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Lemma 3.5. Suppose \( M \) is an \( R \) -module, \( N \) is an \( \left( {R, S}\right) \) -bimodule, and \( P \) is an \( S \) -module. Then there is a canonical isomorphism of abelian groups\n\n\[ \n{\operatorname{Hom}}_{R}\left( {M,{\operatorname{Hom}}_{S}\left( {N, P}\right) }\right) \cong {\operatorname{Hom}}_{S}\... | Proof. Every element \( \alpha \in {\operatorname{Hom}}_{R}\left( {M,{\operatorname{Hom}}_{S}\left( {N, P}\right) }\right) \) determines a map\n\n\[ \n\varphi : M \times N \rightarrow P \n\]\n\nvia \( \varphi \left( {m,\_ }\right) \mathrel{\text{:=}} \alpha \left( m\right) ;\varphi \) is clearly \( \mathbb{Z} \) -bilin... | Yes |
Proposition 3.6. Let \( f : R \rightarrow S \) be a homomorphism of commutative rings. Then, with notation as above, \( {f}_{ * } \) is right-adjoint to \( {f}^{ * } \) and left-adjoint to \( {f}^{!} \) . In particular, \( {f}_{ * } \) is exact, \( {f}^{ * } \) is right-exact, and \( {f}^{!} \) is left-exact. | Proof. Let \( M \), resp., \( N \), be an \( R \) -module, resp., an \( S \) -module. Note that, trivially, \( {\text{Hom}}_{S}\left( {S, N}\right) \) is canonically isomorphic to \( N \) (as an \( S \) -module) and to \( {f}_{ * }\left( N\right) \) (as an \( R \) -module). Thus \( {}^{15} \n\n\[ \n{\operatorname{Hom}}... | Yes |
Lemma 4.2. Let \( \varphi : {M}^{\ell } \rightarrow P \) be an \( R \) -multilinear function.\n\nIf \( \varphi \) is alternating, then for all \( \sigma \in {S}_{\ell } \), and all \( {m}_{1},\ldots ,{m}_{\ell } \), \n\n\[ \varphi \left( {{m}_{\sigma \left( 1\right) },\ldots ,{m}_{\sigma \left( \ell \right) }}\right) =... | Proof. For the first statement, it suffices to show that interchanging any two factors switches the sign of an alternating function (since transpositions generate the symmetric group). Since the other factors have no effect on this operation, this reduces the question to the case \( \ell = 2 \) . Therefore, we only hav... | Yes |
Lemma 4.3. Let \( R \) be a commutative ring, and let \( M \) be a free \( R \) -module of rank \( r \) . Then \( {\Lambda }_{R}^{\ell }\left( M\right) \) is a free \( R \) -module of rank \( \left( \begin{array}{l} r \\ \ell \end{array}\right) \) . | Proof. There are \( \left( \begin{array}{l} r \\ \ell \end{array}\right) \) sequences of indices \( {i}_{1},\ldots ,{i}_{\ell } \) satisfying \( 1 \leq {i}_{1} < \cdots < {i}_{\ell } \leq r \) , so we just need to show that the generators \( {e}_{{i}_{1}} \land \cdots \land {e}_{{i}_{\ell }} \) are linearly independent... | No |
For \( V = {k}^{4},{\Lambda }_{k}^{2}V \) has dimension \( \left( \begin{array}{l} 4 \\ 2 \end{array}\right) = 6 \) . On ’pure wedges’ \( {a}_{1} \land {a}_{2} \) , the isomorphism \( {\Lambda }_{k}^{2}V \rightarrow {k}^{6} \) works as follows. View the vectors \( {a}_{1},{a}_{2} \) as the two columns of a \( 4 \times ... | \[ A = \left( \begin{matrix} {a}_{1}^{1} & {a}_{2}^{1} \\ {a}_{1}^{2} & {a}_{2}^{2} \\ {a}_{1}^{3} & {a}_{2}^{3} \\ {a}_{1}^{4} & {a}_{2}^{4} \end{matrix}\right) \mapsto \left( \begin{matrix} {a}_{1}^{1}{a}_{2}^{2} - {a}_{1}^{2}{a}_{2}^{1} \\ {a}_{1}^{1}{a}_{2}^{3} - {a}_{1}^{3}{a}_{2}^{1} \\ {a}_{1}^{1}{a}_{2}^{4} - {... | Yes |
The polynomial ring \( R\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) over any ring \( R \) carries a natural grading, given by the (ordinary) degree of polynomials. | We may write\n\n\[\nR\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack = R \oplus \left\langle {{x}_{1},\ldots ,{x}_{n}}\right\rangle \oplus \left\langle {{x}_{1}^{2},{x}_{1}{x}_{2},\ldots ,{x}_{n}^{2}}\right\rangle \oplus \cdots .\n\] | Yes |
Lemma 4.9. Let \( S = {\bigoplus }_{i}{S}_{i} \) be a graded ring and let \( I \subseteq S \) be an ideal of \( S \) . Then the following are equivalent:\n\n(i) I is homogeneous;\n\n(ii) if \( s \in S \) and \( s = \mathop{\sum }\limits_{i}{s}_{i} \) is the decomposition of \( s \) into homogeneous elements \( {s}_{i} ... | Proof. \( \left( \mathrm{i}\right) \Leftrightarrow \left( \mathrm{{ii}}\right) \) is the very definition of homogeneous ideal; \( \left( \mathrm{{ii}}\right) \Leftrightarrow \left( \mathrm{{iii}}\right) \) is left to the reader (Exercise 4.10).\n\n(ii) \( \Rightarrow \) (iv): Assuming (ii) holds, define a grading on \(... | No |
The ideal \( I = \left( {y - {x}^{2}}\right) \) is not homogeneous in the ring \( k\left\lbrack {x, y}\right\rbrack \) , if this is given the grading by the usual degree. | indeed, \( y - {x}^{2} \in I \) while \( y \notin I \) , contradicting condition (ii) of Lemma 4.9. | Yes |
Lemma 4.12. Let \( {I}_{\mathbb{S}},{I}_{\mathbb{A}} \subseteq {\mathbb{T}}_{R}^{ * }\left( M\right) \) be the ideals respectively generated by all elements of the form \( \left( {m \otimes n - n \otimes m}\right) \) as \( m, n \in M \) and by elements of the form \( m \otimes m \) as \( m \in M \) . Then\n\n\[ \n{\mat... | Proof. We have observed in \( §{4.3} \) that the kernel of a graded homomorphism is the direct sum of the kernels of the induced homomorphisms in each degree; the statement of the lemma then follows easily from the explicit descriptions of the kernels of the canonical projections from the tensor powers to the symmetric... | No |
Proposition 4.16. Let \( R \) be a commutative ring, and let \( M \) be an \( R \) -module. Then for every \( R \) -algebra \( A \) and every \( R \) -module homomorphism \( \lambda : M \rightarrow A \) such that \( \lambda {\left( m\right) }^{2} = 0\forall m \in M \), there exists a unique homomorphism of \( R \) -alg... | Details may now safely be left to the reader (who may for example establish the first proposition from the universal property of tensor powers and then deduce the second and third from Lemma 4.12). | No |
The free case is particularly easy to understand. For instance,\n\n\[ \n{S}_{R}^{ * }\left( {R}^{\oplus r}\right) \cong R\left\lbrack {{x}_{1},\ldots ,{x}_{r}}\right\rbrack \n\] | Indeed, the polynomial ring satisfies the appropriate universal property with respect to mapping to commutative rings (cf. §III.2.2 and §III.6.4). Likewise, \( {\mathbb{T}}_{R}^{ * }\left( {R}^{\oplus r}\right) \) should be thought of as a ’noncommutative’ polynomial ring, in which the \( r \) inde-terminates do not co... | No |
Proposition 5.2. For every \( R \) -module \( N \), the functor \( {\operatorname{Hom}}_{R}\left( {\_, N}\right) \) is right-adjoint to itself. | Proof. Let \( L, M, N \) denote \( R \) -modules. Recall (cf. the considerations preceding Lemma 2.4) that \( R \) -bilinear maps\n\n\[ \varphi : L \times M \rightarrow N \]\n\nmay be identified with \( R \) -linear maps\n\n\[ L \rightarrow {\operatorname{Hom}}_{R}\left( {M, N}\right) \]\n\nBy the same token, they may ... | Yes |
Proposition 5.5. Let \( M \) be any \( R \) -module, and let \( F \) be a free \( R \) -module of finite rank. Then\n\n\[ \n{\operatorname{Hom}}_{R}\left( {M, F}\right) \cong {M}^{ \vee }{ \otimes }_{R}F.\n\] | Proof. By hypothesis \( F \cong {R}^{\oplus n} \cong {R}^{n} \) ; hence\n\n\( {\mathrm{{Hom}}}_{R}\left( {M, F}\right) \cong {\mathrm{{Hom}}}_{R}\left( {M,{R}^{n}}\right) \cong \mathrm{{Hom}}{\left( M, R\right) }^{n} \cong \mathrm{{Hom}}\left( {M, R}\right) { \otimes }_{R}{R}^{n} \cong {M}^{ \vee }{ \otimes }_{R}F, \)\... | Yes |
Corollary 5.7. The dual of a free module is isomorphic to a product of copies of \( R \) :\n\n\[ \n{\left( {R}^{\oplus S}\right) }^{ \vee } \cong {R}^{S} \n\]\n\nIn particular, \( {\left( {R}^{n}\right) }^{ \vee } \cong {R}^{n} \) : if \( F \) is a free \( R \) -module of finite rank, then \( {F}^{ \vee } \cong F \) . | Proof. This follows from Lemma 5.6 and the fact that \( {R}^{ \vee } = {\operatorname{Hom}}_{R}\left( {R, R}\right) \) is isomorphic to \( R \) . | No |
To see that these isomorphisms do depend on the choice of the basis, consider the standard basis \( \left( {{\mathbf{e}}_{1},{\mathbf{e}}_{2}}\right) \) of \( {R}^{2} \) and the corresponding dual basis \( \left( {{\check{\mathbf{e}}}_{1},{\check{\mathbf{e}}}_{2}}\right) \) of \( {\left( {R}^{2}\right) }^{ \vee } \), a... | By definition\n\n\[ \n{\check{\mathbf{e}}}_{1}^{\prime }\left( {\mathbf{e}}_{2}^{\prime }\right) = 0 \n\] \n\nwhile \n\n\[ \n{\check{\mathbf{e}}}_{1}\left( {\mathbf{e}}_{2}^{\prime }\right) = {\check{\mathbf{e}}}_{1}\left( {{\mathbf{e}}_{1} + {\mathbf{e}}_{2}}\right) = 1 + 0 = 1. \n\] \n\nTherefore, \( {\check{\mathbf{... | Yes |
Lemma 5.12. The duality functor is left-exact: every exact sequence\n\n\[ L \rightarrow M \rightarrow N \rightarrow 0 \]\n\nof \( R \) -modules induces an exact sequence\n\n\[ 0 \rightarrow {N}^{ \vee } \rightarrow {M}^{ \vee } \rightarrow {L}^{ \vee }.\n\] | Proof. This is an immediate consequence of the left-exactness of Hom. | No |
Proposition 5.13. Let\n\n\[ \n0 \rightarrow M\xrightarrow[]{\;\mu \;}N\xrightarrow[]{\;\nu \;}P \rightarrow 0 \n\] \n\nbe an exact sequence of \( R \) -modules, with \( P \) free. Then the induced sequence \n\n\[ \n0 \rightarrow {P}^{ \vee }\overset{{\nu }^{ \vee }}{ \rightarrow }{N}^{ \vee }\overset{{\mu }^{ \vee }}{ ... | Proof. Lemma 5.12 takes care of all but the surjectivity of the map \( {N}^{ \vee } \rightarrow {M}^{ \vee } \) induced from \( M \rightarrow N \) : \n\nThe question is whether every \( R \) -linear \( f : M \rightarrow R \) can be extended to an \( R \) -linear map \( g : N \rightarrow R \) so that \( f = g \circ \mu ... | Yes |
Lemma 5.15. Let \( A \) be the matrix representing a linear map \( \alpha : {R}^{n} \rightarrow {R}^{m} \) with respect to the standard bases. Then the dual map \( {\alpha }^{ \vee } : {\left( {R}^{m}\right) }^{ \vee } \rightarrow {\left( {R}^{n}\right) }^{ \vee } \) is represented by the transpose of \( A \) with resp... | The (easy) verification of this fact is left to the reader (Exercise 5.10). | No |
Proposition 5.16. Let \( R \) be an integral domain, and let \( M \) be an \( R \)-module. Then \( {M}^{ \vee } \) is torsion-free. | Proof. There is a surjection \( {R}^{\oplus S} \rightarrow M \), thus, an exact sequence\n\n\[ \n{R}^{\oplus T} \rightarrow {R}^{\oplus S} \rightarrow M \rightarrow 0 \n\]\n\nDualizing, \( {M}^{ \vee } \) is realized as the kernel of the induced map \( {R}^{S} \rightarrow {R}^{T} \) ; hence \( {M}^{ \vee } \) may be id... | Yes |
For example, assume that \( P \) is projective; then we claim that every exact sequence\n\n\[ 0 \rightarrow L\xrightarrow[]{\;\lambda \;}M\xrightarrow[]{\;\mu \;}P \rightarrow 0 \]\n\nsplits, in the sense that there is a submodule \( {P}^{\prime } \) of \( M \) such that \( \mu \) restricts to an isomorphism \( {P}^{\p... | Indeed, since \( P \) is projective, then the identity \( P\overset{ \equiv }{ \rightarrow }P \) lifts to a homomorphism \( \rho : P \rightarrow M \), and the reader can then verify that \( {P}^{\prime } = \rho \left( P\right) \) fits the requirement. Loosely speaking, in this situation we can simply replace \( M \) by... | No |
Proposition 6.4. An R-module \( P \) is projective if and only if it is a direct summand of a free module, that is, if and only if there exists a free \( R \) -module \( F \) , an \( R \) - module \( K \), and an isomorphism \( K \oplus P \cong F \) . | Proof. Any set \( S \) of generators of \( P \) determines a surjection of the free module \( F = {R}^{\oplus S} \) onto \( P \) and hence an exact sequence\n\n\[ 0 \rightarrow K \rightarrow F \rightarrow P \rightarrow 0. \]\n\nAs observed above, such a sequence necessarily splits if \( P \) is projective; thus \( F \c... | No |
Let \( {P}_{1},{P}_{2} \) be projective \( R \) -modules. Then \( {P}_{1} \oplus {P}_{2} \) and \( {P}_{1}{ \otimes }_{R}{P}_{2} \) are projective. Projective modules are flat. | These statements follow easily from Proposition 6.4, the fact that \( \otimes \) is distributive with respect to \( \oplus \), and the fact that free modules are flat. | No |
Theorem 6.6. An R-module \( Q \) is injective if and only if every \( R \) -linear map \( f \) : \( I \rightarrow Q \), with \( I \) an ideal of \( R \), extends to an \( R \) -linear map \( \widehat{f} : R \rightarrow Q \) . | Proof. The 'only if' part of the statement is immediate from the definition of injective. To verify the ’if’ part, assume \( Q \) satisfies the stated extension condition, let \( L \subseteq M \) be any inclusion of \( R \) -modules, and let \( q : L \rightarrow Q \) be a given \( R \) -linear map:\n\n![ed3b132a-22da-4... | Yes |
Corollary 6.7. Let \( R \) be a PID. Then an \( R \) -module \( Q \) is injective if and only if it is divisible. | Proof. Exercise 6.14. | No |
Viewed as abelian groups (i.e., \( \mathbb{Z} \)-modules), \( \mathbb{Q} \) and \( \mathbb{Q}/\mathbb{Z} \) are injective. More generally, if \( D \) is any divisible abelian group and \( K \subseteq D \), then \( D/K \) is injective. | Indeed, it is trivially divisible! | No |
Lemma 6.9. Let \( f : S \rightarrow R \) be a homomorphism of commutative rings, and let \( Q \) be an injective \( S \) -module. Then \( {f}^{!}\left( Q\right) \) is an injective \( R \) -module. | Proof. By adjunction (Lemma 3.5),\n\n\[{\mathrm{{Hom}}}_{R}\left( {\_ ,{f}^{!}\left( Q\right) }\right) \cong {\mathrm{{Hom}}}_{S}\left( {{f}_{ * }\left( \_ \right), Q}\right)\]\n\nas functors \( R \) -Mod \( \rightarrow R \) -Mod. Since \( {f}_{ * } \) is exact (Proposition 3.6) and \( {\operatorname{Hom}}_{S}\left( {\... | Yes |
Corollary 6.12. Let \( M \) be an \( R \) -module. Then \( M \) can be identified with a submodule of an injective \( R \) -module. | Proof. We claim that it suffices to show that \( \mathbb{Z} \) -Mod has enough injectives, Indeed, this will show that there exists a divisible abelian group \( D \) such that \( M \subseteq D(M \) is in particular an abelian group); since \( R \) -linear maps are in particular \( \mathbb{Z} \) -linear,\n\n\[ M \cong {... | Yes |
Proposition 6.14. An \( R \) -module \( P \) is projective if and only if \( {\operatorname{Ext}}_{R}^{1}\left( {P,\_ }\right) = 0 \), if and only if \( {\operatorname{Ext}}_{R}^{i}\left( {P,\_ }\right) = 0 \) for all \( i > 0 \) . | Proof. The second assertion: we have already noted that \( Q \) is injective if \( {\operatorname{Ext}}_{R}^{1}\left( {\_, Q}\right) = \) 0, and this is trivially the case if \( {\operatorname{Ext}}_{R}^{i}\left( {\_, Q}\right) = 0 \) for all \( i > 0 \) . So we just have to prove that \( {\operatorname{Ext}}_{R}^{i}\l... | No |
Lemma 1.3. A morphism \( \varphi : A \rightarrow B \) in an additive category is a monomorphism if and only if for all \( \zeta : Z \rightarrow A \) ,\n\n\[ \varphi \circ \zeta = 0 \Rightarrow \zeta = 0. \]\n\nIt is an epimorphism if and only if for all \( \beta : B \rightarrow Z \) ,\n\n\[ \beta \circ \varphi = 0 \Rig... | Proof. This is simply because two morphisms with the same source and target are equal if and only if their difference in the corresponding Hom-set (which is an abelian group by hypothesis) is 0. | No |
Lemma 1.4. In any additive category, kernels are monomorphisms and cokernels are epimorphisms. | Proof. Let \( \varphi : A \rightarrow B \) be a morphism in an additive category A, and let \( \operatorname{coker}\varphi \) : \( B \rightarrow C \) be its cokernel. Let \( \gamma : C \rightarrow Z \) be a morphism such that \( \gamma \circ \operatorname{coker}\varphi = 0 \) . The composition \( \left( {\gamma \circ \... | No |
Lemma 1.5. Let \( \varphi : A \rightarrow B \) be a morphism in an additive category. If \( \varphi \) has a kernel, then \( \varphi \) is a monomorphism if and only if \( 0 \rightarrow A \) is its kernel. If \( \varphi \) has a cokernel, then \( \varphi \) is an epimorphism if and only if \( A \rightarrow 0 \) is its ... | Proof. Let's do kernels this time.\n\nFirst assume \( \varphi : A \rightarrow B \) is a monomorphism with a kernel \( \iota : K \rightarrow A \) . In particular, \( \iota \circ \varphi : K \rightarrow A \rightarrow B \) is the zero-morphism; therefore \( \iota = 0 \) by Lemma 1.3. What about \( K \) ? If \( \zeta : Z \... | Yes |
Lemma 1.8. In an abelian category A, every kernel is the kernel of its cokernel; every cokernel is the cokernel of its kernel. | Proof. We 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 \ri... | 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. The kernel of \( \varphi \) exists (because A is abelian), and hence it is \( 0 \rightarrow A \) (by Lemma 1.5, 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 \rightar... | Yes |
Example 1.11. 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 \( B \), co... | The fibered product may be constructed in this context as the kernel of the difference of the two morphisms \n(just as in the particular case of \( R \) -Mod). The reader will prove that these ’fiber squares’ preserv... | 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, we 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\[ \begin{matrix} {K}^{\prime }\xrightarrow[]{\;\operatorname{im}\overline{\v... | No |
For a slightly more interesting example, consider a diagram and the associated sequence obtained by letting \( A \oplus B \) play both roles of product and coproduct. Then - the diagram is commutative if and only if this sequence is a complex; | 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 t... | Yes |
Lemma 2.5. \( z \sim 0 \Leftrightarrow z = 0 \) . Further, a morphism \( \varphi : A \rightarrow B \) in 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, we will propose a division of labor: the reader will prove the first statement (Exercise 2.6), and we 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![ed3b132a-22d... | 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 }{ \leftrighta... | 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 \( \operatorname{im}\varphi = \ker \psi \), 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 \( \mathsf{A} \rightarrow R \) -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. | We 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 ... | No |
Lemma 3.4. 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 \( \mathrm{C}\left( \mathrm{A}\right) \rightarrow \mathrm{A} \) . | Proof. 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}}_{\mathbb{C}\left( \mathsf{A}\right) }\left( {{M}^{ \bullet },{N}^{ \bullet }}\right) \rightarrow {\operatorname{Hom}}_{A}\left( {{H}^{i}\left( {M}^{ \bulle... | Yes |
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 induced by \( {\alpha }^{ \bullet } \) . Chasing the diagram\n\n be a morphism of cochain complexes. Then the induced morphism \( {H}^{ \bullet }\left( {L}^{ \bullet }\right) \rightarrow {H}^{ \bullet }\left( {M}^{ \bullet }\right) \) is an isomorphism if and only if the mapping cone \( {MC}... | ## (Cf. Exercise 3.14.) | No |
The datum of a resolution \( {M}^{ \bullet } \) 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( {\mathcal{F}\... | 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 \( \mathrm{A} \) 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 \( i < 0 \). L... | Proof. We have to construct morphisms \( {h}^{i} : {P}^{i} \rightarrow {L}^{i - 1} \) such that (*) \[ {\alpha }^{i} = {d}_{{L}^{ \bullet }}^{i - 1} \circ {h}^{i} + {h}^{i + 1} \circ {d}_{{P}^{ \bullet }}^{i}. \] Of course \( {h}^{i} = 0 \) necessarily for \( i > 0 \). For \( i = 0 \), use the fact that the morphism in... | 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... | No |
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 \( \mathsf{A} \) be an abelian category, and let \( {L}^{ \bullet } \) be a complex in \( \mathsf{C}\left( \mathsf{A}\right) \) . 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 is (additive and) exact. Then we claim that \( \widehat{\mathrm{A}} \) is sent to \( \widehat{\mathrm{B}} \) by \( \mathrm{L}\mathcal{F} \) . | Indeed, let \( {P}^{ \bullet } \) be a complex in \( \widehat{\mathsf{A}} : {H}^{i}\left( {P}^{ \bullet }\right) = 0 \) for \( i \neq 0 \) . The image \( \operatorname{L}\mathcal{F}\left( {P}^{ \bullet }\right) \) is obtained by choosing a projective resolution \( {P}_{\mathcal{F}\left( {P}^{ \bullet }\right) }^{ \bull... | No |
Proposition 7.3. The left-derived functor LF satisfies the following universal property:\n\n- There is a natural transformation\n\n\[ L\mathcal{F} \circ {\mathcal{P}}_{A} \sim {\mathcal{P}}_{B} \circ K\left( \mathcal{F}\right) \]\n\n- for every functor \( \mathcal{G} : {\mathrm{K}}^{ - }\left( {\mathrm{P}\left( \mathrm... | 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 §VIII.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}}^{ - }\left( {R\text{-... | The reader may note that in §VIII.2.4 we used a free resolution of \( M \) ; free modules are projective, so this was simply a convenient way to choose a projective resolution. The fact that we could use any projective resolution of \( M \) to compute \( {\operatorname{Tor}}_{i}^{R}\left( {M, N}\right) \) was mentioned... | No |
Assuming that \( \mathrm{A} \) has enough projectives and that\n\n\[ 0 \rightarrow L \rightarrow M \rightarrow N \rightarrow 0 \]\n\nis an exact sequence in \( \mathrm{A} \), can we arrange for projective resolutions of \( L, M, N \) to form an exact sequence in \( C\left( A\right) \) ? | Yes. This is often called the 'horseshoe lemma', after the shape of the main diagram appearing in its proof. | No |
Lemma 7.8. Let\n\n(*) \n\n\[ \n0 \rightarrow L \rightarrow M \rightarrow N \rightarrow 0 \n\] \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\(... | 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... | No |
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 } : \\;\\c... | 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} \\) i... | Yes |
Lemma 7.11. Let \( \mathrm{A} \) be an abelian category, and let\n\n(*) \n\n\[ \n0 \rightarrow {L}^{ \bullet } \rightarrow {M}^{ \bullet } \rightarrow {P}^{ \bullet } \rightarrow 0 \n\] \n\nbe an exact sequence of complexes in \( \mathrm{A} \), where \( {P}^{i} \) is projective for all \( i \) . Let \( \mathcal{F} \) :... | Proof. Since \( {P}^{i} \) is projective, the sequence \n\n\[ \n0 \rightarrow {L}^{i} \rightarrow {M}^{i} \rightarrow {P}^{i} \rightarrow 0 \n\] \n\nsplits (see the end of §VIII.6.1). It follows that \n\n\[ \n0 \rightarrow \mathcal{F}\left( {L}^{i}\right) \rightarrow \mathcal{F}\left( {M}^{i}\right) \rightarrow \mathca... | 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\[ 0 \rightarrow L \rightarrow M \rightarrow N \rightarrow 0 \]\n\nin \( \mathrm{A} \) induces a long exact sequence\n\n![ed3b132a-22da-440c-... | 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 \( \mathcal{F} : \mathsf{A} \rightarrow \mathsf{B} \) be a right-exact additive functor. Then \( {\mathsf{L}}_{i}\mathcal{F} = \) 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\n\n... | Applying the (contravariant) \( {\operatorname{Hom}}_{\mathbb{Z}\left\lbrack {x,{x}^{-1}}\right\rbrack }\left( {\_, M}\right) \), we see that \( {H}^{ \bullet }\left( {\mathbb{Z}, M}\right) \) is computed by the cohomology of  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\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, N... | 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}}_{\mathrm{A... | Yes |
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 \( {P}^{i} \) is ... | According to Theorem 8.9, the corresponding total complex is exact; as seen in Example 8.8 (cf. Exercise 8.5), this says that \( {\operatorname{Hom}}_{\mathsf{K}\left( \mathsf{A}\right) }\left( {{P}^{ \bullet }, L{\left\lbrack i\right\rbrack }^{ \bullet }}\right) = 0 \) for all \( i \). In other words, every cochain mo... | Yes |
Theorem 8.12. 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\[ \n\cdots \rightarrow {M}^{-3, \bullet } \rightarrow {M... | Proof. 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\[ \n\cdots \rightarrow {M}^{-2, \bullet } \rightarrow {M}^{-1,... | 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(*)\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 \( {P}_{N}^{j} \) is projec... | 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( {{\mathrm{{Hom}}}_{R}\left( {{P}_{M}^{ \bullet }, N}\right) }\right... | 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 |
Theorem 3.1. Let \( \mathrm{f} : \mathrm{A} \rightarrow \mathrm{B} \) be a function, with \( \mathrm{A} \) nonempty.\n\n(i) \( \mathrm{f} \) is injective if and only if there is a map \( \mathrm{g} : \mathrm{B} \rightarrow \mathrm{A} \) such that \( \mathrm{{gf}} = {1}_{\mathrm{A}} \).\n\n(ii) If \( \mathrm{A} \) is a ... | PROOF. Since every identity map is bijective, (11) and (12) prove the implications \( \left( \Leftarrow \right) \) in (i) and (ii). Conversely if \( f \) is injective, then for each \( {b\varepsilon f}\left( A\right) \) there is a unique \( {a\varepsilon A} \) with \( f\left( a\right) = b \) . Choose a fixed \( {a}_{0}... | Yes |
Theorem 4.1. If \( \mathrm{A} \) is a nonempty set, then the assignment \( \mathrm{R} \mapsto \mathrm{A}/\mathrm{R} \) defines a bijection from the set \( \mathrm{E}\left( \mathrm{A}\right) \) of all equivalence relations on \( \mathrm{A} \) onto the set \( \mathrm{Q}\left( \mathrm{A}\right) \) of all partitions of \( ... | SKETCH OF PROOF. If \( R \) is an equivalence relation on \( A \), then the set \( A/R \) of equivalence classes is a partition of \( A \) by (18),(19), and (21) so that \( R \mapsto A/R \) defines a function \( f : E\left( A\right) \rightarrow Q\left( A\right) \) . Define a function \( g : Q\left( A\right) \rightarrow... | No |
Theorem 5.2. Let \( \left\{ {{\mathrm{A}}_{\mathrm{i}} \mid \mathrm{i} \in \mathrm{I}}\right\} \) be a family of sets indexed by \( \mathrm{I} \) . Then there exists a set \( \mathrm{D} \), together with a family of maps \( \left\{ {{\pi }_{\mathrm{i}} : \mathrm{D} \rightarrow {\mathrm{A}}_{\mathrm{i}} \mid \mathrm{i} ... | PROOF OF 5.2. (Existence) Let \( D = \mathop{\prod }\limits_{{i \in I}}{A}_{i} \) and let the maps \( {\pi }_{i} \) be the projections onto the \( i \) th components. Given \( C \) and the maps \( {\varphi }_{i} \), define \( \varphi : C \rightarrow \mathop{\prod }\limits_{{i \in I}}{A}_{i} \) by \( c \mapsto {f}_{c} \... | Yes |
Theorem 6.1. (Principle of Mathematical Induction) If \( \mathrm{S} \) is a subset of the set \( \mathbf{N} \) of natural numbers such that \( {0\varepsilon }\mathrm{S} \) and either\n\n(i) \( \mathrm{n}\varepsilon \mathrm{S} \Rightarrow \mathrm{n} + {1\varepsilon }\mathrm{S}\; \) for all \( \mathrm{n}\varepsilon \math... | PROOF. If \( \mathbf{N} - S \neq \varnothing \), let \( n \neq 0 \) be its least element. Then for every \( m < n \) , we must have \( m \notin \mathbf{N} - S \) and hence \( {m\varepsilon S} \) . Consequently either (i) or (ii) implies \( {n\varepsilon S} \), which is a contradiction. Therefore \( \mathbf{N} - S = \va... | Yes |
Theorem 6.2. (Recursion Theorem) If \( \mathrm{S} \) is a set, \( \mathrm{a}\varepsilon \mathrm{S} \) and for each \( \mathrm{n}\varepsilon \mathrm{N},{\mathrm{f}}_{\mathrm{n}} : \mathrm{S} \rightarrow \mathrm{S} \) is a function, then there is a unique function \( \varphi : \mathbf{N} \rightarrow \mathbf{S} \) such th... | SKETCH OF PROOF. We shall construct a relation \( R \) on \( \mathbf{N} \times S \) that is the graph of a function \( \varphi : \mathbf{N} \rightarrow S \) with the desired properties. Let \( \mathcal{G} \) be the set of all subsets \( Y \) of \( \mathbf{N} \times S \) such that\n\n\[ \left( {0, a}\right) \in Y;\text{... | Yes |
Theorem 6.3. (Division Algorithm) If \( \mathrm{a},\mathrm{b},\mathrm{e}\mathbf{Z} \) and \( \mathrm{a} \neq 0 \), then there exists unique integers \( \mathrm{q} \) and \( \mathrm{r} \) such that \( \mathrm{b} = \mathrm{{aq}} + \mathrm{r} \), and \( 0 \leq \mathrm{r} < \left| \mathrm{a}\right| \) . | SKETCH OF PROOF. Show that the set \( S = \{ b - {ax} \mid {x\varepsilon }\mathbf{Z}, b - {ax} \geq 0\} \) is a nonempty subset of \( \mathbf{N} \) and therefore contains a least element \( r = b - {aq} \) (for some \( q \in \mathbf{Z} \) ). Thus \( b = {aq} + r \) . Use the fact that \( r \) is the least element in \(... | No |
Theorem 6.5. If \( {\mathrm{a}}_{1},{\mathrm{a}}_{2},\ldots ,{\mathrm{a}}_{\mathrm{n}} \) are integers, not all 0, then \( \left( {{\mathrm{a}}_{1},{\mathrm{a}}_{2},\ldots ,{\mathrm{a}}_{\mathrm{n}}}\right) \) exists. | Furthermore there are integers \( {\mathrm{k}}_{1},{\mathrm{k}}_{2},\ldots ,{\mathrm{k}}_{\mathrm{n}} \) such that\n\nConsider \( \mathrm{{ki}} = 1 \), and others \( = 0 \)\n\n\[ \left( {{a}_{1},{a}_{2},\ldots ,{a}_{n}}\right) = {k}_{1}{a}_{1} + {k}_{2}{a}_{2} + \cdots + {k}_{n}{a}_{n}. \]\n\nSKETCH OF PROOF. Use the D... | No |
Theorem 6.6. If \( \mathrm{a} \) and \( \mathrm{b} \) are relatively prime integers and \( \mathrm{a} \mid \mathrm{{bc}} \), then \( \mathrm{a} \mid \mathrm{c} \) . | SKETCH OF PROOF. By Theorem 6.5 \( 1 = {ra} + {sb} \), whence \( c = {rac} + {sbc} \) . Therefore \( a \mid c \) . | No |
Theorem 6.7. (Fundamental Theorem of Arithmetic) Any positive integer \( \mathrm{n} > 1 \) may be written uniquely in the form \( \mathrm{n} = {\mathrm{p}}_{1}{}^{{\mathrm{t}}_{1}}{\mathrm{p}}_{2}{}^{{\mathrm{t}}_{2}}\cdots {\mathrm{p}}_{\mathrm{k}}{}^{{\mathrm{t}}_{\mathrm{k}}} \), where \( {\mathrm{p}}_{1} < {\mathrm... | The proof, which proceeds by induction, may be found in Shockley [51, p.17]. | No |
Theorem 6.8. Let \( \mathrm{m} > 0 \) be an integer and \( \mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\varepsilon \mathbf{Z} \) .\n\n(i) Congruence modulo \( \mathrm{m} \) is an equivalence relation on the set of integers \( \mathbf{Z} \), which has precisely \( \mathrm{m} \) equivalence classes. | PROOF. (i) The fact that congruence modulo \( m \) is an equivalence relation is an easy consequence of the appropriate definitions. Denote the equivalence class of an integer \( a \) by \( \bar{a} \) and recall property (20), which can be stated in this context as:\n\n\[ \bar{a} = \bar{b} \Leftrightarrow a \equiv b\le... | Yes |
Theorem 7.1. (Principle of Transfinite Induction) If \( \mathbf{B} \) is a subset of a well-ordered set \( \left( {A, \leq }\right) \) such that for every \( {a\varepsilon A} \) ,\n\n\[ \n\\{ {c\\varepsilon A} \\mid c < a\\} \\subset B \\Rightarrow {a\\varepsilon B}, \n\]\n\nthen \( \\mathrm{B} = \\mathrm{A} \) . | PROOF. If \( A - B \neq \varnothing \), then there is a least element \( {a\varepsilon A} - B \) . By the definitions of least element and \( A - B \) we must have \( \\{ c\\varepsilon A \\mid c < a\\} \\subset B \) . By hypothesis then, \( a \in B \) so that \( a \in B \cap \left( {A - B}\\right) = \varnothing \), whi... | Yes |
Theorem 8.1. Equipollence is an equivalence relation on the class \( § \) of all sets. | PROOF. Exercise; note that \( \varnothing \sim \varnothing \) since \( \varnothing \subset \varnothing \times \varnothing \) is a relation that is (vacuously) a bijective function. \( {}^{3} \) | No |
Theorem 8.5. If \( \mathrm{A} \) is a set and \( \mathrm{P}\left( \mathrm{A}\right) \) its power set, then \( \left| \mathrm{A}\right| < \left| {\mathrm{P}\left( \mathrm{A}\right) }\right| \) . | SKETCH OF PROOF. The assignment \( a \mapsto \{ a\} \) defines an injective map \( A \rightarrow P\left( A\right) \) so that \( \left| A\right| \leq \left| {P\left( A\right) }\right| \) . If there were a bijective map \( f : A \rightarrow P\left( A\right) \), then for some \( {a}_{0} \in A, f\left( {a}_{0}\right) = B \... | Yes |
Theorem 8.6. (Schroeder-Bernstein) If \( \mathrm{A} \) and \( \mathrm{B} \) are sets such that \( \left| \mathrm{A}\right| \leq \left| \mathrm{B}\right| \) and \( \left| \mathrm{B}\right| \leq \left| \mathrm{A}\right| \), then \( \left| \mathrm{A}\right| = \left| \mathrm{B}\right| \) . | SKETCH OF PROOF. By hypothesis there are injective maps \( f : A \rightarrow B \) and \( g : B \rightarrow A \) . We shall use \( f \) and \( g \) to construct a bijection \( h : A \rightarrow B \) . This will imply that \( A \sim B \) and hence \( \left| A\right| = \left| B\right| \) . If \( {a\varepsilon A} \), then ... | No |
Theorem 8.7. The class of all cardinal numbers is linearly ordered by \( \leq \) . If \( \alpha \) and \( \beta \) are cardinal numbers, then exactly one of the following is true:\n\n\[ \alpha < \beta ;\;\alpha = \beta ;\;\beta < \alpha \;\text{ (Trichotomy Law). } \] | SKETCH OF PROOF. It is easy to verify that \( \leq \) is a partial ordering. Let \( \alpha ,\beta \) be cardinals and \( A, B \) be sets such that \( \left| A\right| = \alpha ,\left| B\right| = \beta \) . We shall show that \( \leq \) is a linear ordering (that is, either \( \alpha \leq \beta \) or \( \beta \leq \alpha... | Yes |
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