Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
values |
|---|---|---|
Lemma 61.2 (Nulhomotopy lemma). Let \( a \) and \( b \) be points of \( {S}^{2} \) . Let \( A \) be a compact space, and let\n\n\[ f : A \rightarrow {S}^{2} - a - b \]\n\nbe a continuous map. If \( a \) and \( b \) lie in the same component of \( {S}^{2} - f\left( A\right) \), then \( f \) is nulhomotopic. | Proof. One can replace \( {S}^{2} \) by the one-point compactification \( {\mathbb{R}}^{2} \cup \{ \infty \} \) of \( {\mathbb{R}}^{2} \), letting \( a \) and \( b \) correspond to the points \( \mathbf{0} \) and \( \infty \) . Then our lemma reduces to the following: Let \( A \) be a compact space and let \( g : A \ri... | Yes |
Theorem 61.3 (The Jordan separation theorem). Let \( C \) be a simple closed curve in \( {S}^{2} \) . Then \( C \) separates \( {S}^{2} \) . | Proof. Because \( {S}^{2} - C \) is locally path connected, its components and path components are the same. We assume that \( {S}^{2} - C \) is path connected and derive a contradiction.\n\nLet us write \( C \) as the union of two arcs \( {A}_{1} \) and \( {A}_{2} \) that intersect only in their end points \( a \) and... | Yes |
Theorem 61.4 (A general separation theorem). Let \( {A}_{1} \) and \( {A}_{2} \) be closed connected subsets of \( {S}^{2} \) whose intersection consists of precisely two points \( a \) and \( b \) . Then the set \( C = {A}_{1} \cup {A}_{2} \) separates \( {S}^{2} \) . | Proof. We must show first that \( C \) cannot equal all of \( {S}^{2} \) . That fact was obvious in the earlier proof. In the present case, we can see that \( C \neq {S}^{2} \) because \( {S}^{2} - a - b \) is connected and \( C - a - b \) is not. (The sets \( {A}_{i} - a - b \) form a separation of \( C - a - b \) .) ... | No |
Lemma 62.1 (Homotopy extension lemma). Let \( X \) be a space such that \( X \times I \) is normal. Let \( A \) be a closed subspace of \( X \), and let \( f : A \rightarrow Y \) be a continuous map, where \( Y \) is an open subspace of \( {\mathbb{R}}^{n} \) . If \( f \) is nulhomotopic, then \( f \) may be extended t... | Proof. Let \( F : A \times I \rightarrow Y \) be a homotopy between \( f \) and a constant map. Then \( F\left( {a,0}\right) = f\left( a\right) \) and \( F\left( {a,1}\right) = {y}_{0} \) for all \( a \) . Extend \( F \) to the space \( X \times 1 \) by setting \( F\left( {x,1}\right) = {y}_{0} \) for \( x \in X \) . T... | Yes |
Lemma 62.2 (Borsuk lemma). Let \( a \) and \( b \) be points of \( {S}^{2} \) . Let \( A \) be a compact space, and let \( f : A \rightarrow {S}^{2} - a - b \) be a continuous injective map. If \( f \) is nulhomotopic, then \( a \) and \( b \) lie in the same component of \( {S}^{2} - f\left( A\right) \) . | Proof. Because \( A \) is compact and \( {S}^{2} \) is Hausdorff, \( f\left( A\right) \) is a compact subspace of \( {S}^{2} \) that is homeomorphic to \( A \) . Because \( f \) is nulhomotopic, so is the inclusion mapping of \( f\left( A\right) \) into \( {S}^{2} - a - b \) . Hence it suffices to prove the lemma in th... | Yes |
Theorem 62.3 (Invariance of domain). If \( U \) is an open subset of \( {\mathbb{R}}^{2} \) and \( f : U \rightarrow \) \( {\mathbb{R}}^{2} \) is continuous and injective, then \( f\left( U\right) \) is open in \( {\mathbb{R}}^{2} \) and the inverse function \( {f}^{-1} : f\left( U\right) \rightarrow U \) is continuous... | Proof. As usual, we can replace \( {\mathbb{R}}^{2} \) by \( {S}^{2} \) . We show that if \( U \) is an open subset of \( {\mathbb{R}}^{2} \) and \( f : U \rightarrow {S}^{2} \) is continuous and injective, then \( f\left( U\right) \) is open in \( {S}^{2} \) and the inverse function is continuous.\n\nStep 1. We show t... | Yes |
Theorem 63.1. Let \( X \) be the union of two open sets \( U \) and \( V \), such that \( U \cap V \) can be written as the union of two disjoint open sets \( A \) and \( B \). Assume that there is a path \( \alpha \) in \( U \) from a point \( a \) of \( A \) to a point \( b \) of \( B \), and that there is a path \( ... | Proof. The proof is in many ways an imitation of the proof in \( §{54} \) that the fundamental group of the circle is infinite cyclic. As in that proof, the crucial step is to find an appropriate covering space \( E \) for the space \( X \). Step 1. (Construction of \( E \) ). We construct \( E \) by pasting together c... | Yes |
Theorem 63.2 (A nonseparation theorem). Let \( D \) be an arc in \( {S}^{2} \) . Then \( D \) does not separate \( {S}^{2} \) . | Proof. We give two proofs of this theorem. The first uses the results of the preceding section, and the second does not.\n\nFirst proof. Because \( D \) is contractible, the identity map \( i : D \rightarrow D \) is nulhomo-topic. Hence if \( a \) and \( b \) are any two points of \( {S}^{2} \) not in \( D \), the incl... | Yes |
Lemma 64.1. Let \( X \) be a theta space that is a subspace of \( {S}^{2} \) ; let \( A, B \), and \( C \) be the arcs whose union is \( X \) . Then \( X \) separates \( {S}^{2} \) into three components, whose boundaries are \( A \cup B, B \cup C \), and \( A \cup C \), respectively. The component having \( A \cup B \)... | Proof. Let \( a \) and \( b \) be the end points of the arcs \( A, B \), and \( C \) . Consider the simple closed curve \( A \cup B \) ; it separates \( {S}^{2} \) into two components \( U \) and \( {U}^{\prime } \), each of which is open in \( {S}^{2} \) and has boundary \( A \cup B \) . See Figure 64.3.\n\n![f98251b7... | Yes |
Theorem 64.2. Let \( X \) be the utilities graph. Then \( X \) cannot be imbedded in the plane. | Proof. If \( X \) can be imbedded in the plane, then it can be imbedded in \( {S}^{2} \) . So suppose \( X \) is a subspace of \( {S}^{2} \) . We derive a contradiction.\n\nWe use the notation of Example 2, where \( g, w, e,{h}_{1},{h}_{2} \), and \( {h}_{3} \) are the vertices of \( X \) . Let \( A, B \), and \( C \) ... | Yes |
Lemma 64.3. Let \( X \) be a subspace of \( {S}^{2} \) that is a complete graph on four vertices \( {a}_{1} \) , \( {a}_{2},{a}_{3} \), and \( {a}_{4} \) . Then \( X \) separates \( {S}^{2} \) into four components. The boundaries of these components are the sets \( {X}_{1},{X}_{2},{X}_{3} \), and \( {X}_{4} \), where \... | Proof. Let \( Y \) be the union of all the arcs of \( X \) different from the arc \( {a}_{2}{a}_{4} \) . Then we can write \( Y \) as a theta space by setting\n\n\[ A = {a}_{1}{a}_{2}{a}_{3} \]\n\n\[ B = {a}_{1}{a}_{3} \]\n\n\[ C = {a}_{1}{a}_{4}{a}_{3} \]\n\nSee Figure 64.5. The arcs \( A, B \), and \( C \) intersect ... | Yes |
Lemma 65.1. Let \( G \) be a subspace of \( {S}^{2} \) that is a complete graph on four vertices \( {a}_{1},\ldots ,{a}_{4} \). Let \( C \) be the subgraph \( {a}_{1}{a}_{2}{a}_{3}{a}_{4}{a}_{1} \), which is a simple closed curve. Let \( p \) and \( q \) be interior points of the edges \( {a}_{1}{a}_{3} \) and \( {a}_{... | Proof. (a) As in the proof of Lemma 64.3, the theta space \( C \cup {a}_{1}{a}_{3} \) separates \( {S}^{2} \) into three components \( U, V \), and \( W \). One of these, say \( W \), has \( C \) as its boundary; it is the only component whose boundary contains both \( {a}_{2} \) and \( {a}_{4} \). Therefore, \( {a}_{2... | Yes |
Lemma 66.1. Let \( f \) be a loop in \( {\mathbb{R}}^{2} - a \) . (a) If \( \bar{f} \) is the reverse of \( f \), then \( n\left( {\bar{f}, a}\right) = - n\left( {f, a}\right) \) . | Proof. (a) To compute \( n\left( {\bar{f}, a}\right) \), one replace \( s \) by \( 1 - s \) throughout the definition. This has the effect of changing \( \widetilde{g}\left( 1\right) - \widetilde{g}\left( 0\right) \) by a sign. | Yes |
Theorem 66.2. Let \( f \) be a simple loop in \( {\mathbb{R}}^{2} \) . If a lies in the unbounded component of \( {\mathbb{R}}^{2} - f\left( I\right) \), then \( n\left( {f, a}\right) = 0 \) ; while if \( a \) lies in the bounded component, \( n\left( {f, a}\right) = \pm 1 \) . | Proof. Since \( n\left( {f, a}\right) = n\left( {f - a,\mathbf{0}}\right) \), we may restrict ourselves to the case \( a = \mathbf{0} \) . Furthermore, we may assume that the base point of \( f \) lies on the positive \( x \) -axis. For one can gradually rotate \( {\mathbb{R}}^{2} - \mathbf{0} \) until the base point o... | Yes |
Lemma 66.3. Let \( f \) be a piecewise-differentiable loop in the complex plane; let \( a \) be a point not in the image of \( f \) . Then\n\n\[ n\left( {f, a}\right) = \frac{1}{2\pi i}{\int }_{f}\frac{dz}{z - a}. \] | Proof. The proof is a simple exercise in computation. Let \( p : \mathbb{R} \rightarrow {S}^{1} \) be the standard covering map. Let \( r\left( s\right) = \parallel f\left( s\right) - a\parallel \) and \( g\left( s\right) = \left\lbrack {f\left( s\right) - a}\right\rbrack /r\left( s\right) \) . Let \( \widetilde{g} \) ... | Yes |
Theorem 66.4 (Cauchy integral formula-classical version). Let \( C \) be a simple closed piecewise-differentiable curve in the complex plane. Let \( B \) be the bounded component of \( {\mathbb{R}}^{2} - C \) . If \( F\left( z\right) \) is analytic in an open set \( \Omega \) that contains \( B \) and \( C \), then for... | Proof. We derive this formula from the version of it proved in Ahlfors [A], which is the following:\n\nLet \( F \) be analytic in a region \( \Omega \) . Let \( f \) be a piecewise-differentiable loop in \( \Omega \) . Assume that \( n\left( {f, b}\right) = 0 \) for each \( b \) not in \( \Omega \) . If \( a \in \Omega... | Yes |
Lemma 67.1. Let \( G \) be an abelian group; let \( \left\{ {G}_{\alpha }\right\} \) be a family of subgroups of \( G \) . If \( G \) is the direct sum of the groups \( {G}_{\alpha } \), then \( G \) satisfies the following condition:\n\nGiven any abelian group \( H \) and any family of homomorphisms\n\n\( \left( *\rig... | Proof. We show first that if \( G \) has the stated extension property, then \( G \) is the direct sum of the \( {G}_{\alpha } \) . Suppose \( x = \sum {x}_{\alpha } = \sum {y}_{\alpha } \) ; we show that for any particular index \( \beta \) , we have \( {x}_{\beta } = {y}_{\beta } \) . Let \( H \) denote the group \( ... | Yes |
Corollary 67.2. Let \( G = {G}_{1} \oplus {G}_{2} \) . Suppose \( {G}_{1} \) is the direct sum of subgroups \( {H}_{\alpha } \) for \( \alpha \in J \), and \( {G}_{2} \) is the direct sum of subgroups \( {H}_{\beta } \) for \( \beta \in K \), where the index sets \( J \) and \( K \) are disjoint. Then \( G \) is the di... | Proof. If \( {h}_{\alpha } : {H}_{\alpha } \rightarrow H \) and \( {h}_{\beta } : {H}_{\beta } \rightarrow H \) are families of homomorphisms, they extend to homomorphisms \( {h}_{1} : {G}_{1} \rightarrow H \) and \( {h}_{2} : {G}_{2} \rightarrow H \) by the preceding lemma. Then \( {h}_{1} \) and \( {h}_{2} \) extend ... | No |
Corollary 67.3. If \( G = {G}_{1} \oplus {G}_{2} \), then \( G/{G}_{2} \) is isomorphic to \( {G}_{1} \) . | Proof. Let \( H = {G}_{1} \), let \( {h}_{1} : {G}_{1} \rightarrow H \) be the identity homomorphism, and let \( {h}_{2} : {G}_{2} \rightarrow H \) be the trivial homomorphism. Let \( h : G \rightarrow H \) be their extension to \( G \) . Then \( h \) is surjective with kernel \( {G}_{2} \) . | Yes |
Theorem 67.4. Given a family of abelian groups \( {\left\{ {G}_{\alpha }\right\} }_{\alpha \in J} \), there exists an abelian group \( G \) and a family of monomorphisms \( {i}_{\alpha } : {G}_{\alpha } \rightarrow G \) such that \( G \) is the direct sum of the groups \( {i}_{\alpha }\left( {G}_{\alpha }\right) \) . | Proof. Consider first the cartesian product\n\n\[\n\mathop{\prod }\limits_{{\alpha \in J}}{G}_{\alpha }\n\]\n\nit is an abelian group if we add two \( J \) -tuples by adding them coordinate-wise. Let \( G \) denote the subgroup of the cartesian product consisting of those tuples \( {\left( {x}_{\alpha }\right) }_{\alph... | Yes |
Lemma 67.5. Let \( {\left\{ {G}_{\alpha }\right\} }_{\alpha \in J} \) be an indexed family of abelian groups; let \( G \) be an abelian group; let \( {i}_{\alpha } : {G}_{\alpha } \rightarrow G \) be a family of homomorphisms. If each \( {i}_{\alpha } \) is a monomorphism and \( G \) is the direct sum of the groups \( ... | Proof. The only part that requires proof is the statement that if the extension condition holds, then each \( {i}_{\alpha } \) is a monomorphism. That is proved as follows. Given an index \( \beta \), set \( H = {G}_{\beta } \) and let \( {h}_{\alpha } : {G}_{\alpha } \rightarrow H \) be the identity homomorphism if \(... | Yes |
Theorem 67.6 (Uniqueness of direct sums). Let \( {\left\{ {G}_{\alpha }\right\} }_{\alpha \in J} \) be a family of abelian groups. Suppose \( G \) and \( {G}^{\prime } \) are abelian groups and \( {i}_{\alpha } : {G}_{\alpha } \rightarrow G \) and \( {i}_{\alpha }^{\prime } : {G}_{\alpha } \rightarrow {G}^{\prime } \) ... | Proof. We apply the preceding lemma (four times!). Since \( G \) is the external direct sum of the \( {G}_{\alpha } \) and \( \left\{ {i}_{\alpha }^{\prime }\right\} \) is a family of homomorphisms, there exists a unique homomorphism \( \phi : G \rightarrow {G}^{\prime } \) such that \( \phi \circ {i}_{\alpha } = {i}_{... | Yes |
Lemma 67.7. Let \( G \) be an abelian group; let \( {\left\{ {a}_{\alpha }\right\} }_{\alpha \in J} \) be a family of elements of \( G \) that generates \( G \) . Then \( G \) is a free abelian group with basis \( \left\{ {a}_{\alpha }\right\} \) if and only if for any abelian group \( H \) and any family \( \left\{ {y... | Proof. Let \( {G}_{\alpha } \) denote the subgroup of \( G \) generated by \( {a}_{\alpha } \) . Suppose first that the extension property holds. We show first that each group \( {G}_{\alpha } \) is infinite cyclic. Suppose that for some index \( \beta \), the element \( {a}_{\beta } \) generates a finite cyclic subgro... | Yes |
Theorem 67.8. If \( G \) is a free abelian group with basis \( \left\{ {{a}_{1},\ldots ,{a}_{n}}\right\} \), then \( n \) is uniquely determined by \( G \) . | Proof. The group \( G \) is isomorphic to the \( n \) -fold product \( \mathbb{Z} \times \cdots \times \mathbb{Z} \) ; the subgroup \( {2G} \) corresponds to the product \( \left( {2\mathbb{Z}}\right) \times \cdots \times \left( {2\mathbb{Z}}\right) \) . Then the quotient group \( G/{2G} \) is in bijective corresponden... | Yes |
Theorem 68.2. Given a family \( {\left\{ {G}_{\alpha }\right\} }_{\alpha \in J} \) of groups, there exists a group \( G \) and a family of monomorphisms \( {i}_{\alpha } : {G}_{\alpha } \rightarrow G \) such that \( G \) is the free product of the groups \( {i}_{\alpha }\left( {G}_{\alpha }\right) \) . | Proof. For convenience, we assume that the groups \( {G}_{\alpha } \) are disjoint as sets. (This can be accomplished by replacing \( {G}_{\alpha } \) by \( {G}_{\alpha } \times \{ \alpha \} \) for each index \( \alpha \), if necessary.)\n\nThen as before, we define a word (of length \( n \) ) in the elements of the gr... | Yes |
Lemma 68.5. Let \( {\left\{ {G}_{\alpha }\right\} }_{\alpha \in J} \) be a family of groups; let \( G \) be a group; let \( {i}_{\alpha } : {G}_{\alpha } \rightarrow G \) be a family of homomorphisms. If the extension condition of Lemma 68.3 holds, then each \( {i}_{\alpha } \) is a monomorphism and \( G \) is the free... | Proof. We first show that each \( {i}_{\alpha } \) is a monomorphism. Given an index \( \beta \), let us set \( H = {G}_{\beta } \) . Let \( {h}_{\alpha } : {G}_{\alpha } \rightarrow H \) be the identity if \( \alpha = \beta \), and the trivial homomorphism if \( \alpha \neq \beta \) . Let \( h : G \rightarrow H \) be ... | Yes |
Corollary 68.6. Let \( G = {G}_{1} * {G}_{2} \), where \( {G}_{1} \) is the free product of the subgroups \( {\left\{ {H}_{\alpha }\right\} }_{\alpha \in J} \) and \( {G}_{2} \) is the free product of the subgroups \( {\left\{ {H}_{\beta }\right\} }_{\beta \in K} \) . If the index sets \( J \) and \( K \) are disjoint,... | Proof. The proof is almost a copy of the proof of Corollary 67.2. | No |
Theorem 68.7. Let \( G = {G}_{1} * {G}_{2} \) . Let \( {N}_{i} \) be a normal subgroup of \( {G}_{i} \), for \( i = 1,2 \) . If \( N \) is the least normal subgroup of \( G \) that contains \( {N}_{1} \) and \( {N}_{2} \), then\n\n\[ G/N \cong \left( {{G}_{1}/{N}_{1}}\right) * \left( {{G}_{2}/{N}_{2}}\right) \] | Proof. The composite of the inclusion and projection homomorphisms\n\n\[ {G}_{1} \rightarrow {G}_{1} * {G}_{2} \rightarrow \left( {{G}_{1} * {G}_{2}}\right) /N \]\ncarries \( {N}_{1} \) to the identity element, so that it induces a homomorphism\n\n\[ {i}_{1} : {G}_{1}/{N}_{1} \rightarrow \left( {{G}_{1} * {G}_{2}}\righ... | Yes |
Lemma 68.9. Let \( S \) be a subset of the group \( G \). If \( N \) is the least normal subgroup of \( G \) containing \( S \), then \( N \) is generated by all conjugates of elements of \( S \). | Proof. Let \( {N}^{\prime } \) be the subgroup of \( G \) generated by all conjugates of elements of \( S \). We know that \( {N}^{\prime } \subset N \); to verify the reverse inclusion, we need merely show that \( {N}^{\prime } \) is normal in \( G \). Given \( x \in {N}^{\prime } \) and \( c \in G \), we show that \(... | Yes |
Lemma 69.1. Let \( G \) be a group; let \( {\left\{ {a}_{\alpha }\right\} }_{\alpha \in J} \) be a family of elements of \( G \). If \( G \) is a free group with system of free generators \( \left\{ {a}_{\alpha }\right\} \), then \( G \) satisfies the following condition:\n\n\( \left( *\right) \)\n\nGiven any group \( ... | Proof. If \( G \) is free, then for each \( \alpha \), the group \( {G}_{\alpha } \) generated by \( {a}_{\alpha } \) is infinite cyclic, so there is a homomorphism \( {h}_{\alpha } : {G}_{\alpha } \rightarrow H \) with \( {h}_{\alpha }\left( {a}_{\alpha }\right) = {y}_{\alpha } \). Then Lemma 68.1 applies. To prove th... | Yes |
Lemma 69.3. Given \( G \), the subgroup \( \left\lbrack {G, G}\right\rbrack \) is a normal subgroup of \( G \) and the quotient group \( G/\left\lbrack {G, G}\right\rbrack \) is abelian. If \( h : G \rightarrow H \) is any homomorphism from \( G \) to an abelian group \( H \), then the kernel of \( h \) contains \( \le... | Proof. Step 1. First we show that any conjugate of a commutator is in \( \left\lbrack {G, G}\right\rbrack \) . We compute as follows:\n\n\[ g\left\lbrack {x, y}\right\rbrack {g}^{-1} = g\left( {{xy}{x}^{-1}{y}^{-1}}\right) {g}^{-1} \]\n\n\[ = \left( {{gxy}{x}^{-1}}\right) \left( 1\right) \left( {{y}^{-1}{g}^{-1}}\right... | Yes |
Theorem 69.4. If \( G \) is a free group with free generators \( {a}_{\alpha } \), then \( G/\left\lbrack {G, G}\right\rbrack \) is a free abelian group with basis \( \left\lbrack {a}_{\alpha }\right\rbrack \), where \( \left\lbrack {a}_{\alpha }\right\rbrack \) denotes the coset of \( {a}_{\alpha } \) in \( G/\left\lb... | Proof. We apply Lemma 67.7. Given any family \( \left\{ {y}_{\alpha }\right\} \) of elements of the abelian group \( H \), there exists a homomorphism \( h : G \rightarrow H \) such that \( h\left( {a}_{\alpha }\right) = {y}_{\alpha } \) for each \( \alpha \) . Because \( H \) is abelian, the kernel of \( h \) contains... | Yes |
Corollary 69.5. If \( G \) is a free group with \( n \) free generators, then any system of free generators for \( G \) has \( n \) elements. | Proof. The free abelian group \( G/\left\lbrack {G, G}\right\rbrack \) has rank \( n \) . | No |
Theorem 70.1 (Seifert-van Kampen theorem). Let \( X = U \cup V \), where \( U \) and \( V \) are open in \( X \) ; assume \( U, V \), and \( U \cap V \) are path connected; let \( {x}_{0} \in U \cap V \) . Let \( H \) be a group, and let\n\n\[{\phi }_{1} : {\pi }_{1}\left( {U,{x}_{0}}\right) \rightarrow H\;\text{ and }... | Proof. Uniqueness is easy. Theorem 59.1 tells us that \( {\pi }_{1}\left( {X,{x}_{0}}\right) \) is generated by the images of \( {j}_{1} \) and \( {j}_{2} \) . The value of \( \Phi \) on the generator \( {j}_{1}\left( {g}_{1}\right) \) must equal \( {\phi }_{1}\left( {g}_{1}\right) \), and its value on \( {j}_{2}\left(... | No |
Theorem 70.2 (Seifert-van Kampen theorem, classical version). Assume the hypotheses of the preceding theorem. Let\n\n\[ j : {\pi }_{1}\left( {U,{x}_{0}}\right) * {\pi }_{1}\left( {V,{x}_{0}}\right) \rightarrow {\pi }_{1}\left( {X,{x}_{0}}\right) \]\n\nbe the homomorphism of the free product that extends the homomorphis... | Proof. The fact that \( {\pi }_{1}\left( {X,{x}_{0}}\right) \) is generated by the images of \( {j}_{1} \) and \( {j}_{2} \) implies that \( j \) is surjective.\n\nWe show that \( N \subset \ker j \) . Since \( \ker j \) is normal, it is enough to show that \( {i}_{1}{\left( g\right) }^{-1}{i}_{2}\left( g\right) \) bel... | Yes |
Theorem 71.1. Let \( X \) be the wedge of the circles \( {S}_{1},\ldots ,{S}_{n} \) ; let \( p \) be the common point of these circles. Then \( {\pi }_{1}\left( {X, p}\right) \) is a free group. If \( {f}_{i} \) is a loop in \( {S}_{i} \) that represents a generator of \( {\pi }_{1}\left( {{S}_{i}, p}\right) \), then t... | Proof. The result is immediate if \( n = 1 \) . We proceed by induction on \( n \) . The proof is similar to the one given in Example 1 of the preceding section.\n\nLet \( X \) be the wedge of the circles \( {S}_{1},\ldots ,{S}_{n} \), with \( p \) the common point of these circles. Choose a point \( {q}_{i} \) of \( {... | Yes |
Lemma 71.2. Let \( X \) be the wedge of the circles \( {S}_{\alpha } \), for \( \alpha \in J \) . Then \( X \) is normal. Furthermore, any compact subspace of \( X \) is contained in the union of finitely many circles \( {S}_{\alpha } \). | Proof. It is clear that one-point sets are closed in \( X \) . Let \( A \) and \( B \) be disjoint closed subsets of \( X \) ; assume that \( B \) does not contain \( p \) . Choose disjoint subsets \( {U}_{\alpha } \) and \( {V}_{\alpha } \) of \( {S}_{\alpha } \) that are open in \( {S}_{\alpha } \) and contain \( \{ ... | Yes |
Theorem 71.3. Let \( X \) be the wedge of the circles \( {S}_{\alpha } \), for \( \alpha \in J \) ; let \( p \) be the common point of these circles. Then \( {\pi }_{1}\left( {X, p}\right) \) is a free group. If \( {f}_{\alpha } \) is a loop in \( {S}_{\alpha } \) representing a generator of \( {\pi }_{1}\left( {{S}_{\... | Proof. Let \( {i}_{\alpha } : {\pi }_{1}\left( {{S}_{\alpha }, p}\right) \rightarrow {\pi }_{1}\left( {X, p}\right) \) be the homomorphism induced by inclusion; let \( {G}_{\alpha } \) be the image of \( {i}_{\alpha } . \n\nNote that if \( f \) is any loop in \( X \) based at \( p \), then the image set of \( f \) is c... | Yes |
Theorem 72.1. Let \( X \) be a Hausdorff space; let \( A \) be a closed path-connected subspace of \( X \) . Suppose that there is a continuous map \( h : {B}^{2} \rightarrow X \) that maps Int \( {B}^{2} \) bijectively onto \( X - A \) and maps \( {S}^{1} = \operatorname{Bd}{B}^{2} \) into \( A \) . Let \( p \in {S}^{... | Proof. Step 1. The origin \( \mathbf{0} \) is the center point of \( {B}^{2} \) ; let \( {x}_{0} \) be the point \( h\left( \mathbf{0}\right) \) of \( X \) . If \( U \) is the open set \( U = X - {x}_{0} \) of \( X \), we show that \( A \) is a deformation retract of \( U \) . See Figure 72.1. Let \( C = h\left( {B}^{2... | Yes |
Theorem 73.1. The fundamental group of the torus has a presentation consisting of two generators \( \alpha ,\beta \) and a single relation \( {\alpha \beta }{\alpha }^{-1}{\beta }^{-1} \) . | Proof. Let \( X = {S}^{1} \times {S}^{1} \) be the torus, and let \( h : {I}^{2} \rightarrow X \) be obtained by restricting the standard covering map \( p \times p : \mathbb{R} \times \mathbb{R} \rightarrow {S}^{1} \times {S}^{1} \) . Let \( p \) be the point \( \left( {0,0}\right) \) of Bd \( {I}^{2} \), let \( a = h... | Yes |
Lemma 73.3. Let \( \pi : E \rightarrow X \) be a closed quotient map. If \( E \) is normal, then so is \( X \) . | Proof. Assume \( E \) is normal. One-point sets are closed in \( X \) because one-point sets are closed in \( E \) . Now let \( A \) and \( B \) be disjoint closed sets of \( X \) . Then \( {\pi }^{-1}\left( A\right) \) and \( {\pi }^{-1}\left( B\right) \) are disjoint closed sets of \( E \) . Choose disjoint open sets... | Yes |
Theorem 73.4. The fundamental group of the \( n \) -fold dunce cap is a cyclic group of order \( n \) . | Proof. Let \( h : {B}^{2} \rightarrow X \) be the quotient map, where \( X \) is the \( n \) -fold dunce cap. Set \( A = h\left( {S}^{1}\right) \) . Let \( p = \left( {1,0}\right) \in {S}^{1} \) and let \( a = h\left( p\right) \) . Then \( h \) maps the arc \( C \) of \( {S}^{1} \) running from \( p \) to \( r\left( p\... | Yes |
Lemma 79.1 (The general lifting lemma). Let \( p : E \rightarrow B \) be a covering map; let \( p\left( {e}_{0}\right) = {b}_{0} \) . Let \( f : Y \rightarrow B \) be a continuous map, with \( f\left( {y}_{0}\right) = {b}_{0} \) . Suppose \( Y \) is path connected and locally path connected. The map \( f \) can be lift... | Proof. If the lifting \( \widetilde{f} \) exists, then \[ {f}_{ * }\left( {{\pi }_{1}\left( {Y,{Y}_{0}}\right) }\right) = {p}_{ * }\left( {{\widetilde{f}}_{ * }\left( {{\pi }_{1}\left( {Y,{y}_{0}}\right) }\right) }\right) \subset {p}_{ * }\left( {{\pi }_{1}\left( {E,{e}_{0}}\right) }\right) . \] This proves the \ | No |
Theorem 79.2. Let \( p : E \rightarrow B \) and \( {p}^{\prime } : {E}^{\prime } \rightarrow B \) be covering maps; let \( p\left( {e}_{0}\right) = \) \( {p}^{\prime }\left( {e}_{0}^{\prime }\right) = {b}_{0} \) . There is an equivalence \( h : E \rightarrow {E}^{\prime } \) such that \( h\left( {e}_{0}\right) = {e}_{0... | Proof. We prove the \ | No |
Lemma 79.3. Let \( p : E \rightarrow B \) be a covering map. Let \( {e}_{0} \) and \( {e}_{1} \) be points of \( {p}^{-1}\left( {b}_{0}\right) \) , and let \( {H}_{i} = {p}_{ * }\left( {{\pi }_{1}\left( {E,{e}_{i}}\right) }\right) \). (a) If \( \gamma \) is a path in \( E \) from \( {e}_{0} \) to \( {e}_{1} \), and \( ... | Proof. (a) First, we show that \( \left\lbrack \alpha \right\rbrack * {H}_{1} * {\left\lbrack \alpha \right\rbrack }^{-1} \subset {H}_{0} \). Given an element \( \left\lbrack h\right\rbrack \) of \( {H}_{1} \), we have \( \left\lbrack h\right\rbrack = {p}_{ * }\left( \left\lbrack \widetilde{h}\right\rbrack \right) \) f... | Yes |
Theorem 79.4. Let \( p : E \rightarrow B \) and \( {p}^{\prime } : {E}^{\prime } \rightarrow B \) be covering maps; let \( p\left( {e}_{0}\right) = \) \( {p}^{\prime }\left( {e}_{0}^{\prime }\right) = {b}_{0} \) . The covering maps \( p \) and \( {p}^{\prime } \) are equivalent if and only if the subgroups\n\n\[ \n{H}_... | Proof. If \( h : E \rightarrow {E}^{\prime } \) is an equivalence, let \( {e}_{1}^{\prime } = h\left( {e}_{0}\right) \), and let \( {H}_{1}^{\prime } = {p}_{ * }\left( {{\pi }_{1}\left( {{E}^{\prime },{e}_{1}^{\prime }}\right) }\right) \) . Theorem 79.2 implies that \( {H}_{0} = {H}_{1}^{\prime } \), while the precedin... | Yes |
Lemma 80.1. Let \( B \) be path connected and locally path connected. Let \( p : E \rightarrow B \) be a covering map in the former sense (so that \( E \) is not required to be path connected). If \( {E}_{0} \) is a path component of \( E \), then the map \( {p}_{0} : {E}_{0} \rightarrow B \) obtained by restricting \(... | Proof. We first show \( {p}_{0} \) is surjective. Since the space \( E \) is locally homeomorphic to \( B \), it is locally path connected. Therefore \( {E}_{0} \) is open in \( E \) . It follows that \( p\left( {E}_{0}\right) \) is open in \( B \) . We show that \( p\left( {E}_{0}\right) \) is also closed in \( B \), ... | Yes |
(a) If \( p \) and \( r \) are covering maps, so is \( q \) . | Proof. By our convention, \( X, Y \), and \( Z \) are path connected and locally path connected. Let \( {x}_{0} \in X \) ; set \( {y}_{0} = q\left( {x}_{0}\right) \) and \( {z}_{0} = p\left( {x}_{0}\right) \) . (a) Assume that \( p \) and \( r \) are covering maps. We show first that \( q \) is surjective. Given \( y \... | Yes |
Theorem 80.3. Let \( p : E \rightarrow B \) be a covering map, with \( E \) simply connected. Given any covering map \( r : Y \rightarrow B \), there is a covering map \( q : E \rightarrow Y \) such that \( r \circ q = p \) . | Proof. Let \( {b}_{0} \in B \) ; choose \( {e}_{0} \) and \( {y}_{0} \) so that \( p\left( {e}_{0}\right) = {b}_{0} \) and \( r\left( {y}_{0}\right) = {b}_{0} \) . We apply Lemma 79.1 to construct \( q \) . The map \( r \) is a covering map, and the condition\n\n\[ \n{p}_{ * }\left( {{\pi }_{1}\left( {E,{e}_{0}}\right)... | Yes |
Lemma 80.4. Let \( p : E \rightarrow B \) be a covering map; let \( p\left( {e}_{0}\right) = {b}_{0} \) . If \( E \) is simply connected, then \( {b}_{0} \) has a neighborhood \( U \) such that inclusion \( i : U \rightarrow B \) induces the trivial homomorphism\n\n\[ \n{i}_{ * } : {\pi }_{1}\left( {U,{b}_{0}}\right) \... | Proof. Let \( U \) be a neighborhood of \( {b}_{0} \) that is evenly covered by \( p \) ; break \( {p}^{-1}\left( U\right) \) up into slices; let \( {U}_{\alpha } \) be the slice containing \( {e}_{0} \) . Let \( f \) be a loop in \( U \) based at \( {b}_{0} \) . Because \( p \) defines a homeomorphism of \( {U}_{\alph... | Yes |
Lemma 81.1. The image of the map \( \Psi \) equals the image under \( \Phi \) of the subgroup \( N\left( {H}_{0}\right) /{H}_{0} \) of \( {\pi }_{1}\left( {B,{b}_{0}}\right) /{H}_{0} \) . | Proof. Recall that the lifting correspondence \( \phi : {\pi }_{1}\left( {B,{b}_{0}}\right) \rightarrow F \) is defined as follows: Given a loop \( \alpha \) in \( B \) at \( {b}_{0} \), let \( \gamma \) be its lift to \( E \) beginning at \( {e}_{0} \) ; let \( {e}_{1} = \gamma \left( 1\right) \) ; and define \( \phi ... | Yes |
Theorem 81.2. The bijection\n\n\[ \n{\Phi }^{-1} \circ \Psi : \mathcal{C}\left( {E, p, B}\right) \rightarrow N\left( {H}_{0}\right) /{H}_{0} \n\]\n\nis an isomorphism of groups. | Proof. We need only show that \( {\Phi }^{-1} \circ \Psi \) is a homomorphism. Let \( h, k : E \rightarrow E \) be covering transformations. Let \( h\left( {e}_{0}\right) = {e}_{1} \) and \( k\left( {e}_{0}\right) = {e}_{2} \) ; then\n\n\[ \n\Psi \left( h\right) = {e}_{1}\;\text{ and }\;\Psi \left( k\right) = {e}_{2}, ... | No |
Corollary 81.3. The group \( {H}_{0} \) is a normal subgroup of \( {\pi }_{1}\left( {B,{b}_{0}}\right) \) if and only if for every pair of points \( {e}_{1} \) and \( {e}_{2} \) of \( {p}^{-1}\left( {b}_{0}\right) \), there is a covering transformation \( h : E \rightarrow \) \( E \) with \( h\left( {e}_{1}\right) = {e... | \[ {\Phi }^{-1} \circ \Psi : \mathcal{C}\left( {E, p, B}\right) \rightarrow {\pi }_{1}\left( {B,{b}_{0}}\right) /{H}_{0}. \] | No |
Theorem 81.5. Let \( X \) be path connected and locally path connected; let \( G \) be a group of homeomorphisms of \( X \) . The quotient map \( \pi : X \rightarrow X/G \) is a covering map if and only if the action of \( G \) is properly discontinuous. In this case, the covering map \( \pi \) is regular and \( G \) i... | Proof. We show \( \pi \) is an open map. If \( U \) is open in \( X \), then \( {\pi }^{-1}\pi \left( U\right) \) is the union of the open sets \( g\left( U\right) \) of \( X \), for \( g \in G \) . Hence \( {\pi }^{-1}\pi \left( U\right) \) is open in \( X \), so that \( \pi \left( U\right) \) is open in \( X/G \) by ... | Yes |
Theorem 81.6. If \( p : X \rightarrow B \) is a regular covering map and \( G \) is its group of covering transformations, then there is a homeomorphism \( k : X/G \rightarrow B \) such that \( p = k \circ \pi \), where \( \pi : X \rightarrow X/G \) is the projection. | Proof. If \( g \) is a covering transformation, then \( p\left( {g\left( x\right) }\right) = p\left( x\right) \) by definition. Hence \( p \) is constant on each orbit, so it induces a continuous map \( k \) of the quotient space \( X/G \) into \( B \) . On the other hand, \( p \) is a quotient map because it is contin... | Yes |
Theorem 74.1. Let \( X \) be the space obtained from a finite collection of polygonal regions by pasting edges together according to some labelling scheme. Then \( X \) is a compact Hausdorff space. | Proof. For simplicity, we treat the case where \( X \) is formed from a single polygonal region. The general case is similar.\n\nIt is immediate that \( X \) is compact, since the quotient map is continuous. To show \( X \) is Hausdorff, it suffices to show that the quotient map \( \pi \) is a closed map. (See Lemma 73... | Yes |
Theorem 74.2. Let \( P \) be a polygonal region; let\n\n\[ w = {\left( {a}_{{i}_{1}}\right) }^{{\epsilon }_{1}}\cdots {\left( {a}_{{i}_{n}}\right) }^{{\epsilon }_{n}} \]\n\nbe a labelling scheme for the edges of \( P \) . Let \( X \) be the resulting quotient space; let \( \pi : P \rightarrow X \) be the quotient map. ... | Proof. The proof is similar to the proof we gave for the torus in \( §{73} \) . Because \( \pi \) maps all vertices of \( P \) to a single point of \( X \), the space \( A = \pi \left( {\operatorname{Bd}P}\right) \) is a wedge of \( k \) circles. For each \( i \), choose an edge of \( P \) that is labelled \( {a}_{i} \... | Yes |
Theorem 74.3. Let \( X \) denote the \( n \) -fold torus. Then \( {\pi }_{1}\left( {X,{x}_{0}}\right) \) is isomorphic to the quotient of the free group on the \( {2n} \) generators \( {\alpha }_{1},{\beta }_{1},\ldots ,{\alpha }_{n},{\beta }_{n} \) by the least normal subgroup containing the element\n\n\[ \left\lbrack... | Proof. In order to apply Theorem 74.2, one must show that under the labelling scheme for \( X \), all the vertices of the polygonal region belong to the same equivalence class. We leave this to you to check. | No |
Theorem 74.4. Let \( X \) denote the \( m \) -fold projective plane. Then \( {\pi }_{1}\left( {X,{x}_{0}}\right) \) is isomorphic to the quotient of the free group on \( m \) generators \( {\alpha }_{1},\ldots ,{\alpha }_{m} \) by the least normal subgroup containing the element\n\n\[{\left( {\alpha }_{1}\right) }^{2}{... | Proof. One needs only to check that under the labelling scheme for \( X \), all the vertices of the polygonal region belong to the same equivalence class. This we leave to you. - | No |
Theorem 75.1. Let \( F \) be a group; let \( N \) be a normal subgroup of \( F \) ; let \( q : F \rightarrow F/N \) be the projection. The projection homomorphism \[ p : F \rightarrow F/\left\lbrack {F, F}\right\rbrack \] induces an isomorphism \[ \phi : q\left( F\right) /\left\lbrack {q\left( F\right), q\left( F\right... | Proof. One has projection homomorphisms \( p, q, r, s \), as in the following diagram, where \( q\left( F\right) = F/N \) and \( p\left( F\right) = F/\left\lbrack {F, F}\right\rbrack \) .  Because \( r \circ p \) map... | Yes |
Corollary 75.2. Let \( F \) be a free group with free generators \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) ; let \( N \) be the least normal subgroup of \( F \) containing the element \( x \) of \( F \) ; let \( G = F/N \) . Let \( p : F \rightarrow F/\left\lbrack {F, F}\right\rbrack \) be projection. Then \( G/\left\l... | Proof. Note that because \( N \) is generated by \( x \) and all its conjugates, the group \( p\left( N\right) \) is generated by \( p\left( x\right) \) . The corollary then follows from the preceding theorem. | No |
Theorem 75.3. If \( X \) is the \( n \) -fold connected sum of tori, then \( {H}_{1}\left( X\right) \) is a free abelian group of rank \( {2n} \) . | Proof. In view of the preceding corollary, Theorem 74.3 implies that \( {H}_{1}\left( X\right) \) is isomorphic to the quotient of the free abelian group \( {F}^{\prime } \) on the set \( {\alpha }_{1},{\beta }_{1},\ldots ,{\alpha }_{n},{\beta }_{n} \) by the subgroup generated by the element \( \left\lbrack {{\alpha }... | Yes |
Theorem 75.4. If \( X \) is the \( m \) -fold connected sum of projective planes, then the torsion subgroup \( T\left( X\right) \) of \( {H}_{1}\left( X\right) \) has order 2, and \( {H}_{1}\left( X\right) /T\left( X\right) \) is a free abelian group of rank \( m - 1 \) . | Proof. In view of the preceding corollary, Theorem 74.4 implies that \( {H}_{1}\left( X\right) \) is isomorphic to the quotient of the free abelian group \( {F}^{\prime } \) on the set \( {\alpha }_{1},\ldots ,{\alpha }_{m} \) by the subgroup generated by \( {\left( {\alpha }_{1}\right) }^{2}\cdots {\left( {\alpha }_{m... | Yes |
Lemma 77.1. Let \( w \) be a proper scheme of the form\n\n\[ w = \left\lbrack {y}_{0}\right\rbrack a\left\lbrack {y}_{1}\right\rbrack a\left\lbrack {y}_{2}\right\rbrack \]\n\nwhere some of the \( {y}_{i} \) may be empty. Then one has the equivalence\n\n\[ w \sim {aa}\left\lbrack {{y}_{0}{y}_{1}^{-1}{y}_{2}}\right\rbrac... | Proof. Step 1. We first consider the case where \( {y}_{0} \) is empty. We show that\n\n\[ a\left\lbrack {y}_{1}\right\rbrack a\left\lbrack {y}_{2}\right\rbrack \sim {aa}\left\lbrack {{y}_{1}^{-1}{y}_{2}}\right\rbrack \]\n\nIf \( {y}_{1} \) is empty, this result is immediate, while if \( {y}_{2} \) is empty, it follows... | No |
Corollary 77.2. If \( w \) is a scheme of projective type, then \( w \) is equivalent to a scheme of the same length having the form\n\n\[ \left( {{a}_{1}{a}_{1}}\right) \left( {{a}_{2}{a}_{2}}\right) \cdots \left( {{a}_{k}{a}_{k}}\right) {w}_{1}, \]\n\nwhere \( k \geq 1 \) and \( {w}_{1} \) is either empty or of torus... | Proof. The scheme \( w \) can be written in the form\n\n\[ w = \left\lbrack {y}_{0}\right\rbrack a\left\lbrack {y}_{1}\right\rbrack a\left\lbrack {y}_{2}\right\rbrack \]\n\nthen the preceding lemma implies that \( w \) is equivalent to a scheme of the form \( {w}^{\prime } = \) \( {aa}{w}_{1} \) that has the same lengt... | Yes |
Lemma 77.4. Let \( w \) be a proper scheme of the form\n\n\[ w = {w}_{0}\left( {cc}\right) \left( {{ab}{a}^{-1}{b}^{-1}}\right) {w}_{1}. \]\n\nThen \( w \) is equivalent to the scheme\n\n\[ {w}^{\prime } = {w}_{0}\left( {aabbcc}\right) {w}_{1}. \] | Proof. Recall Lemma 77.1, which states that for proper schemes we have\n\n(*) \n\n\[ \left\lbrack {y}_{0}\right\rbrack a\left\lbrack {y}_{1}\right\rbrack a\left\lbrack {y}_{2}\right\rbrack \sim {aa}\left\lbrack {{y}_{0}{y}_{1}^{-1}{y}_{2}}\right\rbrack . \]\n\nWe proceed as follows:\n\n\[ w \sim \left( {cc}\right) \lef... | Yes |
Theorem 78.1. If \( X \) is a compact triangulable surface, then \( X \) is homeomorphic to the quotient space obtained from a collection of disjoint triangular regions in the plane by pasting their edges together in pairs. | Proof. Let \( {A}_{1},\ldots ,{A}_{n} \) be a triangulation of \( X \), with corresponding homeomorphisms \( {h}_{i} : {T}_{i} \rightarrow {A}_{i} \) . We assume the triangles \( {T}_{i} \) are disjoint; then the maps \( {h}_{i} \) combine to define a map \( h : E = {T}_{1} \cup \cdots \cup {T}_{n} \rightarrow X \) tha... | No |
Theorem 78.2. If \( X \) is a compact connected triangulable surface, then \( X \) is homeomorphic to a space obtained from a polygonal region in the plane by pasting the edges together in pairs. | Proof. It follows from the preceding theorem that there is a collection \( {T}_{1},\ldots ,{T}_{n} \) of triangular regions in the plane, and orientations and a labelling of the edges of these regions, where each label appears exactly twice in the total labelling scheme, such that \( X \) is homeomorphic to the quotien... | Yes |
Proposition 1.1.4. Given \( \varphi \) as above, it induces a continuous map\n\n\[ \n{\varphi }^{\sharp } : \operatorname{Spec}\left( {R}_{2}\right) \rightarrow \operatorname{Spec}\left( {R}_{1}\right) \n\]\n\n\[ \n\mathfrak{p} \mapsto {\varphi }^{-1}\left( \mathfrak{p}\right) \n\]\n\nwith respect to the Zariski topolo... | Proof. Clearly, \( {ab} \in {\varphi }^{-1}\left( \mathfrak{p}\right) \) is equivalent to \( \varphi \left( a\right) \varphi \left( b\right) \in \mathfrak{p} \), which is in turn equivalent to \( \left( {\varphi \left( a\right) \in \mathfrak{p}}\right) \vee \left( {\varphi \left( b\right) \in \mathfrak{p}}\right) \) wh... | Yes |
Proposition 1.1.5 (Prime avoidance). Let \( I \) and \( {\mathfrak{p}}_{1},\ldots ,{\mathfrak{p}}_{n} \) be ideals of \( R \) such that \( I \subset \) \( \mathop{\bigcup }\limits_{{i = 1}}^{n}{\mathfrak{p}}_{i} \) . Suppose that\n\n\( \diamond \) either \( R \) contains an infinite field, or\n\n- at most two of the id... | Proof. If \( R \) contains an infinite field \( F \), the ideals are automatically \( F \) -vector subspaces of \( R \) . Since \( I = \mathop{\bigcup }\limits_{{i = 1}}^{r}I \cap {\mathfrak{p}}_{i} \) whereas an \( F \) -vector space cannot be covered by finitely many proper subspaces, there must exist some \( i \) wi... | Yes |
Proposition 1.2.5 (Exactness of localization). Let \( S \) be any multiplicative subset of \( R \) . If\n\n\[ \cdots \rightarrow {M}_{i}\overset{{f}_{i}}{ \rightarrow }{M}_{i + 1} \rightarrow \cdots \]\n\nis an exact sequence of \( R \) -modules, then\n\n\[ \cdots \rightarrow {M}_{i}\left\lbrack {S}^{-1}\right\rbrack \... | Proof. By homological common sense, we are reduced to the exact sequences (i) \( 0 \rightarrow \) \( {M}^{\prime } \rightarrow M \rightarrow {M}^{\prime \prime } \) (i.e. left exactness),(ii) \( {M}^{\prime } \rightarrow M \rightarrow {M}^{\prime \prime } \rightarrow 0 \) (i.e. right exactness). The case (ii) is known ... | Yes |
Lemma 1.2.6. Let \( M \) be an \( R \) -module. The localizations \( M \rightarrow {M}_{\mathfrak{m}} \) for various maximal ideals \( \mathfrak{m} \) assemble into an injection | Proof. Let \( m \in M \) be such that \( m \mapsto 0 \in {M}_{\mathfrak{m}} \) for all \( \mathfrak{m} \) . This means that for all \( \mathfrak{m} \) there exists \( s \in R \smallsetminus \mathfrak{m} \) such that \( {sm} = 0 \) . Hence the annihilator ideal \( {\operatorname{ann}}_{R}\left( m\right) \mathrel{\text{:... | Yes |
Lemma 1.2.7. Let \( R \) be an integral domain, then \( R = \mathop{\bigcap }\limits_{{\mathfrak{m} \in \operatorname{MaxSpec}\left( R\right) }}{R}_{\mathfrak{m}} \) as subrings of \( \operatorname{Frac}\left( R\right) \) . | Proof. Only the inclusion \( \supset \) requires proof. Let \( x \in \operatorname{Frac}\left( R\right) \) and define \( D \mathrel{\text{:=}} \{ r \in R \) : \( {rx} \in R\} \) (the ideal of denominators). Suppose \( x \in {R}_{\mathfrak{m}} \) for all maximal \( \mathfrak{m} \), then \( D ⊄ \mathfrak{m} \) for all ma... | Yes |
Proposition 1.3.2. Let \( I \) be a proper ideal of \( R \) . We have\n\n\[ \sqrt{I} = \mathop{\bigcap }\limits_{\substack{{\mathfrak{p} \in \operatorname{Spec}\left( R\right) } \\ {\mathfrak{p} \supset I} }}\mathfrak{p} \] | Proof. By replacing \( R \) by \( R/I \), this is easily reduced to the case \( I = \{ 0\} \) . If \( r \) is nilpotent and \( \mathfrak{p} \in \operatorname{Spec}\left( R\right) \), then \( {r}^{n} = 0 \in \mathfrak{p} \) implies \( r \in \mathfrak{p} \) . Conversely, suppose that \( r \) is not nilpotent. There exist... | Yes |
Lemma 1.3.4. Suppose that \( I \subset R \) is an ideal, \( M \) is an \( R \) -module with generators \( {x}_{1},\ldots ,{x}_{n} \) and \( \varphi \in {\operatorname{End}}_{R}\left( M\right) \) satisfies \( \varphi \left( M\right) \subset {IM} \), then there exists a polynomial \( P\left( X\right) = {X}^{n} + {a}_{n -... | Proof. Write \( \varphi \left( {x}_{i}\right) = \mathop{\sum }\limits_{{j = 1}}^{n}{a}_{ij}{x}_{j} \) where \( {a}_{ij} \in I \) . Set \( A \mathrel{\text{:=}} {\left( {a}_{ij}\right) }_{1 \leq i, j \leq n} \in {\operatorname{Mat}}_{n}\left( R\right) \) . Regard \( M \) as an \( R\left\lbrack X\right\rbrack \) -module ... | Yes |
Theorem 1.3.5 (Nakayama's Lemma). Suppose that \( M \) is a finitely generated \( R \) -module and \( I \) is an ideal of \( R \) such that \( {IM} = M \) . Then there exists \( a \in I \) such that \( \left( {1 + a}\right) M = 0 \) . If \( I \subset \operatorname{rad}\left( R\right) \), then we have \( M = \{ 0\} \) u... | Proof. Write \( M = R{x}_{1} + \cdots + R{x}_{n} \) . Plug \( \varphi = {\mathrm{{id}}}_{M} \) into Lemma 1.3.4 to deduce that \( P\left( {\operatorname{id}}_{M}\right) = 1 + \underset{ = : a \in I}{\underline{{a}_{n - 1} + \cdots + {a}_{0}}} \) acts as 0 on \( M \) . This proves the first part. Assume furthermore that... | Yes |
Corollary 1.3.6. Let \( M \) be a finitely generated \( R \) -module, and let \( I \subset \operatorname{rad}\left( R\right) \) be an ideal of \( R \) . If the images of \( {x}_{1},\ldots ,{x}_{n} \in M \) in \( M/{IM} \) form a set of generators, then \( {x}_{1},\ldots ,{x}_{n} \) generate \( M \) . | Proof. Apply Theorem 1.3.5 to \( N \mathrel{\text{:=}} M/\left( {R{x}_{1} + \cdots + R{x}_{n}}\right) \) ; our assumption \( M = {IM} + \) \( R{x}_{1} + \cdots + R{x}_{n} \) entails that \( {IN} = N \), thus \( N = 0 \) . | Yes |
Proposition 1.3.7. Let \( M \) be a finitely generated \( R \) -module and \( \psi \in {\operatorname{End}}_{R}\left( M\right) \) . If \( \psi \) is surjective then \( \psi \) is an automorphism. | Proof. Introduce a variable \( Y \) . Make \( M \) into an \( R\left\lbrack Y\right\rbrack \) -module by letting \( Y \) act as \( \psi \) . Put \( I \mathrel{\text{:=}} \left( Y\right) \) so that \( {IM} = M \) . Theorem 1.3.5 yields some \( Q\left( Y\right) \in R\left\lbrack Y\right\rbrack \) satisfying \( (1 - \) \(... | Yes |
Corollary 1.4.3. A ring \( R \) is Artinian if and only if it is Noetherian and every prime ideal of \( R \) is maximal. | Proof. If \( R \) is Artinian, then \( R \) is of finite length, hence is Noetherian as well. For every prime ideal \( \mathfrak{p} \), we have \( \mathfrak{p} \supset \{ 0\} = {\left( {\mathfrak{m}}_{1}\cdots {\mathfrak{m}}_{n}\right) }^{k} \) in the notations of the proof above, therefore \( \mathfrak{p} \supset {\ma... | Yes |
Proposition 2.1.2. If \( M \) is finitely generated then \( \operatorname{Supp}\left( M\right) = V\left( {{\operatorname{ann}}_{R}\left( M\right) }\right) \) ; in particular it is Zariski-closed in \( \operatorname{Spec}\left( R\right) \) . | Proof. Suppose \( M = R{x}_{1} + \cdots + R{x}_{n} \) . Set \( {I}_{i} \mathrel{\text{:=}} {\operatorname{ann}}_{R}\left( {x}_{i}\right) \) and observe that \( \mathop{\bigcap }\limits_{{i = 1}}^{n}{I}_{i} = \) \( {\operatorname{ann}}_{R}\left( M\right) \) . Then \( \mathfrak{p} \in \operatorname{Supp}\left( M\right) \... | No |
Proposition 2.1.3. For an exact sequence \( 0 \rightarrow {M}^{\prime } \rightarrow M \rightarrow {M}^{\prime \prime } \rightarrow 0 \) we have \( \operatorname{Supp}\left( M\right) = \) \( \operatorname{Supp}\left( {M}^{\prime }\right) \cup \operatorname{Supp}\left( {M}^{\prime \prime }\right) \) . For arbitrary direc... | Proof. Again, we use the exactness of localization for the first assertion. For \( \mathfrak{p} \in \operatorname{Spec}\left( R\right) \) we have an exact \( 0 \rightarrow {M}_{\mathfrak{p}}^{\prime } \rightarrow {M}_{\mathfrak{p}} \rightarrow {M}_{\mathfrak{p}}^{\prime \prime } \rightarrow 0 \), hence \( {M}_{\mathfra... | Yes |
Lemma 2.2.3. Consider the set \( \mathcal{S} \mathrel{\text{:=}} \left\{ {{\operatorname{ann}}_{R}\left( x\right) : x \in M, x \neq 0}\right\} \) of ideals, partially ordered by inclusion. Every maximal element in \( \mathcal{S} \) is prime. | Proof. Let \( \mathfrak{p} = {\operatorname{ann}}_{R}\left( x\right) \) be a maximal element of \( \mathcal{S} \) and suppose \( {ab} \in \mathfrak{p} \) . If \( b \notin \mathfrak{p} \), then\n\n\[ \n{bx} \neq 0,\;{abx} = 0,\;R \neq {\operatorname{ann}}_{R}\left( {bx}\right) \supset {\operatorname{ann}}_{R}\left( x\ri... | Yes |
Theorem 2.2.5. Let \( M \) be an \( R \) -module.\n\n(i) We have \( M = \{ 0\} \) if and only if \( \operatorname{Ass}\left( M\right) = \varnothing \) . | Proof. (i) Clearly \( \operatorname{Ass}\left( {\{ 0\} }\right) = \varnothing \) . If \( M \neq 0 \), the set \( \mathcal{S} \) in Lemma 2.2.3 is then nonempty, hence contains a maximal element \( \mathfrak{p} \) because \( R \) is Noetherian; this yields \( \mathfrak{p} \in \operatorname{Ass}\left( M\right) \) . | Yes |
Theorem 2.2.7. For every \( R \) -module \( M \) we have \( \operatorname{Supp}\left( M\right) = \mathop{\bigcup }\limits_{{\mathfrak{p} \in \operatorname{Ass}\left( M\right) }}V\left( \mathfrak{p}\right) \), in particular \( \operatorname{Ass}\left( M\right) \subset \operatorname{Supp}\left( M\right) \) . Furthermore,... | Proof. For any prime \( \mathfrak{q} \) we have \( {M}_{\mathfrak{q}} \neq 0 \Leftrightarrow \operatorname{Ass}\left( {M}_{\mathfrak{q}}\right) \neq \varnothing \), and the latter condition holds precisely when there exists \( \mathfrak{p} \in \operatorname{Ass}\left( M\right) \) with \( \mathfrak{p} \cap \left( {R \sm... | Yes |
Theorem 2.2.9. Let \( M \) be a finitely generated \( R \) -module. There exists a chain \( M = {M}_{n} \supset \) \( {M}_{n - 1} \supset \cdots \supset {M}_{0} = \{ 0\} \) of submodules such that for every \( 0 < i \leq n \), the subquotient \( {M}_{i}/{M}_{i - 1} \) is isomorphic to \( R/{\mathfrak{p}}_{i} \) for som... | Proof. Assume \( M \supsetneq {M}_{0} \mathrel{\text{:=}} \{ 0\} \) . There exists \( {\mathfrak{p}}_{1} \in \operatorname{Ass}\left( M\right) \) together with a submodule \( {M}_{1} \subset M \) isomorphic to \( R/{\mathfrak{p}}_{1} \) . Furthermore \( \operatorname{Ass}\left( {M}_{1}\right) = \operatorname{Ass}\left(... | Yes |
Proposition 2.3.2. The following are equivalent for a nonzero R-module M.\n\n(i) \( M \) is coprimary;\n\n(ii) for every zero divisor \( r \in R \) for \( M \) and every \( x \in M \), there exists \( n \geq 1 \) such that \( {r}^{n}x = 0 \) . | Proof. (i) \( \Rightarrow \) (ii): Suppose Ass \( \left( M\right) = \{ \mathfrak{p}\} \) and \( x \in M \smallsetminus \{ 0\} \) . From \( \varnothing \neq \operatorname{Ass}\left( {Rx}\right) \subset \) Ass \( \left( M\right) \) we infer that \( \operatorname{Ass}\left( {Rx}\right) = \{ \mathfrak{p}\} \), thus by Theo... | Yes |
Lemma 2.3.5. Let \( \mathfrak{p} \in \operatorname{Spec}\left( R\right) \) and \( {N}_{1},{N}_{2} \subset M \) are \( \mathfrak{p} \) -primary submodules. Then \( {N}_{1} \cap {N}_{2} \) is a \( \mathfrak{p} \) -primary submodule of \( M \) . | Proof. We have \( M/{N}_{1} \cap {N}_{2} \hookrightarrow M/{N}_{1} \oplus M/{N}_{2} \) . Since \( {N}_{1} \cap {N}_{2} \neq M \), we have\n\n\[\n\varnothing \neq \operatorname{Ass}\left( {M/{N}_{1} \cap {N}_{2}}\right) \subset \operatorname{Ass}\left( {M/{N}_{1}}\right) \cup \operatorname{Ass}\left( {M/{N}_{2}}\right) ... | Yes |
Theorem 2.4.1 (Lasker-Noether). Let \( N \subsetneq M \) be an R-submodule. Then we can express \( N \) as\n\n\[ N = {M}_{1} \cap \cdots \cap {M}_{n} \]\n\nfor some \( n \geq 1 \) and primary R-submodules \( {M}_{i} \), say with \( \operatorname{Ass}\left( {M/{M}_{i}}\right) = \left\{ {\mathfrak{p}}_{i}\right\} \) for ... | Proof. Establish the existence of primary decompositions first. Replacing \( M \) by \( M/N \) , we may assume \( N = \{ 0\} \) from the outset. We claim that\n\n\[ \forall \mathfrak{p} \in \operatorname{Ass}\left( M\right) ,\exists Q\left( \mathfrak{p}\right) \subset M,\left\{ \begin{array}{l} Q\left( \mathfrak{p}\rig... | Yes |
Corollary 2.4.2. Let \( N = {M}_{1} \cap \cdots \cap {M}_{n} \) be a minimal primary decomposition of \( N \subsetneq M \) . Suppose that \( {M}_{1} \) is \( \mathfrak{p} \) -primary where \( \mathfrak{p} \mathrel{\text{:=}} {\mathfrak{p}}_{1} \) is a minimal element in \( \operatorname{Ass}\left( {M/N}\right) \), then... | Proof. Recall the proof of Theorem 2.4.1, especially the part (iii); here we localize with respect to \( S \mathrel{\text{:=}} R \smallsetminus \mathfrak{p} \) . The minimality assumption entails\n\n\[ {N}_{\mathfrak{p}} = {M}_{1,\mathfrak{p}} \subset {M}_{\mathfrak{p}} \]\n\nIt remains to show that the preimage of \( ... | Yes |
Claim: this gives two minimal primary decompositions of \( I \). | The ideal \( \left( X\right) \) is prime, hence primary. In fact, \( \left( {{X}^{2},{XY},{Y}^{2}}\right) = {\left( X, Y\right) }^{2} \) and \( \left( {{X}^{2}, Y}\right) \) are both primary ideals associated to the maximal ideal \( \left( {X, Y}\right) \). This follows either by direct arguments or by Exercise 2.3.4, ... | No |
An element \( x \in A \) is integral over \( R \) if and only if there exists an \( R \) - submodule \( M \subset A \) such that\n\n\( \diamond M \) is a finitely generated \( R \) -module;\n\n\( \diamond {xM} \subset M \), thus \( M \) is an \( R\left\lbrack x\right\rbrack \) -module;\n\n\( \diamond M \) is a faithful... | Proof. This is a familiar application of Cayley-Hamilton theorem, which we recall below. If \( x \) is integral, say \( {x}^{n} + {a}_{n - 1}{x}^{n - 1} + \cdots + {a}_{0} = 0 \), a straightforward induction shows that\n\n\[ R\left\lbrack x\right\rbrack = \mathop{\sum }\limits_{{i = 0}}^{{n - 1}}R{x}^{i} \]\n\nIn parti... | Yes |
Corollary 3.1.3. The integral elements in an R-subalgebra A form a subalgebra. In particular, \( A \) is integral over \( R \) if and only if it has a set of integral generators. | Proof. Let \( a, b \in A \) be integral elements. One readily checks that\n\n- \( R\left\lbrack {a, b}\right\rbrack \) is finitely generated as an \( R \) -module (say by certain monomials \( {a}^{i}{b}^{j} \) );\n\n- \( R\left\lbrack {a, b}\right\rbrack \) is faithful (as an \( R\left\lbrack {a, b}\right\rbrack \) -mo... | Yes |
Proposition 3.1.4. Consider ring homomorphisms \( R \rightarrow A \rightarrow B \) such that \( A \) is integral over R. If \( y \in B \) is integral over \( A \), then it is integral over \( R \) . | Proof. Assume \( {y}^{n} + {a}_{n - 1}{y}^{n - 1} + \cdots + {a}_{0} = 0 \) with \( {a}_{0},\ldots ,{a}_{n - 1} \in A \) integral over \( R \) . The \( R\left\lbrack {{a}_{0},\ldots ,{a}_{n - 1}}\right\rbrack \) -module \( R\left\lbrack {{a}_{0},\ldots ,{a}_{n - 1}}\right\rbrack \left\lbrack y\right\rbrack \) is also f... | Yes |
Proposition 3.1.6. Unique factorization domains are normal. | Proof. Given \( x = r/s \in K \) with coprime \( r, s \in R \) . If there is an integral dependence relation \( {x}^{n} + {a}_{n - 1}{x}^{n - 1} + \cdots + {a}_{0} = 0 \), we will have \( {r}^{n} + {a}_{n - 1}{r}^{n - 1}s + \cdots + {a}_{0}{s}^{n} = 0 \), hence \( s \mid {r}^{n} \) . As \( r \) is coprime to \( s \), w... | Yes |
Lemma 3.1.7. Let \( A \) be an R-algebra, \( S \) be a multiplicative subset of \( R \) . Denote by \( \widetilde{R} \) the integral closure of \( R \) in \( A \) . Then the integral closure of \( R\left\lbrack {S}^{-1}\right\rbrack \) in \( A\left\lbrack {S}^{-1}\right\rbrack \) equals \( \widetilde{R}\left\lbrack {S}... | Proof. Suppose that \( x \in A \) satisfies \( {x}^{n} + \mathop{\sum }\limits_{{i = 0}}^{{n - 1}}{b}_{i}{x}^{i} = 0 \) with \( {b}_{0},\ldots ,{b}_{n - 1} \in R \) . For all \( s \in S \) we deduce \( {\left( \frac{x}{s}\right) }^{n} + \mathop{\sum }\limits_{{i = 0}}^{{n - 1}}\frac{{b}_{i}}{{s}^{n - i}} \cdot {\left( ... | Yes |
Theorem 3.2.3 (E. Snapper). For any \( R \), the polynomial algebra \( R\left\lbrack X\right\rbrack \) satisfies \( \operatorname{rad}\left( {R\left\lbrack X\right\rbrack }\right) = \) \( \operatorname{nil}\left( {R\left\lbrack X\right\rbrack }\right) \) . | Proof. To show that \( \operatorname{nil}\left( {R\left\lbrack X\right\rbrack }\right) \supset \operatorname{rad}\left( {R\left\lbrack X\right\rbrack }\right) \), let \( f\left( X\right) = \mathop{\sum }\limits_{i}{a}_{i}{X}^{i} \in \operatorname{rad}\left( {R\left\lbrack X\right\rbrack }\right) \), then \( 1 + {Xf}\le... | Yes |
Lemma 3.2.4. Let \( R \subset A \) be integral domains such that \( A \) is a finitely generated \( R \) -algebra. If \( \operatorname{rad}\left( R\right) = \{ 0\} \), then \( \operatorname{rad}\left( A\right) = \{ 0\} \) . | Proof. We may assume that \( A \) is generated by a single element \( a \in A \) over \( R \) . If \( a \) is transcendental over \( K \mathrel{\text{:=}} \operatorname{Frac}\left( R\right) \), Theorem 3.2.3 above can be applied as nil \( \left( {R\left\lbrack X\right\rbrack }\right) = \) \( \{ 0\} \) . Let us assume t... | Yes |
Theorem 3.2.5 (Nullstellensatz). Let \( A \) be a finitely generated R-algebra. Assume that \( R \) is a Jacobson ring, then the following statements hold.\n\n(i) \( A \) is a Jacobson ring.\n\n(ii) Let \( \mathfrak{n} \in \operatorname{Spec}\left( A\right) \) be maximal, then its image \( \mathfrak{m} \in \operatornam... | Proof. We start with (i). One may assume \( R \subset A \) from the outset. Condition (3-1) for \( A \) amounts to \( \operatorname{rad}\left( {A/\mathfrak{p}}\right) = 0 \) for every \( \mathfrak{p} \in \operatorname{Spec}\left( A\right) \) . Apply the previous Lemma to the integral domains \( R/R \cap \mathfrak{p} \s... | Yes |
Corollary 3.2.6. Let \( \mathbb{k} \) be an algebraically closed field, and \( A \mathrel{\text{:=}} \mathbb{k}\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) . The maximal ideals of \( A \) are in bijection with \( {\mathbb{k}}^{n} \) by attaching to each \( x \mathrel{\text{:=}} \left( {{x}_{1},\ldots ,{x}_{n}... | Proof. Since \( A/{\mathfrak{m}}_{x}\overset{ \sim }{ \rightarrow }\mathbb{k} \) by evaluation at \( x \), we see \( {\mathfrak{m}}_{x} \) is indeed maximal. It is routine to show that \( x = y \Leftrightarrow {\mathfrak{m}}_{x} = {\mathfrak{m}}_{y} \) . It remains to show that every maximal ideal \( \mathfrak{n} \) co... | Yes |
Corollary 3.2.7. Keep the notations above and set\n\n\[ Z\\left( \\mathfrak{a}\\right) \\mathrel{\\text{:=}} \\left\\{ {x \\in {\\mathbb{k}}^{n} : \\forall f \\in \\mathfrak{a}, f\\left( x\\right) = 0}\\right\\} \]\n\n\[ I\\left( X\\right) \\mathrel{\\text{:=}} \\{ f \\in A : \\forall x \\in X, f\\left( x\\right) = 0\\... | Proof. If \\( f \\in A \\) and \\( {f}^{n} \\in \\mathfrak{a} \\) for some \\( n \\), the vanishing of \\( {f}^{n} \\) on \\( Z\\left( \\mathfrak{a}\\right) \\) will entail that of \\( f \\), hence the inclusion \\( \\supset \\) holds. Assume conversely that \\( f \\in A \\) vanishes on \\( Z\\left( \\mathfrak{a}\\righ... | Yes |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.