof. As abelian group \( \mathbb{Z}{ \otimes }_{N}\mathbb{Z}G \) is free on the set of right cosets \( G/N \cong Q \) . It is easy to see that the right action of \( G \) induced by the product in \( \mathbb{Z}G \) is the right \( G \) -action in \( \mathbb{Z}Q \) via \( p \) . | Yes |
Lemma 6.2. If \( N \rightarrow G \rightarrow Q \) is an exact sequence of groups and if \( A \) is a left \( G \) -module, then \( {\operatorname{Tor}}_{n}^{N}\left( {\mathbb{Z}, A}\right) \cong {\operatorname{Tor}}_{n}^{G}\left( {\mathbb{Z}Q, A}\right) \) . | Proof. The argument that follows applies, in generalized form, to a change of rings (see Proposition IV. 12.2). Let \( \mathbf{X} \) be a \( G \) -projective resolution of \( A \), hence by Corollary 1.4 also an \( N \) -projective resolution of \( A \) . By Lemma \( {6.1},\mathbb{Z}{ \otimes }_{N}X \cong \mathbb{Z}{ \... | No |
Theorem 6.3. Let \( N \rightarrowtail G \rightarrow Q \) be an exact sequence of groups. Then\n\n\[ 0 \rightarrow {N}_{ab}\overset{\kappa }{ \rightarrow }\mathbb{Z}Q{ \otimes }_{G}{IG}\overset{v}{ \rightarrow }{IQ} \rightarrow 0 \]\n\n(6.2)\n\nis an exact sequence of \( Q \) -modules. | For our applications of (6.2) we shall need an explicit description of the \( Q \) -module structure in \( {N}_{ab} = N/{N}^{\prime } \), as well as of the map\n\n\[ \kappa : {N}_{ab} \rightarrow \mathbb{Z}Q{ \otimes }_{G}{IG}. \]\n\nFor that we compute \( {\operatorname{Tor}}_{1}^{N}\left( {\mathbb{Z},\mathbb{Z}}\righ... | Yes |
Corollary 6.4. Let \( R \rightarrow F \rightarrow Q \) be an exact sequence of groups with \( F \) a free group, i.e. a free presentation of \( Q \) . Then\n\n\[ 0 \rightarrow {R}_{ab}\overset{\kappa }{ \rightarrow }\mathbb{Z}Q{ \otimes }_{F}{IF}\overset{v}{ \rightarrow }{IQ} \rightarrow 0 \]\n\nis a \( Q \) -free pres... | Proof. By Theorem 5.5 IF is \( F \) -free, therefore \( \mathbb{Z}Q{ \otimes }_{F}{IF} \) is \( Q \) -free. | No |
Let \( R \rightarrow F \rightarrow Q \) be a free presentation of \( Q \). Then for any \( Q \)-modules \( A, B \) and all \( n \geqq 3 \)\n\n\[ \n{H}_{n}\left( {Q, B}\right) \cong {\operatorname{Tor}}_{n - 1}^{Q}\left( {B,{IQ}}\right) \cong {\operatorname{Tor}}_{n - 2}^{Q}\left( {B,{R}_{ab}}\right) ,\n\]\n\n\[ \n{H}^{... | Proof. The exact sequences \( {IQ} \rightarrowtail \mathbb{Z}Q \rightarrow \mathbb{Z} \) and (6.6) together with (2.7) give the result. | No |
Theorem 8.1. Let \( N \rightarrowtail G \rightarrow Q \) be an exact sequence of groups. For Q-modules \( A, B \) the following sequences are exact (and natural)\n\n\[ \n{H}_{2}\left( {G, B}\right) \rightarrow {H}_{2}\left( {Q, B}\right) \rightarrow B{ \otimes }_{Q}{N}_{ab} \rightarrow B{ \otimes }_{G}{IG} \rightarrow ... | Proof. We only prove the first of the two sequences, the cohomology sequence being proved similarly, using in addition the natural isomorphisms \( \operatorname{Der}\left( {G, A}\right) \cong {\operatorname{Hom}}_{G}\left( {{IG}, A}\right) ,\operatorname{Der}\left( {Q, A}\right) \cong {\operatorname{Hom}}_{Q}\left( {{I... | No |
Theorem 9.1. Let \( f : G \rightarrow H \) be a group homomorphism such that the induced homomorphism \( {f}_{ * } : {G}_{ab} \rightarrow {H}_{ab} \) is an isomorphism, and that\n\n\[ \n{f}_{ * } : {H}_{2}\left( G\right) \rightarrow {H}_{2}\left( H\right)\n\]\n\nis an epimorphism. Then \( f \) induces isomorphisms\n\n\... | Proof. We proceed by induction. For \( n = 0,1 \) the assertion is trivial or part of the hypotheses. For \( n \geqq 2 \) consider the exact sequences\n\n\[ \n{G}_{n - 1} \rightarrowtail G \rightarrow G/{G}_{n - 1},\;{H}_{n - 1} \rightarrowtail H \rightarrow H/{H}_{n - 1}\n\]\n\nand the associated 5-term sequences in h... | Yes |
Corollary 9.2. Let \( f : G \rightarrow H \) satisfy the hypotheses of Theorem 9.1. Suppose further that \( G, H \) are nilpotent. Then \( f \) is an isomorphism, \( f : G\widetilde{ \rightarrow }H \) . | Proof. The assertion follows from Theorem 9.1 and the remark that there exists \( n \geqq 0 \) such that \( {G}_{n} = \{ 1\} \) and \( {H}_{n} = \{ 1\} \) . | No |
Proposition 10.1. The map \( \Delta : M\left( {G, A}\right) \rightarrow {H}^{2}\left( {G, A}\right) \) is surjective. | Proof. Since \( \sigma \) in (10.6) is surjective, it suffices to show that every \( G \) -module homomorphism \( \varphi : {R}_{ab} \rightarrow A \) arises from a diagram of the form (10.5). In other words we have to fill in the diagram\n\n \) and the set \( M\left( {G, A}\right) \) of equivalence classes of extensions of \( G \) by \( A \) . The set \( M\left( {G, A}\right) \) has therefore a natural abelian group structure and\n\n\[ M\left( {G, - }\right) : {\mathf... | Note that, if \( A \) is a trivial \( G \) -module, then \( M\left( {G, A}\right) \) is the set of equivalence classes of central extensions of \( G \) by \( A \), i.e., extensions \( A \rightarrow E \rightarrow G \) with \( A \) a central subgroup of \( E \) .\n\nWe conclude this section with the observation that the ... | Yes |
Proposition 11.1. If \( B \) is an induced \( G \) -module, then \( {H}_{n}\left( {G, B}\right) = 0 \) for \( n \geqq 1 \) . | Proof. Let \( \mathbf{P} \) be a \( G \) -projective resolution of \( \mathbb{Z} \) . The homology of \( G \) with coefficients in \( B \) is the homology of the complex \( B{ \otimes }_{G}\mathbf{P} \) . Since \( B \cong X \otimes \mathbb{Z}G \) for a certain abelian group \( X \), we have \( B{ \otimes }_{G}\mathbf{P... | Yes |
Proposition 11.2. A direct sum \( B = {\bigoplus }_{i \in I}{B}_{i} \) is relative projective if and only if each \( {B}_{i}, i \in I \), is relative projective. | The proof is immediate from the definition. | No |
Lemma 11.7. Let \( A \) be a \( G \) -module. Then the \( G \) -modules \( {A}^{\prime } = \mathbb{Z}G \otimes A \) and \( {A}^{\prime \prime } = \mathbb{Z}G \otimes {A}_{0} \) are isomorphic. | Proof. We define a homomorphism \( \varphi : {A}^{\prime } \rightarrow {A}^{\prime \prime } \) by\n\n\[ \varphi \left( {x \otimes a}\right) = x \otimes \left( {{x}^{-1}a}\right) ,\;x \in G,\;a \in A. \]\n\nPlainly, \( \varphi \) respects the \( G \) -module structures and has a two-sided inverse \( \psi : {A}^{\prime \... | Yes |
Lemma 11.9. Let \( A \) be a left \( G \) -module. Then the \( G \) -modules\n\n\[ \n{A}^{\prime } = \operatorname{Hom}\left( {\mathbb{Z}G, A}\right)\n\]\n\nand \( {A}^{\prime \prime } = \operatorname{Hom}\left( {\mathbb{Z}G,{A}_{0}}\right) \) are isomorphic. | Proof. We define \( \varphi : {A}^{\prime } \rightarrow {A}^{\prime \prime } \) by \( \left( {\varphi \left( \alpha \right) }\right) \left( x\right) = {x}^{-1}\left( {\alpha \left( x\right) }\right), x \in G,\alpha : \mathbb{Z}G \rightarrow A \) . We verify that \( \varphi \) is a homomorphism of \( G \) -modules:\n\n\... | Yes |
Theorem 12.1. For \( n \geqq 2 \) we have\n\n\[ \n{H}_{n}\left( {G, B}\right) \cong {H}_{n - 1}\left( {G, B \otimes {IG}}\right) ,\n\]\n\n\[ \n{H}^{n}\left( {G, A}\right) \cong {H}^{n - 1}\left( {G,\operatorname{Hom}\left( {{IG}, A}\right) }\right) ,\n\]\n\nwhere \( B \otimes {IG} \) and \( \operatorname{Hom}\left( {{I... | Proof. We only prove the cohomology part of this theorem. Consider the short exact sequence of \( G \) -module homomorphisms (see Exercise 11.7)\n\n\[ \n\operatorname{Hom}\left( {\mathbb{Z}, A}\right) \rightarrowtail \operatorname{Hom}\left( {\mathbb{Z}G, A}\right) \rightarrow \operatorname{Hom}\left( {{IG}, A}\right) ... | No |
Theorem 12.2. Let \( G \cong F/R \) with \( F \) free. For \( n \geqq 3 \), we have\n\n\[ \n{H}_{n}\left( {G, B}\right) \cong {H}_{n - 2}\left( {G, B \otimes {R}_{ab}}\right) ,\n\]\n\n\[ \n{H}^{n}\left( {G, A}\right) \cong {H}^{n - 2}\left( {G,\operatorname{Hom}\left( {{R}_{ab}, A}\right) }\right) ,\n\]\n\nwhere \( B \... | Proof. Again we only prove the cohomology part. By Corollary 6.4 we have the following short exact sequence of \( G \) -module homomorphisms\n\n\[ \n\operatorname{Hom}\left( {{IG}, A}\right) \rightarrowtail \operatorname{Hom}\left( {\mathbb{Z}G{ \otimes }_{F}{IF}, A}\right) \rightarrow \operatorname{Hom}\left( {{R}_{ab... | Yes |
Lemma 14.1. Let \( G = {G}_{1} * {G}_{2} \) . Then there is a natural isomorphism\n\n\[ \n{IG} \cong \left( {\mathbb{Z}G{ \otimes }_{{G}_{1}}I{G}_{1}}\right) \oplus \left( {\mathbb{Z}G{ \otimes }_{{G}_{2}}I{G}_{2}}\right) .\n\] | Proof. First we claim that for all \( G \) -modules \( A \) there is a natural isomorphism\n\n\[ \n\operatorname{Der}\left( {G, A}\right) \cong \operatorname{Der}\left( {{G}_{1}, A}\right) \oplus \operatorname{Der}\left( {{G}_{2}, A}\right) .\n\]\n\nClearly, by restriction, a derivation \( d : G \rightarrow A \) gives ... | Yes |
Theorem 14.2. Let \( G = {G}_{1} * {G}_{2}, A \) a left \( G \) -module, \( B \) a right \( G \) -module. Then for \( n \geqq 2 \)\n\n\[ \n{H}^{n}\left( {G, A}\right) \cong {H}^{n}\left( {{G}_{1}, A}\right) \oplus {H}^{n}\left( {{G}_{2}, A}\right) ,\n\]\n\n\[ \n{H}_{n}\left( {{G}_{1}, B}\right) \oplus {H}_{n}\left( {{G... | Proof. We only prove the cohomology part of the assertion. For \( n \geqq 2 \) we have, by (6.7),\n\n\[ \n{H}^{n}\left( {G, A}\right) \cong {\operatorname{Ext}}_{G}^{n - 1}\left( {{IG}, A}\right)\n\]\n\n\[ \n\cong {\operatorname{Ext}}_{G}^{n - 1}\left( {\mathbb{Z}G{ \otimes }_{{G}_{1}}I{G}_{1}, A}\right) \oplus {\opera... | Yes |
Theorem 15.1. Let \( G \) be a group and let \( C \) be an abelian group considered as a trivial G-module. Then the following sequences are exact and natural, for every \( n \geqq 0 \) , \[ {H}_{n}\left( G\right) \otimes C \rightarrowtail {H}_{n}\left( {G, C}\right) \rightarrow \operatorname{Tor}\left( {{H}_{n - 1}\lef... | Proof. Let \( \mathbf{P} \) be a \( G \) -free (or \( G \) -projective) resolution of \( \mathbb{Z} \) . Tensoring over \( G \) with \( \mathbb{Z} \) yields \( {\mathbf{P}}_{G} = \mathbf{P}{ \otimes }_{G}\mathbb{Z} \), which is a complex of free abelian groups. Also, plainly, \( \mathbf{P}{ \otimes }_{G}C \cong {\mathb... | Yes |
Theorem 15.2. Let \( {G}_{i}, i = 1,2 \) be two groups, and let \( G = {G}_{1} \times {G}_{2} \) be their direct product. Then the following sequence is exact: | \[ {\bigoplus }_{p + q = n}{H}_{p}\left( {G}_{1}\right) \otimes {H}_{q}\left( {G}_{2}\right) \rightarrow {H}_{n}\left( G\right) \rightarrow {\bigoplus }_{p + q = n - 1}\operatorname{Tor}\left( {{H}_{p}\left( {G}_{1}\right) ,{H}_{q}\left( {G}_{2}\right) }\right) . \] Moreover the sequence splits by an unnatural splittin... | No |
Proposition 16.1. Let \( U \) be a subgroup of \( G \), and let \( A \) be a \( G \) -module. Denote by \( K \) the kernel of \( {\varepsilon }^{\prime } : \mathbb{Z}G{ \otimes }_{U}\mathbb{Z} \rightarrow \mathbb{Z} \) in (16.2). Then the following sequence is exact :\n\n\[ \cdots \rightarrow {\operatorname{Ext}}_{G}^{... | Note that, in case \( U \) is normal in \( G \) with quotient group \( Q \), the module \( \mathbb{Z}G{ \otimes }_{U}\mathbb{Z} \) is isomorphic to \( \mathbb{Z}Q \) by Lemma 6.1. Hence \( K \cong {IQ} \), the augmentation ideal of \( Q \) . | No |
Proposition 16.2. Let \( \left( {f,\alpha }\right) : \left( {G, A}\right) \rightarrow \left( {G, A}\right) \) be defined as in (16.3). Then \( {\left( f,\alpha \right) }^{ * } : {H}^{n}\left( {G, A}\right) \rightarrow {H}^{n}\left( {G, A}\right), n \geqq 0 \), is the identity. | Proof. We proceed by induction on \( n \) . For \( n = 0,{H}^{0}\left( {G, A}\right) = {A}^{G} \), and the assertion is trivial. If \( n \geqq 1 \) we choose an injective presentation \( A \rightarrowtail I \rightarrow {A}^{\prime } \), and consider the long exact cohomology sequence\n\n\[ \begin{array}{l} \cdots \righ... | Yes |
Proposition 16.3. Let \( U \) be a subgroup of finite index \( m \) in \( G \), and let\n\n\[ G = \mathop{\bigcup }\limits_{{i = 1}}^{m}U{x}_{i} \]\n\nbe a coset decomposition. Then the map \( \theta : {\operatorname{Hom}}_{U}\left( {\mathbb{Z}G, A}\right) \rightarrow A \), defined by\n\n\[ \theta \left( \varphi \right... | Now since, by Proposition IV. 12.3, \( {H}^{n}\left( {G,{\operatorname{Hom}}_{U}\left( {\mathbb{Z}G, A}\right) }\right) \cong {H}^{n}\left( {U, A}\right) \) , \( n \geqq 0 \), we may define the corestriction map (from \( U \) to \( G \) )\n\n\[ \text{Cor} : {H}^{n}\left( {U, A}\right) \rightarrow {H}^{n}\left( {G, A}\r... | No |
Theorem 16.4. Let \( U \) be a subgroup of finite index \( m \) in the group \( G \) , and let \( A \) be a \( G \) -module. Then \( \operatorname{Cor} \circ \operatorname{Res} : {H}^{n}\left( {G, A}\right) \rightarrow {H}^{n}\left( {G, A}\right), n \geqq 0 \), is just multiplication by \( m \) . | Proof. We proceed by induction on \( n \) . For \( n = 0 \) the restriction Res : \( {H}^{0}\left( {G, A}\right) \rightarrow {H}^{0}\left( {U, A}\right) \) simply embeds \( {A}^{G} \) in \( {A}^{U} \) . The corestriction Cor \( : {A}^{U} \rightarrow {\left( {\operatorname{Hom}}_{U}\left( \mathbb{Z}G, A\right) \right) }... | Yes |
Corollary 16.5. Let \( G \) be a finite group of order \( m \) . Then \( m{H}^{n}\left( {G, A}\right) = 0 \) for all \( n \geqq 1 \) . | Proof. Use Theorem 16.4 with \( U = \{ 1\} \) and observe that \( {H}^{n}\left( {\{ 1\}, A}\right) = 0 \) for \( n \geqq 1 \) . | No |
Theorem 16.6 (Maschke). Let \( G \) be a group \( \cdot \) of order \( m \), and let \( K \) be a field, whose characteristic does not divide \( m \) . Then the \( K \) -representations of \( G \) are completely reducible. | Proof. We have to show that every short exact sequence\n\n\[ \n{V}^{\prime }\overset{\alpha }{ \rightarrow }V\overset{\beta }{ \rightarrow }{V}^{\prime \prime }\n\]\n\n(16.7)\n\nof \( {KG} \) -modules splits. This is equivalent to the assertion that the induced sequence\n\n\[ \n0 \rightarrow {\operatorname{Hom}}_{G}\le... | Yes |
Lemma 16.7. Under the hypotheses of Theorem 16.6 we have\n\n\[ \n{H}^{n}\left( {G, W}\right) = 0 \n\] \n\nfor \( n \geqq 1 \) and any \( {KG} \) -module \( W \) . | Proof. Consider the map \( m : W \rightarrow W \), multiplication by \( m \) . This clearly is a \( G \) -module homomorphism. Since the characteristic of \( K \) does not divide \( m \), the map \( m : W \rightarrow W \) is in fact an isomorphism, having \( 1/m : W \rightarrow W \) as its inverse. Hence the induced ma... | Yes |
Theorem 1.2 (Birkhoff-Witt). Let \( \left\{ {e}_{i}\right\}, i \in J \), be a \( K \) -basis of \( \mathfrak{g} \) . Then the elements \( {e}_{I} \) corresponding to all finite increasing sequences \( I \) (including the empty one) form a K-basis of \( U\mathfrak{g} \) . | For a proof of this theorem we refer the reader to N. Jacobson [29, p. 159]; J.-P. Serre [42, LA. 3]. | No |
Corollary 1.4. Let \( \mathfrak{h} \) be a Lie subalgebra of \( \mathfrak{g} \) . Then \( U\mathfrak{g} \) is free as an h-module. | Proof. Choose \( \left\{ {e}_{i}^{\prime }\right\}, i \in {J}^{\prime } \), a basis in \( \mathfrak{h} \) and expand it by \( \left\{ {e}_{i}\right\}, j \in J \) , to a basis in \( \mathfrak{g} \) . Let both \( {J}^{\prime }, J \) be simply ordered. Make \( {J}^{\prime } \cup J \) simply ordered by setting\n\n\[ i \leq... | Yes |
Corollary 1.6. If \( \mathfrak{n} \rightarrow \mathfrak{g} \rightarrow \mathfrak{h} \) is an exact sequence of Lie algebras, then \( K{ \otimes }_{U\mathfrak{n}}U\mathfrak{g} \cong U\mathfrak{h} \) as right \( \mathfrak{g} \) -modules. | The proof is left to the reader. | No |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.