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Theorem 2.13 Every space \( X \) is the quotient of a Hausdorff space \( H \) . | Proof. Omitted. See Shimrat (1956). | No |
Example 2.5 Metric spaces are Hausdorff. | To see this, let \( x \) and \( y \) be points in a metric space. If \( x \neq y \), then \( d \mathrel{\text{:=}} d\left( {x, y}\right) > 0 \) and \( B\left( {x,\frac{d}{2}}\right) \) and \( B\left( {y,\frac{d}{2}}\right) \) are disjoint open sets separating \( x \) and \( y \) . | Yes |
Theorem 2.14 A space \( X \) is Hausdorff if and only if the diagonal map \( \Delta : X \rightarrow X \times X \) is closed. | Proof. Exercise. | No |
Theorem 2.15 If \( X \) is compact and \( f : X \rightarrow Y \) is continuous, then \( {fX} \) is compact. | Proof. Exercise. | No |
The Bolzano-Weierstrass Theorem Every infinite set in a compact space has a limit point. | Proof. Suppose that \( F \) is an infinite subset with no limit points. If \( x \) is not a limit point of \( F \) and \( x \notin F \), there is an open set \( {U}_{x} \) around \( x \) that misses \( F \) . If \( x \) is not a limit point of \( F \) and \( x \in F \), then there is an open set \( {U}_{x} \) with \( {... | Yes |
Theorem 2.16 A space \( X \) is compact if and only if every collection of closed subsets of \( X \) with the FIP has nonempty intersection. | Proof. Exercise. | No |
Theorem 2.17 Closed subsets of compact spaces are compact. | Proof. Let \( X \) be compact with \( C \subseteq X \) closed and suppose \( \mathcal{U} = {\left\{ {U}_{\alpha }\right\} }_{\alpha \in A} \) is an open cover of \( C \) . Then \( X \smallsetminus C \) together with \( \mathcal{U} \) forms an open cover of \( X \) . Since \( X \) is compact, there are finitely many set... | Yes |
Theorem 2.18 Let \( X \) be Hausdorff. For any point \( x \in X \) and any compact set \( K \subseteq X \smallsetminus \{ x\} \) there exist disjoint open sets \( U \) and \( V \) with \( x \in U \) and \( K \subseteq V \) . | Proof. Let \( x \in X \) and let \( K \subsetneq X \) be compact. For each \( y \in K \), there are disjoint open sets \( {U}_{y} \) and \( {V}_{y} \) with \( x \in {U}_{y} \) and \( y \in {V}_{y} \) . The collection \( \left\{ {V}_{y}\right\} \) is an open cover of \( K \) ; hence there is a finite subcover \( \left\{... | Yes |
Corollary 2.18.1 Compact subsets of Hausdorff spaces are closed. | Proof. Exercise. | No |
Corollary 2.18.2 If \( X \) is compact and \( Y \) is Hausdorff, then every map \( f : X \rightarrow Y \) is closed. | Proof. Let \( f : X \rightarrow Y \) be a map from a compact space to a Hausdorff space, and let \( C \subseteq X \) be closed. Then \( C \) is compact, so \( {fC} \) is compact, so \( {fC} \) is closed. | Yes |
Corollary 2.18.3 (Heine-Borel Theorem) A subset of \( {\mathbb{R}}^{n} \) is compact if and only if it is closed and bounded. | Proof. Suppose that \( K \subset {\mathbb{R}}^{n} \) is compact. Since the cover of \( K \) consisting of open balls centered at the origin of all possible radii must have a finite subcover, \( K \) must be bounded. Since \( {\mathbb{R}}^{n} \) is Hausdorff and all compact subsets of a Hausdorff space must be closed, \... | Yes |
Corollary 2.18.4 Continuous functions from compact spaces to \( \mathbb{R} \) have both a global maximum and a global minimum. | Proof. Exercise. | No |
The Tube Lemma For any open set \( U \subseteq X \times Y \) and any set \( K \times \{ y\} \subseteq U \) with \( K \subseteq X \) compact, there exist open sets \( V \subseteq X \) and \( W \subseteq Y \) with \( K \times \{ y\} \subseteq V \times W \subseteq U \) . | Proof. For each point \( \left( {x, y}\right) \in K \times \{ y\} \), there are open sets \( {V}_{x} \subseteq X \) and \( {W}_{x} \subseteq Y \) with \( \left( {x, y}\right) \in \) \( {V}_{x} \times {W}_{x} \subseteq U \) . Then, \( {\left\{ {V}_{x}\right\} }_{x \in K} \) is an open cover of \( K \) ; take a finite su... | Yes |
Theorem 2.19 Suppose \( X \) is locally compact and Hausdorff. Then for every point \( x \in X \) and every neighborhood \( U \) of \( x \), there exists a neighborhood \( V \) of \( x \) such that the closure \( \bar{V} \) is compact and \( x \in V \subseteq \bar{V} \subseteq U \) . | Proof. This is a corollary of theorem 2.18 and the definition of local compactness. | No |
Theorem 2.20 If \( {X}_{1} \rightarrow {Y}_{1} \) and \( {X}_{2} \rightarrow {Y}_{2} \) are quotient maps and \( {Y}_{1} \) and \( {X}_{2} \) are locally compact and Hausdorff, then \( {X}_{1} \times {X}_{2} \rightarrow {Y}_{1} \times {Y}_{2} \) is a quotient map. | Proof. We postpone the proof until theorem 5.7 in chapter 5. | No |
Consider \( \mathbb{Z} \) with the cofinite topology. For any \( m \in \mathbb{Z} \), the constant sequence \( m, m, m,\ldots \) converges to \( m \) and only to \( m \). | Indeed if \( l \neq m \), then the set \( \mathbb{R} \smallsetminus m \) is an open set around \( l \) containing no elements of the sequence. | Yes |
Example 3.3 Consider \( \mathbb{R} \) with the usual topology. If \( \left\{ {x}_{n}\right\} \rightarrow x \), then \( \left\{ {x}_{n}\right\} \) does not converge to any number \( y \neq x \) . | To prove it, note we can always find disjoint open sets \( U \) and \( V \) with \( x \in U \) and \( y \in V \) . (We can be explicit if necessary: \( U = \left( {x - c, x + c}\right) \) and \( V = \left( {y - c, y + c}\right) \) where \( c = \frac{1}{2}\left| {x - y}\right| \) .) Then there is a number \( N \) so tha... | Yes |
Theorem 3.1 A space \( X \) is \( {T}_{1} \) if and only if for any \( x \in X \), the constant sequence \( x, x, x,\ldots \) converges to \( x \) and only to \( x \) . | Proof. Suppose \( X \) is \( {T}_{1} \) and \( x \in X \) . It’s clear that \( x, x, x,\ldots \rightarrow x \) . Let \( y \neq x \) . Then there exists an open set \( U \) with \( y \in U \) and \( x \notin U \) . Therefore, \( x, x, x,\ldots \) cannot converge to \( y \) .\n\nFor the converse, suppose \( X \) is not \... | Yes |
Theorem 3.2 If \( X \) is Hausdorff, then sequences in \( X \) have at most one limit. | Proof. Let \( X \) be Hausdorff and suppose \( \left\{ {x}_{n}\right\} \) is a sequence such that \( \left\{ {x}_{n}\right\} \rightarrow x \) and \( \left\{ {x}_{n}\right\} \rightarrow y \) . If \( x \neq y \), then there are disjoint open sets \( U \) and \( V \) with \( x \in U \) and \( y \in V \) . Since \( \left\{... | Yes |
Theorem 3.3 If \( \left\{ {x}_{n}\right\} \) is a sequence in \( A \) that converges to \( x \), then \( x \in \bar{A} \) . | Proof. Exercise. | No |
Theorem 3.5 Let \( X \) be a first countable space. Then \( X \) is Hausdorff if and only if every sequence has at most one limit. | Proof. Suppose that \( X \) is first countable. If \( X \) is not Hausdorff, there exist points \( x \) and \( y \) that cannot be separated by open sets. Let \( {U}_{1},{U}_{2},\ldots \) be a neighborhood base of \( x \) and \( {V}_{1},{V}_{2},\ldots \) be a neighborhood base for \( y \) . For every \( n \) choose a p... | Yes |
Theorem 3.6 Let \( X \) be a first countable space and let \( A \subseteq X \) . A point \( x \in \bar{A} \) if and only if there exists a sequence \( \left\{ {x}_{n}\right\} \) in \( A \) with \( \left\{ {x}_{n}\right\} \rightarrow x \) . | Proof. Exercise. | No |
Theorem 3.7 Suppose \( X \) and \( Y \) are first countable. A function \( f : X \rightarrow Y \) is continuous if and only if for every sequence \( \left\{ {x}_{n}\right\} \) in \( X \) with \( \left\{ {x}_{n}\right\} \rightarrow x \), the sequence \( \left\{ {f{x}_{n}}\right\} \rightarrow {fx} \). | Proof. Exercise. | No |
Given a topological space \( \left( {X,\mathcal{T}}\right) \), the open neighborhoods \( {\mathcal{T}}_{x} \) of a point \( x \) form a filterbase, though they generally do not form a filter. | The reason is simply that (usually) not every set containing an open neighborhood of \( x \) is open. But there is some ambiguity in the mathematics community on this point. Kelley (1955) defines \ | No |
Example 3.14 The real-valued sequence \( \left\{ {x}_{n}\right\} \mathrel{\text{:=}} \{ 1, - 1,\frac{1}{2}, - 1,\frac{1}{4}, - 1,\frac{1}{8},\ldots \} \) does not converge, whereas the subsequence \( \left\{ {x}_{2n}\right\} = \left\{ {1,\frac{1}{2},\frac{1}{4},\frac{1}{8},\ldots }\right\} \) does. | We can see this by reasoning with eventuality filters, which isn't so different from reasoning with sequences. Here's the thing to notice: the eventuality filter \( {\mathcal{E}}_{{x}_{2n}} \) is the set of all subsets \( A \subset \mathbb{R} \) for which there exists an \( N \) so that \( \frac{1}{{2}^{n}} \in A \) fo... | Yes |
Theorem 3.8 A space is Hausdorff if and only if limits of convergent proper filters are unique. | Proof. Suppose \( X \) is Hausdorff and that a proper filter \( \mathcal{F} \) converges to both \( x \) and \( y \) with \( x \neq y \) . Then there are open neighborhoods \( U \) of \( x \) and \( V \) of \( y \) with \( U \cap V = \varnothing \) . By convergence, \( U, V \in \mathcal{F} \) . Since \( \mathcal{F} \) ... | Yes |
Theorem 3.9 Let \( X \) be a space with \( A \subseteq X \) . A point \( x \in \bar{A} \) if and only if there exists a proper filter \( \mathcal{F} \) containing \( A \) with \( \mathcal{F} \rightarrow x \) . | Proof. First recall that \( x \in \bar{A} \) if and only if every neighborhood of \( x \) nontrivially intersects \( A \) or equivalently if and only if the filterbase \( \mathcal{B} = \{ U \cap A{\} }_{U \in {\mathcal{T}}_{x}} \) does not contain the empty set. So if \( x \in \bar{A} \), then simply generate a proper ... | Yes |
Theorem 3.10 A function \( f : X \rightarrow Y \) is continuous if and only if for every filter \( \mathcal{F} \) on \( X \) , if \( \mathcal{F} \rightarrow x \), then \( {f}_{ * }\mathcal{F} \rightarrow {fx} \) . | Proof. Let \( \mathcal{F} \) be a filter on \( X \) with \( \mathcal{F} \rightarrow x \), and suppose \( f : X \rightarrow Y \) is continuous. We want to show \( {\mathcal{T}}_{fx} \subseteq {f}_{ * }\mathcal{F} \) ; that is, for any \( B \in {\mathcal{T}}_{fx} \) there exists a set \( A \in \mathcal{F} \) with \( {fA}... | Yes |
Proposition 3.1 A filter \( \mathcal{U} \) on a set \( X \) is an ultrafilter if and only if for every subset \( A \subseteq X \) the following condition holds: \( A \notin \mathcal{U} \) if and only if there exists \( B \in \mathcal{U} \) with \( A \cap B = \varnothing \) . | Proof. Let \( \mathcal{U} \) be an ultrafilter. Then \( A \notin \mathcal{U} \) if and only if the filter generated by \( \mathcal{U} \cup \{ A\} \) is the powerset \( {2}^{X} \) . Since the generated filter consists of all sets containing an intersection of the form \( B \cap A \) for some \( B \in \mathcal{U} \), thi... | Yes |
Example 3.18 Let \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) be a bounded function and let \( \left( {\mathcal{P}, \preccurlyeq }\right) \) be the poset of partitions of \( \left\lbrack {a, b}\right\rbrack \) ordered by refinement. Any partition \( P = \left\{ {a = {x}_{0} < {x}_{1} < \cdots < {x}... | So ultrafilters and the natural ordering of partitions allow us to replace the typical \ | No |
Theorem 3.11 A filter on \( X \) is maximal if and only if it is prime. | Proof. Suppose \( \mathcal{F} \) is an ultrafilter on \( X \) and fails to be prime. Then there are \( A, B \subseteq X \) such that \( A \cup B \in \mathcal{F} \), but neither \( A \) nor \( B \) are in \( \mathcal{F} \) . The latter holds if and only if there exist sets \( {A}^{\prime },{B}^{\prime } \in \mathcal{F} ... | Yes |
Corollary 3.11.1 Any infinite set has a non-principal ultrafilter. | Proof. Consider the Fréchet filter \( \mathcal{F} \mathrel{\text{:=}} \{ A \subseteq X \mid X \smallsetminus A \) is finite \( \} \) and appeal to the Ultrafilter Lemma to extend \( \mathcal{F} \) to an ultrafilter \( \mathcal{U} \) . Were \( \mathcal{U} \) to contain any finite set, it would contain its (cofinite) com... | Yes |
Theorem 3.12 A space \( X \) is compact if and only if every prime filter converges. | Proof. Suppose \( \mathcal{F} \) is a prime filter that fails to converge to any \( x \in X \) . Equivalently, suppose for all \( x \) there exists \( {U}_{x} \in {\mathcal{T}}_{x} - \mathcal{F} \) . The set \( {\left\{ {U}_{x}\right\} }_{x \in X} \) is an open cover. By compactness, choose a finite subcover \( {\left\... | Yes |
Theorem 3.13 Let \( \mathcal{U} \) be an ultrafilter on \( X \) and let \( f : X \rightarrow Y \) . The pushforward \( {f}_{ * }\mathcal{U} \) is an ultrafilter on \( Y \) . | Proof. Exercise. | No |
Theorem 3.14 Tychonoff’s theorem is equivalent to the axiom of choice. | Proof. We used Zorn's lemma to prove Tychonoff's theorem. Although we don't prove it, the axiom of choice implies Zorn's lemma (see exercise 3.14 at the end of the chapter), from which it follows that Tychonoff's theorem is implied by the axiom of choice.\n\nTo prove that Tychonoff’s theorem implies the axiom of choice... | No |
In linear algebra, the colimit of \( \mathbb{N} \) copies of \( \mathbb{R} \) is the set of sequences of real numbers for which all but finitely many are zero and is denoted \( { \oplus }_{n \in \mathbb{N}}\mathbb{R} \) . This is not the same as the colimit of \( \mathbb{N} \) copies of \( \mathbb{R} \) in Top, which i... | Specifically, \( X = { \oplus }_{n \in \mathbb{N}}\mathbb{R} \) is the colimit of the diagram of (vector and topological) spaces\n\n\[ \mathbb{R} \rightarrow {\mathbb{R}}^{2} \rightarrow {\mathbb{R}}^{3} \rightarrow \cdots \]\n\nwhere the map \( {\mathbb{R}}^{n} \rightarrow {\mathbb{R}}^{n + 1} \) is given by \( \left(... | Yes |
If a category has products and equalizers, then it is complete. If it has coproducts and coequalizers, then it is cocomplete. | Here's how to construct the colimit of a diagram in a category with coproducts and coequalizers. Proceed in two steps. First, take the coproduct \( Y \) of all the objects \( {X}_{\alpha } \) in the diagram that have morphisms \( {X}_{\alpha } \rightarrow {X}_{\beta } \) from them (there may be multiple copies of \( {X... | No |
For any sets \( X, Y \), and \( Z \), the bijection \( {Y}^{X \times Z}\overset{ \cong }{ \rightarrow }{\left( {Y}^{X}\right) }^{Z} \) arises from an adjunction. | The functor \( X \times - : \) Set \( \rightarrow \) Set is left adjoint to the functor \( \operatorname{Set}\left( {X, - }\right) : \) Set \( \rightarrow \) Set. To see it clearly, fix a set \( X \), and define two functors\n\n\[ L \mathrel{\text{:=}} X \times - : \text{ Set } \rightarrow \text{ Set }\;R \mathrel{\tex... | Yes |
Example 5.2 Let’s look at the unit and counit of the product-hom adjunction \( L \) : Set \( \rightleftarrows \) Set: \( R \) in Set where\n\n\[ L = X \times - : \text{ Set } \rightarrow \text{ Set }\;\text{ and }\;R = \operatorname{Set}\left( {X, - }\right) : \text{ Set } \rightarrow \text{ Set } \] | The counit of this adjunction is the evaluation map eval: \( X \times {Y}^{X} \rightarrow Y \) defined by \( \operatorname{eval}\left( {x, f}\right) = \) \( f\left( x\right) \) . The unit is the map \( Z \rightarrow {\left( X \times Z\right) }^{X} \) defined by \( z \mapsto \left( {-, z}\right) \), where \( \left( {-, ... | Yes |
Theorem 5.1 If \( L : \mathrm{C} \rightarrow \mathrm{D} \) has a right adjoint, then \( L \) is cocontinuous. If \( R : \mathrm{D} \rightarrow \mathrm{C} \) has a left adjoint, then \( R \) is continuous. | Proof. Recall from exercise 4.8 at the end of chapter 4 that there is a natural isomorphism:\n\n\[ \mathrm{C}\left( {\operatorname{colim}F, Y}\right) \; \cong \;\lim \mathrm{C}\left( {F\left( -\right), Y}\right) \]\n\nfor any functor \( F : \mathrm{B} \rightarrow \mathrm{C} \) . Therefore,\n\n\[ \mathrm{D}\left( {L\lef... | No |
Corollary 5.1.1 Right adjoints preserve products. | Proof. Immediate. Products are limits. | No |
If a compactification \( Y \) of a space \( X \) is obtained by adding a single point to \( X \), then \( X \hookrightarrow Y \) is called a one-point compactification-also sometimes called the Alexandroff one-point compactification. A space \( X \) has one-point compactification if and only if \( X \) is Hausdorff and... | To see this, suppose \( X \hookrightarrow {X}^{ * } \) is a compactification and \( {X}^{ * } \smallsetminus X = \{ p\} \) . The open neighborhoods of \( p \) are precisely the complements of compact subsets of \( X \) : the complement of an open set containing \( p \) is a closed subset of a compact space and so is co... | Yes |
Theorem 5.2 Suppose \( X \) is locally compact, Hausdorff, and not compact, and let \( i : X \rightarrow \) \( {X}^{ * } \) be the one-point compactification of \( X \) . If \( e : X \rightarrow Y \) is any other compactification of \( X \), then there exists a unique quotient map \( q : Y \rightarrow {X}^{ * } \) with... | Proof. The idea is that the quotient of \( Y \) obtained by identifying \( Y \smallsetminus {eX} \) to one point is homeomorphic to \( {X}^{ * } \) . The details are left as an exercise. | No |
Lemma 5.1 A topology on \( \operatorname{Top}\left( {X, Y}\right) \) is conjoining if and only if the evaluation map eval : \( X \times \operatorname{Top}\left( {X, Y}\right) \rightarrow Y \) is continuous. | Proof. Assume we have a topology on \( \operatorname{Top}\left( {X, Y}\right) \) for which the evaluation map is continuous. Consider a continuous map \( \widehat{g} : Z \rightarrow \operatorname{Top}\left( {X, Y}\right) \), and look at the following diagram\n\n\[ X \times Z\overset{\text{ id } \times \widehat{g}}{ \ri... | Yes |
Lemma 5.2 Every splitting topology on \( \operatorname{Top}\left( {X, Y}\right) \) is coarser than every conjoining topology. | Proof. Let \( \mathcal{T},{\mathcal{T}}^{\prime } \) be topologies on \( \operatorname{Top}\left( {X, Y}\right) \) . If \( {\mathcal{T}}^{\prime } \) is conjoining, then the evaluation map \( X \times \left( {\operatorname{Top}\left( {X, Y}\right) ,{\mathcal{T}}^{\prime }}\right) \rightarrow Y \) is continuous. If in a... | Yes |
Theorem 5.3 If there exists an exponential topology on \( \operatorname{Top}\left( {X, Y}\right) \), then it is unique. | Proof. Suppose \( \mathcal{T} \) and \( {\mathcal{T}}^{\prime } \) are exponential topologies on \( \operatorname{Top}\left( {X, Y}\right) \) . Since \( \mathcal{T} \) is splitting and \( {\mathcal{T}}^{\prime } \) is conjoining, we have \( \mathcal{T} \subseteq {\mathcal{T}}^{\prime } \) . And vice versa: we have \( {... | Yes |
Lemma 5.3 Let \( X \) be a metric space and let \( U \) be open. For every compact set \( K \subseteq U \) , there is an \( \varepsilon > 0 \) so that for any \( x \in K \) and any \( y \in X \smallsetminus U, d\left( {x, y}\right) > \varepsilon \) . | Proof. This is a straightforward argument using the definition of compactness. | No |
Theorem 5.5 For any spaces \( X \) and \( Y \), the compact-open topology on \( \operatorname{Top}\left( {X, Y}\right) \) is splitting. | Proof. Let \( Z \) be any space, and suppose \( g : X \times Z \rightarrow Y \) is continuous. To show that the adjunct \( \widehat{g} : Z \rightarrow \operatorname{Top}\left( {X, Y}\right) \) is continuous, consider a subbasic open set \( S\left( {K, U}\right) \) in \( \operatorname{Top}\left( {X, Y}\right) \) . We ne... | Yes |
Theorem 5.6 If \( X \) is locally compact and Hausdorff and \( Y \) is any space, then the compact-open topology on \( \operatorname{Top}\left( {X, Y}\right) \) is exponential. | Proof. We only need to check that the compact-open topology is conjoining, and this is equivalent to showing that the evaluation map eval: \( X \times \operatorname{Top}\left( {X, Y}\right) \rightarrow Y \) is continuous at every point \( \left( {x, f}\right) \) . Let \( \left( {x, f}\right) \in X \times \operatorname{... | Yes |
Lemma 5.4 If \( f : X \rightarrow Y \) is a quotient map and \( Z \) is locally compact and Hausdorff, then \( f \times {\operatorname{id}}_{Z} : X \times Z \rightarrow Y \times Z \) is a quotient map. | Proof. Let \( f : X \rightarrow Y \) be a quotient map. We want to prove that the product \( Y \times Z \) has the quotient topology inherited from the map \( f \times {\mathrm{{id}}}_{Z} \) . So consider \( Y \times Z \) with two possibly distinct topologies: \( {\left( Y \times Z\right) }_{p} \) will denote the produ... | Yes |
Theorem 5.7 If \( {X}_{1} \rightarrow {Y}_{1} \) and \( {X}_{2} \rightarrow {Y}_{2} \) are quotient maps and \( {Y}_{1} \) and \( {X}_{2} \) are locally compact and Hausdorff, then \( {X}_{1} \times {X}_{2} \rightarrow {Y}_{1} \times {Y}_{2} \) is a quotient map. | Proof. Suppose \( {Y}_{1} \) and \( {X}_{2} \) are locally compact and Hausdorff and that \( {f}_{1} : {X}_{1} \rightarrow {Y}_{1} \) and \( {f}_{2} : {X}_{2} \rightarrow {Y}_{2} \) are quotient maps. By the lemma, the two maps \( {f}_{1} \times {\mathrm{{id}}}_{{X}_{2}} : {X}_{1} \times {X}_{2} \rightarrow {Y}_{1} \ti... | Yes |
Theorem 5.8 If \( X \) is any space and \( Y \) is Hausdorff, then a subset \( A \subseteq \operatorname{Top}\left( {X, Y}\right) \) has compact closure in the product topology if and only if for each \( x \in X \), the set \( {A}_{x} = \left\{ {{fx} \in }\right. \) \( Y \mid f \in A\} \) has compact closure in \( Y \)... | Proof. This was exercise 2.19 at the end of chapter 2. | No |
Theorem 5.9 If \( X \) and \( Z \) are locally compact Hausdorff, then for any space \( Y \), the isomorphism of sets \( \operatorname{Top}\left( {Z \times X, Y}\right) \rightarrow \operatorname{Top}\left( {Z,\operatorname{Top}\left( {X, Y}\right) }\right) \) is a homeomorphsim of spaces. | Proof. Throughout this proof, remember that for any spaces \( A \) and \( C \), the compact open topology on \( \operatorname{Top}\left( {A, C}\right) \) is splitting. This means that the adjunct of a continuous map \( A \times \) \( B \rightarrow C \), which is a function \( B \rightarrow \operatorname{Top}\left( {A, ... | Yes |
Theorem 5.10 The following setup is an adjunction: \( U : \mathrm{{CG}} \rightleftarrows \) Top: \( k \), where \( U \) is the inclusion of \( \mathrm{{CG}} \rightarrow \) Top and \( k \) is the \( k \) -ification functor. | Proof. Proved as theorem 3.2 in Steenrod (1967). | Yes |
Theorem 5.12 There is an adjunction \( q : \mathrm{{CG}} \rightleftarrows \mathrm{{CGWH}} : U \) where \( U \) is the inclusion of \( \mathrm{{CGWH}} \rightarrow \mathrm{{CG}} \) . | Proof. See appendix A of Lewis (1978). | No |
Theorem 5.13 CGWH is a complete category. | Proof. See proposition 2.22 in Strickland (2009). | No |
Theorem 5.14 If \( X \) and \( Y \) are CGWH, then \( {Y}^{X} \) is in CGWH. For a fixed \( X \), the assignment \( Y \mapsto {Y}^{X} \) defines a functor \( - {}^{X} \) : CGWH \( \rightarrow \) CGWH that fits into the adjunction\n\n\[ X \times - : \mathrm{{CGWH}} \rightleftarrows \mathrm{{CGWH}} : - {}^{X} \]\n\ninduc... | Proof. See Lewis (1978). | No |
Theorem 6.1 The map \( \pi : {X}^{I} \rightarrow X \) defined by \( \gamma \mapsto ⓾ \) and the map \( i : X \rightarrow {X}^{I} \) defined by \( x \mapsto {c}_{x} \), the constant path at \( x \), are homotopy inverses. | Proof. Note that \( {i\pi } : {X}^{I} \rightarrow {X}^{I} \) is the map that sends a path \( \gamma \) to \( {c}_{⓿} \), the constant path at \( ⓿ \) . Let \( h : {X}^{I} \times I \rightarrow {X}^{I} \) by \( h\left( {\gamma, t}\right) = {\gamma }_{t} \) where \( {\gamma }_{t} : I \rightarrow X \) is given by \( {\gamm... | Yes |
Theorem 6.2 The map \( i : X \rightarrow X \times I \) defined by \( x \mapsto \left( {x,1}\right) \) and the projection \( p : X \times I \rightarrow X \) are homotopy inverses. | Proof. Exercise. | No |
Theorem 6.4 Let \( X \) be a pointed space. Then \( {\pi }_{n}X \cong {\pi }_{n - 1}{\Omega X} \) for each \( n \geq 1 \) . | Proof. By corollary 6.3.1 and the suspension-loop adjunction,\n\n\[ \n{\pi }_{n}X = \left\lbrack {{S}^{n}, X}\right\rbrack \]\n\n\[ \n\cong \;\left\lbrack {\sum {S}^{n - 1}, X}\right\rbrack \]\n\n\[ \n\cong \;\left\lbrack {{S}^{n - 1},{\Omega X}}\right\rbrack \]\n\n\[ \n= {\pi }_{n - 1}{\Omega X} \]\n\nSince this const... | Yes |
For any based space \( X \), the map \( p : \mathcal{P}X \rightarrow X \) sending a path \( \gamma \) to its endpoint \( \gamma \left( 1\right) \) is a fibration. | Suppose we have the commuting square\n\n\n\nwhere \( Z \) is any pointed space. Notice that for a fixed \( z \in Z,{gz} \) is a path in \( X \) ending at the point \( h\left( {z,0}\right) \), which, with \( z \) stil... | Yes |
Proposition 6.1 \( \mathcal{P}X \) is contractible. | Proof. Let \( * \) denote the one-point space. The composition \( * \rightarrow \mathcal{P}X \rightarrow * \) is equal to \( {\mathrm{{id}}}_{ * } \), so to prove \( \mathcal{P}X \) is homotopy equivalent to \( * \), we need only show the composition \( \mathcal{P}X \rightarrow * \rightarrow \mathcal{P}X \) that sends ... | Yes |
The map \( p \) from \( \mathbb{R} \) to \( {S}^{1} \) given by \( y \mapsto {e}^{2\pi iy} \) is a fibration with fiber \( \mathbb{Z} \) . That is, if the following diagram commutes\n\n\n\nthen there is a lift of the... | The key here is that \( p \) is a local homeomorphism: for every \( y \in \mathbb{R} \) there is an open neighborhood of \( y \) that maps homeomorphically onto its image-for instance, the interval \( \left( {y - \frac{1}{2}, y + \frac{1}{2}}\right) \) would work. So there is an open cover \( \mathcal{U} \) of \( {S}^{... | Yes |
Theorem 6.5 Suppose \( p \) and \( q \) are fibrations with base space \( B \) and \( f \) is a map of total spaces causing the diagram to commute:\n\n\n\nIf \( f \) is a homotopy equivalence, then \( f \) induces a ... | Proof. By assumption, there is a homotopy \( {h}^{\prime } \) from \( f{f}^{\prime } \) to \( {\mathrm{{id}}}_{D} \) . Postcomposing it with the fibration \( q \) gives homotopy \( h : D \times I \rightarrow B \) from \( p{f}^{\prime } \) to \( q \), and the outer square commutes by commutativity of the triangle:\n\n![... | Yes |
Corollary 6.5.1 The loop space of the circle \( \Omega {S}^{1} \) is homotopy equivalent to \( \mathbb{Z} \) | Proof. Both \( \mathbb{R} \rightarrow {S}^{1} \) and \( \mathcal{P}{S}^{1} \rightarrow {S}^{1} \) are fibrations by earlier examples. Moreover, both \( {\mathcal{{PS}}}^{1} \) and \( \mathbb{R} \) are contractible (proposition 6.1 and example 1.21). There is thus a homotopy equivalence between them that commutes with t... | Yes |
Corollary 6.5.3 The fundamental group of \( {S}^{1} \) is isomorphic to \( \mathbb{Z} \) . | Proof. By the previous corollary, \( \Omega {S}^{1} \simeq \mathbb{Z} \) which implies\n\n\[{\pi }_{0}\Omega {S}^{1} \cong {\pi }_{0}\mathbb{Z}\]\n\nThe left-hand side is \( {\pi }_{1}{S}^{1} \) by theorem 6.4. The right-hand side is the set of path components of \( \mathbb{Z} \), which is simply \( \mathbb{Z} \). \n\n... | Yes |
Corollary 6.5.4 The \( n \) th homotopy group of the circle is trivial for \( n \geq 2 \) . | Proof. If \( n \geq 2 \), then\n\n\[ \n{\pi }_{n}{S}^{1} = {\pi }_{n - 1}\Omega {S}^{1} = {\pi }_{n - 1}\mathbb{Z} = \left\lbrack {{S}^{n - 1},\mathbb{Z}}\right\rbrack \]\n\nThe result follows since \( {S}^{n - 1} \) is connected for \( n > 1 \) so any basepoint-preserving map from it to \( \mathbb{Z} \) must be consta... | Yes |
Theorem 6.6 For any space \( X \) the \( n \) th homotopy group \( {\pi }_{n}X \) is abelian for \( n \geq 2 \) . | The proof is left to exercise 6.4 at the end of the chapter. The picture below gives a hint when \( n = 2 \) . | No |
The degree of the identity map on \( {S}^{1} \) is 1. | The degree of the map sending \( z \) to \( {iz} \) is also 1 since rotation by \( {90}^{ \circ } \) is homotopic to the identity map. And for any \( n \geq 1 \) , the degree of the map \( z \mapsto {z}^{n} \) is \( n \) . | No |
Theorem 6.7 If \( f : {S}^{1} \rightarrow {S}^{1} \) has degree \( n \neq 1 \), then \( f \) has a fixed point. | Proof. If \( f \) does not have a fixed point, define \( h : {S}^{1} \times I \rightarrow {S}^{1} \) by\n\n\[ h\left( {x, t}\right) = \frac{\left( {1 - t}\right) {fx} + {tx}}{\left| \left( 1 - t\right) fx + tx\right| }\n\]\n\nThen \( h \) gives a homotopy between \( f \) and \( {\operatorname{id}}_{{S}^{1}} \) and so \... | Yes |
Proposition 6.2 Suppose \( U \) and \( V \) are open subsets of a topological space \( X = U \cup V \) , and suppose \( {x}_{0} \in U \cap V \) . Then one has the following diagram of spaces and continuous functions:\n\n is path connected.\n\nTo see why \( U \cap V \) must be path connected, consider the circle \( X = {S}^{1} \), and let \( U \) and \( V \)... | No |
Example 6.4 Suppose \( X = {S}^{2} \) is the sphere. Let \( U \) be all of \( {S}^{2} \) except for the point \( \left( {0,0,1}\right) \), and let \( V \) be all of \( {S}^{2} \) except for \( \left( {0,0, - 1}\right) \) . Then \( U \cap V \) is homotopy equivalent to a circle, and thus its fundamental group is isomorp... | The fundamental group of \( {S}^{2} \) is then trivial. | No |
Example 6.6 Recall from example 1.17 that we can view the torus \( T \) as the quotient of a square with opposite sides identified: | All four corners of the square are identified to a single point, say, \( {t}_{0} \) . So let’s consider the pointed torus \( \left( {T,{t}_{0}}\right) \) . Now suppose \( p \) is any other point in \( T \) . Set \( V = T \smallsetminus \{ p\} \), and let \( U \) be a \ | No |
We can compute its fundamental group in the same way as for the torus. Let \( {k}_{0} \) be the single vertex of the square, and let \( p \) be another point in \( K \) . If \( U \) is an open disc containing \( p \) and \( V = K \smallsetminus \{ p\} \), then by the same arguments as in the previous example, we have t... | by Seifert van Kampen. The conclusion is that \( {\pi }_{1}\left( {K,{k}_{0}}\right) \) is isomorphic to a group with a presentation given in two generators and one relation, namely, \( \left\langle {\alpha ,\beta \mid {\alpha \beta \alpha }{\beta }^{-1}}\right\rangle \) . | Yes |
Proposition 1. Let \( \mathrm{K} \) be a field, and let \( v : {\mathrm{K}}^{ * } \rightarrow \mathbf{Z} \) be a homomorphism having properties a) and b) above. Then the set \( \mathrm{A} \) of \( x \in \mathrm{K} \) such that \( v\left( x\right) \geq 0 \) is a discrete valuation ring having \( v \) as its associated v... | Indeed, let \( \pi \) be an element such that \( v\left( \pi \right) = 1 \) . Every \( x \in A \) can be written in the form \( x = {\pi }^{n}u \), with \( n = v\left( x\right) \), and \( v\left( u\right) = 0 \), i.e., \( u \) invertible. Every nonzero ideal of \( \mathrm{A} \) is therefore of the form \( {\pi }^{n}\ma... | Yes |
Proposition 2. Let \( \mathrm{A} \) be a commutative ring. In order that \( \mathrm{A} \) be a discrete valuation ring, it is necessary and sufficient that it be a Noetherian local ring, and that its maximal ideal be generated by a non-nilpotent element. | It is clear that a discrete valuation ring has the stated properties. Conversely, suppose that A has these properties, and let \( \pi \) be a generator of the maximal ideal \( \mathfrak{m}\left( \mathrm{A}\right) \) of \( \mathrm{A} \) . Let \( \mathfrak{u} \) be the ideal of the ring \( \mathrm{A} \) formed by the ele... | Yes |
Proposition 3. Let A be a Noetherian integral domain. In order that A be a discrete valuation ring, it is necessary and sufficient that it satisfy the two following conditions:\n\n(i) \( \mathrm{A} \) is integrally closed.\n\n(ii) A has a unique non-zero prime ideal. | It is clear that a discrete valuation ring satisfies (ii). Let us show that it satisfies \( \left( i\right) \) . Let \( \mathrm{K} \) be the field of fractions of \( \mathrm{A} \), and let \( x \) be an element of \( \mathrm{K} \) satisfying an equation of type \( \left( *\right) \), and suppose \( x \) were not in A. ... | No |
Lemma 1. Let \( \mathrm{A} \) be a discrete valuation ring, and let \( {x}_{i} \) be elements of the field of fractions of \( \mathrm{A} \) such that \( v\left( {x}_{i}\right) > v\left( {x}_{1}\right) \) for \( i \geq 2 \) . One then has \[ {x}_{1} + {x}_{2} + \cdots + {x}_{n} \neq 0. \] | One can assume \( {x}_{1} = 1 \) (dividing by \( {x}_{1} \) if necessary), whence \( v\left( {x}_{i}\right) \geq 1 \) for \( i \geq 2 \), i.e., \( {x}_{i} \in \mathfrak{m}\left( \mathrm{A}\right) \) ; as \( {x}_{1} \notin \mathfrak{m}\left( \mathrm{A}\right) \), it follows that \( {x}_{1} + \cdots + {x}_{n} \notin \mat... | Yes |
Proposition 4. If \( \mathrm{A} \) is a Noetherian integral domain, the following two properties are equivalent:\n\n(i) For every prime ideal \( \mathfrak{p} \neq 0 \) of \( \mathrm{A},{\mathrm{A}}_{\mathfrak{p}} \) is a discrete valuation ring.\n\n(ii) A is integrally closed and of dimension \( \leq 1 \) . | (i) implies (ii): If \( \mathfrak{p} \subset {\mathfrak{p}}^{\prime } \), then \( {\mathrm{A}}_{{\mathfrak{p}}^{\prime }} \) contains the prime ideal \( \mathfrak{p}{\mathrm{A}}_{{\mathfrak{p}}^{\prime }} \), which implies \( \mathfrak{p} = 0 \) or \( \mathfrak{p} = {\mathfrak{p}}^{\prime } \) (cf. prop. 3,(ii)). On th... | Yes |
Proposition 5. In a Dedekind domain, every non-zero fractional ideal is invertible. | [If \( \mathrm{K} \) is the field of fractions of \( \mathrm{A} \), a fractional ideal \( \mathfrak{a} \) of \( \mathrm{A} \) is a sub- \( \mathrm{A} \) - module of \( \mathrm{K} \) finitely generated over \( \mathrm{A} \) . One says \( \mathfrak{a} \) is invertible if there exists \( \left. {{\mathfrak{a}}^{\prime } \... | Yes |
Proposition 6. If \( x \in \mathrm{A}, x \neq 0 \), then only finitely many prime ideals contain \( x \) . | Indeed, the ideals containing \( x \) satisfy the descending chain condition: if \( \mathrm{A}x \subset \mathfrak{a} \subset {\mathfrak{a}}^{\prime } \subset \mathrm{A} \), one has \( \mathrm{A}{x}^{-1} \supset {\mathfrak{a}}^{-1} \supset {\mathfrak{a}}^{\prime - 1} \supset \mathrm{A} \), and \( \mathrm{A} \) is Noethe... | Yes |
Proposition 7. Every fractional ideal \( \mathfrak{a} \) of \( \mathrm{A} \) can be written uniquely in the form:\n\n\[ \mathfrak{a} = \prod {\mathfrak{p}}^{{v}_{\mathfrak{p}}\left( \mathfrak{a}\right) } \]\n\nwhere the \( {v}_{\mathfrak{p}}\left( \mathfrak{a}\right) \) are integers almost all zero. | The following formulas are immediate:\n\n\[ {v}_{\mathfrak{p}}\left( {\mathfrak{a}.\mathfrak{b}}\right) = {v}_{\mathfrak{p}}\left( \mathfrak{a}\right) + {v}_{\mathfrak{p}}\left( \mathfrak{b}\right) \]\n\n\[ {v}_{\mathfrak{p}}\left( \left( {\mathrm{b} : \mathrm{a}}\right) \right) = {v}_{\mathfrak{p}}\left( {\mathrm{b} \... | No |
Proposition 8. Hypothesis (F) is satisfied when \( \\mathrm{L}/\\mathrm{K} \) is a separable extension. | Let \( \\operatorname{Tr} : \\mathrm{L} \\rightarrow \\mathrm{K} \) be the trace map (Bourbaki, Alg., Chap. V, \( §{10} \), no. 6). One knows (loc. cit., prop. 12) that \( \\operatorname{Tr}\left( {xy}\\right) \) is a symmetric non-degenerate \( \\mathbf{K} \) - bilinear form on \( \\mathrm{L} \) . If \( x \\in \\mathr... | Yes |
Lemma 2. Let \( \mathrm{A} \subset \mathrm{B} \) be rings, with \( \mathrm{B} \) integral over \( \mathrm{A} \) . If \( \mathfrak{P} \subset \mathfrak{Q} \) are prime ideals of \( \mathrm{B} \) such that \( \mathfrak{P} \cap \mathrm{A} = \mathfrak{Q} \cap \mathrm{A} \), then \( \mathfrak{P} = \mathfrak{Q} \) . | Passing to the quotient by \( \mathfrak{P} \), one may assume \( \mathfrak{P} = 0 \) . If \( \mathfrak{Q} \neq \mathfrak{P} \), there is a non-zero \( x \in \mathfrak{Q} \) . Let\n\n\[ \n{x}^{n} + {a}_{n - 1}{x}^{n - 1} + \cdots + {a}_{0} = 0,\;{a}_{i} \in \mathrm{A}, \n\]\n\nbe its minimal equation over A. One has \( ... | Yes |
Proposition 10. Let \( \mathfrak{p} \) be a non-zero prime ideal of \( \mathrm{A} \), the ring \( \mathrm{B}/\mathfrak{p}\mathrm{B} \) is an \( \mathrm{A}/\mathfrak{p} \) - algebra of degree \( n = \left\lbrack {\mathrm{L} : \mathrm{K}}\right\rbrack \), isomorphic to the product \( \mathop{\prod }\limits_{{\mathfrak{P}... | \[ n = \mathop{\sum }\limits_{{\mathfrak{P} \mid \mathfrak{p}}}{e}_{\mathfrak{P}}{f}_{\mathfrak{P}} \] Let \( \mathrm{S} = \mathrm{A} - \mathfrak{p},{\mathrm{A}}^{\prime } = {\mathrm{S}}^{-1}\mathrm{\;A} \), and \( {\mathrm{B}}^{\prime } = {\mathrm{S}}^{-1}\mathrm{\;B} \) . The ring \( {\mathrm{A}}^{\prime } = {\mathrm... | Yes |
Proposition 11. Let \( w \) be a discrete valuation of \( \mathrm{L} \) which prolongs \( {v}_{\mathfrak{p}} \) with index e. Then there is a prime divisor \( \mathfrak{P} \) of \( \mathfrak{p} \) with \( w = {v}_{\mathfrak{P}} \) and \( e = {e}_{\mathfrak{P}} \) . | Let \( \mathrm{W} \) be the ring of \( w \), and let \( \mathfrak{Q} \) be its maximal ideal. This ring is integrally closed with field of fractions \( \mathbf{L} \), and contains \( \mathbf{A} \) ; hence it contains \( \mathbf{B} \) . Let \( \mathfrak{P} = \mathfrak{Q} \cap \mathrm{B} \) . Obviously \( \mathfrak{P} \c... | Yes |
Proposition 12. If \( \mathrm{M} \) is a B-module of finite length, then \( {\chi }_{\mathrm{A}}\left( \mathrm{M}\right) = \mathrm{N}\left( {{\chi }_{\mathrm{B}}\left( \mathrm{M}\right) }\right) \) . | By linearity, it suffices to consider the case \( \mathrm{M} = \mathrm{B}/\mathfrak{P} \), which case follows from the definition of the norm. | No |
Proposition 13. If \( \mathbf{M} \) is an A-module of finite length, then \( {\chi }_{\mathbf{B}}\left( {\mathbf{M}}_{\mathbf{B}}\right) = i\left( {{\chi }_{\mathbf{A}}\left( \mathbf{M}\right) }\right) \) . | By linearity, it suffices to consider the case \( \mathrm{M} = \mathrm{A}/\mathfrak{p} \), whence \( {\mathrm{M}}_{\mathrm{B}} = \) \( \mathrm{B}/\mathrm{{pB}} \), and the proposition is clear. | Yes |
Proposition 14. If \( x \in \mathrm{L} \), then \( \mathrm{N}\left( {x\mathrm{\;B}}\right) = {\mathrm{N}}_{\mathrm{L}/\mathrm{K}}\left( x\right) \mathrm{A} \) . | One may assume \( x \) integral over \( \mathrm{A} \), and, by localising, that \( \mathrm{A} \) is principal. The ring B is then a free A-module of rank \( n \) . Let \( {u}_{x} \) be multiplication by \( x \) in B. One has \( {\mathrm{N}}_{\mathrm{L}/\mathrm{K}}\left( x\right) = \det \left( {u}_{x}\right) \) and \( \... | No |
Lemma 3. Let \( \\mathrm{A} \) be a principal ideal domain and \( u : {\\mathrm{A}}^{n} \\rightarrow {\\mathrm{A}}^{n} \) a linear map with \( \\det \\left( u\\right) \\neq 0 \) . Then \( \\det \\left( u\\right) \\mathrm{A} = {\\chi }_{\\mathrm{A}}\\left( {\\text{Coker}u}\\right) \) . | The ideal \( \\det \\left( u\\right) \) A does not change when one multiplies \( u \) by an invertible linear map; hence one may reduce by the theory of elementary divisors to the case where \( u \) is diagonal (Bourbaki, Alg., Chap. VII,§4, no. 5, prop. 4). The proof is then carried out by induction on \( n \), the ca... | Yes |
Lemma 4. Let \( {\mathfrak{m}}_{i} = \left( {\mathfrak{m},{g}_{i}}\right) \) be the ideal of \( {\mathrm{B}}_{f} \) generated by \( \mathfrak{m} \) and the canonical image of \( {g}_{i} \) in \( {\mathrm{B}}_{f} \) ; the ideals \( {\mathfrak{m}}_{i}, i \in \mathrm{I} \), are maximal and distinct, and every maximal idea... | By definition, \( {m}_{i} \) is the inverse image in \( {\mathbf{B}}_{f} \) of the ideal \( {\bar{m}}_{i} \) of \( {\overline{\mathbf{B}}}_{f} \) generated by \( {\varphi }_{i} \) ; as \( {\overline{\mathbf{B}}}_{f}/\left( {\varphi }_{i}\right) = {k}_{i} = k\left\lbrack \mathrm{X}\right\rbrack /\left( {\varphi }_{i}\ri... | Yes |
Proposition 15. If \( \mathrm{A} \) is a discrete valuation ring, and if \( \bar{f} \) is irreducible, then \( {\mathrm{B}}_{f} \) is a discrete valuation ring with maximal ideal \( {\mathrm{{mB}}}_{f} \) and residue field \( k\left\lbrack \mathrm{X}\right\rbrack /\left( \bar{f}\right) \) . | By lemma 4, \( {\mathbf{B}}_{f} \) is local with maximal ideal \( m{\mathbf{B}}_{f} \) and residue field \( k\left\lbrack \mathbf{X}\right\rbrack /\left( \bar{f}\right) \) . Moreover, if \( \pi \) generates \( m \), the image of \( \pi \) in \( {\mathrm{B}}_{f} \) generates \( m{\mathrm{\;B}}_{f} \) and is not nilpoten... | Yes |
If \( \mathrm{K} \) is the field of fractions of \( \mathrm{A} \), the polynomial \( f \) is irreducible in \( \mathrm{K}\left\lbrack \mathrm{X}\right\rbrack \) . If \( L \) denotes the field \( \mathrm{K}\left\lbrack \mathrm{X}\right\rbrack /\left( f\right) \), then the ring \( {\mathrm{B}}_{f} \) is the integral clos... | One has \( \mathrm{K}\left\lbrack \mathrm{X}\right\rbrack /\left( f\right) = {\mathrm{B}}_{f}{ \otimes }_{\mathrm{A}}\mathrm{K} \) . As \( {\mathrm{B}}_{f} \) is an integral domain, so is \( {\mathrm{B}}_{f}{ \otimes }_{\mathrm{A}}\mathrm{K} \) , hence \( \mathrm{K}\left\lbrack \mathrm{X}\right\rbrack /\left( f\right) ... | Yes |
Proposition 16. Let \( \mathrm{A} \) be a discrete valuation ring, \( \mathrm{K} \) its field of fractions, and let \( \mathrm{L} \) be an extension of \( \mathrm{K} \) of finite degree \( n \) . Let \( \mathrm{B} \) be the integral closure of \( \mathrm{A} \) in \( \mathbf{L} \) . Suppose that \( \mathbf{B} \) is a di... | The coefficients of \( f \) are integral over \( \mathrm{A} \) and belong to \( \mathrm{K} \) ; as \( \mathrm{A} \) is integrally closed, they belong to A. Furthermore, the equation \( f\left( x\right) = 0 \) shows that the map \( \mathrm{A}\left\lbrack \mathrm{X}\right\rbrack \rightarrow \mathrm{B} \) factors into \( ... | Yes |
Proposition 17. Suppose A is a discrete valuation ring and that \( f \) has the following form:\n\n\[ f = {\mathrm{X}}^{n} + {a}_{1}{\mathrm{X}}^{n - 1} + \cdots + {a}_{n},\;{a}_{i} \in \mathfrak{m},{a}_{n} \notin {\mathfrak{m}}^{2}. \]\n\nThen \( {\mathbf{B}}_{f} \) is a discrete valuation ring, with maximal ideal gen... | One has \( \bar{f} = {\mathrm{X}}^{n} \) . Lemma 4 then shows that \( {\mathrm{B}}_{f} \) is local with maximal ideal generated by \( \left( {\mathfrak{m}, x}\right) \) . Furthermore, the element \( \pi = {a}_{n} \) uniformizes A. Since:\n\n\[ - \pi = {x}^{n} + {a}_{1}{x}^{n - 1} + \cdots + {a}_{n - 1}x \]\n\none sees ... | Yes |
Proposition 18. Let \( \mathrm{A} \) be a discrete valuation ring, \( \mathrm{K} \) its field of fractions, and let \( \mathrm{L} \) be a finite extension of \( \mathrm{K} \) of degree \( n \) . Let \( \mathrm{B} \) be the integral closure of \( \mathrm{A} \) in \( \mathrm{L} \) . Suppose that \( \mathrm{B} \) is a dis... | One sees as in case (i) that the coefficients of \( f \) belong to A. Write \( f \) in the form:\n\n\[ f = {a}_{0}{\mathrm{X}}^{n} + \cdots + {a}_{n},\;{a}_{i} \in \mathrm{A},{a}_{0} = 1. \]\n\nSince \( f\left( x\right) = 0 \), one has:\n\n\[ {a}_{0}{x}^{n} + \cdots + {a}_{n} = 0. \]\n\nLet \( w \) be the discrete valu... | Yes |
Proposition 19. The group \( \mathrm{G}\left( {\mathrm{L}/\mathrm{K}}\right) \) acts transitively on the set of prime ideals \( \mathfrak{P} \) of \( \mathrm{B} \) dividing a given prime ideal \( \mathfrak{p} \) of \( \mathrm{A} \) . | Let \( \mathfrak{P} \mid \mathfrak{p} \), and suppose there were a prime ideal \( {\mathfrak{P}}^{\prime } \) of \( \mathrm{B} \) over \( \mathfrak{p} \) distinct from all the \( s\left( \mathfrak{P}\right), s \in \mathrm{G}\left( {\mathrm{L}/\mathrm{K}}\right) \) . By the approximation lemma, there exists \( a \in {\m... | Yes |
Proposition 20. The residue extension \( \overline{\mathrm{L}}/\overline{\mathrm{K}} \) is normal and the homomorphism\n\n\[ \varepsilon : \mathrm{D} \rightarrow \mathrm{G}\left( {\overline{\mathrm{L}}/\overline{\mathrm{K}}}\right) \]\n\ndefines an isomorphism of \( \mathrm{D}/\mathrm{T} \) onto \( \mathrm{G}\left( {\o... | We first show that \( \overline{\mathrm{L}}/\overline{\mathrm{K}} \) is normal. Let \( \bar{a} \in \overline{\mathrm{L}} \), and let \( a \in \mathrm{B} \) represent \( \bar{a} \) . Let \( \mathrm{P}\left( \mathrm{X}\right) = \prod \left( {\mathrm{X} - s\left( a\right) }\right) \), where \( s \) runs through \( \mathrm... | Yes |
Proposition 21. With notation as above, let \( w,{w}_{\mathrm{T}},{w}_{\mathrm{D}}, v \) be the discrete valuations defined by the ideals \( \mathfrak{P},{\mathfrak{P}}_{\mathrm{T}},{\mathfrak{P}}_{\mathrm{D}},\mathfrak{p} \) . Then:\n\na) \( \left\lbrack {\mathrm{L} : {\mathrm{K}}_{\mathrm{T}}}\right\rbrack = e{p}^{\m... | We know that the order of \( \mathrm{D} \) is \( {ef} \), and we’ve just seen that the order of \( \mathrm{D}/\mathrm{T} \) is \( {f}_{0} \) ; the order of \( \mathrm{T} \) is thus \( e{p}^{s} \), which proves \( a \) ).\n\nOn the other hand, we can apply prop. 20 to the group \( T \) : it tells us that \( \mathrm{L} \... | Yes |
Proposition 23. Let \( \mathrm{E} \) be a subfield of \( \mathrm{L} \) containing \( \mathrm{K} \), and let \( {\mathfrak{P}}_{\mathrm{E}} = \mathfrak{P} \cap \mathrm{E} \) . Then:\n\na) \( \left( {\mathfrak{P},\mathrm{L}/\mathrm{E}}\right) = {\left( \mathfrak{P},\mathrm{L}/\mathrm{K}\right) }^{f} \), with \( f = \left... | Immediate. | No |
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