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Lemma 4.9. Let \( L \) be a distributive lattice. Define\n\n\[ \nV\left( a\right) = \{ P \in X \mid a \notin P\} \text{, where}a \in L\text{.} \]\n\nThen \( V \) is injective; \( a \leqq b \) if and only if \( V\left( a\right) \subseteq V\left( b\right) ;V\left( {a \land b}\right) = V\left( a\right) \cap V\left( b\righ... | Proof. If \( P \) is a prime ideal, then \( a \land b \notin P \) if and only if \( a \notin P \) and \( b \notin P \) ; therefore \( V\left( {a \land b}\right) = V\left( a\right) \cap V\left( b\right) \) . If \( P \) is any ideal, then \( a \vee b \notin P \) if and only if \( a \notin P \) or \( b \notin P \) ; there... | Yes |
Proposition 5.1. In a distributive lattice with a least element and a greatest element: (1) an element has at most one complement; (2) if \( {a}^{\prime } \) is the complement of \( a \) and \( {b}^{\prime } \) is the complement of \( b \), then \( {a}^{\prime } \vee {b}^{\prime } \) is the complement of \( a \land b \... | Proof. (1). If \( b \) and \( c \) are complements of \( a \), then\n\n\[ b = b \land \left( {a \vee c}\right) = \left( {b \land a}\right) \vee \left( {b \land c}\right) = b \land c \leqq c; \]\n\nexchanging \( b \) and \( c \) then yields \( c \leqq b \) . (2). By distributivity,\n\n\[ \left( {a \land b}\right) \land ... | Yes |
Lemma 5.2. A Boolean ring is commutative and has characteristic 2. | Proof. If \( R \) is Boolean, then\n\n\[ x + x = \left( {x + x}\right) \left( {x + x}\right) = {x}^{2} + {x}^{2} + {x}^{2} + {x}^{2} = x + x + x + x \]\n\nand \( x + x = 0 \), for all \( x \in R \) . Then\n\n\[ x + y = \left( {x + y}\right) \left( {x + y}\right) = {x}^{2} + {xy} + {yx} + {y}^{2} = x + {xy} + {yx} + y, ... | Yes |
Proposition 5.3. If \( R \) is a Boolean ring, then \( R \), partially ordered by \( x \leqq y \) if and only if \( {xy} = x \), is a Boolean lattice \( \mathrm{L}\left( R\right) \), in which \( x \land y = {xy}, x \vee y = \) \( x + y + {xy} \), and \( {x}^{\prime } = 1 - x \) . | The proof is an exercise. | No |
Proposition 5.4 (Stone [1936]). If \( L \) is a Boolean lattice, then \( L \), with addition and multiplication\n\n\[ x + y = \left( {{x}^{\prime } \land y}\right) \vee \left( {x \land {y}^{\prime }}\right) ,{xy} = x \land y, \]\n\nis a Boolean ring \( \mathrm{R}\left( L\right) \) . Moreover, \( \mathrm{L}\left( {\math... | Proof. The addition on \( L \) is commutative; readers who love computation will delight in showing that it is associative. Moreover, \( x + 0 = x \) and \( x + x = 0 \) for all \( x \in L \) ; hence \( \left( {L, + }\right) \) is an abelian group.\n\nThe multiplication on \( L \) is commutative, associative, and idemp... | No |
Theorem 5.5. A finite lattice \( L \) is Boolean if and only if \( L \cong {2}^{X} \) for some finite set \( X \) . | Proof. The lattice \( {2}^{X} \) is always Boolean. Conversely, let \( L \) be Boolean. By 4.5, \( L \cong \operatorname{Id}\left( S\right) \), where \( S = \operatorname{Irr}\left( L\right) \) is the partially ordered set of all atoms (irreducible elements) of \( L \) . Since the atoms of \( L \) satisfy no strict ine... | Yes |
Theorem 5.6 (Birkhoff [1933]). A lattice \( L \) is Boolean if and only if it is isomorphic to a Boolean sublattice of \( {2}^{X} \) for some set \( X \) . | Proof. Let \( L \) be Boolean and let \( X \) be the set of all prime ideals of \( L \) . Define \( V : L \rightarrow {2}^{X} \) by \( V\left( x\right) = \{ P \in X \mid x \notin P\} \) . By 4.9 and 5.7 below, \( V \) is a homomorphism of Boolean lattices, so that \( \operatorname{Im}V \) is a Boolean sublattice of \( ... | No |
Lemma 5.7. Let \( L \) be a Boolean lattice. The sets \[ V\left( a\right) = \{ P \in X \mid a \notin P\} \text{, where } a \in L \text{,} \] constitute a basis for a topology on the set \( X \) of all prime ideals of \( L \). Moreover, \( V\left( 0\right) = \varnothing, V\left( 1\right) = X \), and \( V\left( {a}^{\pri... | Proof. By 4.9, \( V\left( {a \land b}\right) = V\left( a\right) \cap V\left( b\right) \) for all \( a, b \in L \) ; hence the sets \( V\left( a\right) \) with \( a \in L \) constitute a basis of open sets for a topology on \( X \). If \( P \) is a prime ideal of \( L \), then \( 0 \in P \), since \( P \neq \varnothing ... | Yes |
Proposition 5.8. The Stone space of a Boolean lattice is compact Hausdorff and totally disconnected. | Proof. Let \( X \) be the Stone space of a Boolean lattice \( L \) . If \( P \neq Q \) in \( X \), then, say, \( a \in P, a \notin Q \) for some \( a \in L \), and then \( Q \in V\left( a\right) \) and \( P \in V\left( {a}^{\prime }\right) = V{\left( a\right) }^{\prime } \) . Therefore \( X \) is Hausdorff. Moreover, e... | Yes |
Theorem 5.9 (Stone [1934]). Every Boolean lattice is isomorphic to the lattice of closed and open subsets of its Stone space. | Proof. Let \( L \) be a Boolean lattice and let \( X \) be its Stone space. For every \( a \in L, V\left( a\right) \) is open in \( X \), and is closed in \( X \) since \( X \smallsetminus V\left( a\right) = V\left( {a}^{\prime }\right) \) is open. Conversely, if \( U \in \mathrm{L}\left( X\right) \) is a closed and op... | Yes |
Proposition 1.3. Let \( A \) be a universal algebra of type \( T \) . For an equivalence relation \( \mathcal{E} \) on \( A \) the following conditions are equivalent:\n\n(1) there exists a type- \( T \) algebra structure on \( A/\mathcal{E} \) such that the canonical projection \( \pi : A \rightarrow A/\mathcal{E} \) ... | Proof. (1) implies (2); that (2) implies (3) follows from the definitions.\n\n(3) implies (1). Let \( Q = A/\mathcal{E} \) and let \( \pi : A \rightarrow Q \) be the projection. For every \( \omega \in T \) of arity \( n \) and every equivalence classes \( {E}_{1},\ldots ,{E}_{n} \), the set\n\n\[{\omega }_{A}\left( {{... | Yes |
Theorem 1.8 (Homomorphism Theorem). If \( \varphi : A \rightarrow B \) is a homomorphism of universal algebras, then \( \ker \varphi \) is a congruence on \( A,\operatorname{Im}\varphi \) is a subalgebra of \( B \), and \[ A/\ker \varphi \cong \operatorname{Im}\varphi \] in fact, there is an isomorphism \( \theta : A/\... | Proof. First, \( \ker \varphi \) is a congruence on \( A \) by definition, and it is clear that \( \operatorname{Im}\varphi \) is a subalgebra of \( B \) . Let \( \theta : A/\ker \varphi \rightarrow \operatorname{Im}\varphi \) be the bijection that sends an equivalence class \( E \) of \( \ker \varphi \) to the sole el... | Yes |
Proposition 2.1. Let \( X \) be a subset of a universal algebra \( A \) of type \( T \) . Define \( {S}_{k} \subseteq A \) for every integer \( k \geqq 0 \) by \( {S}_{0} = X \) ; if \( k > 0 \), then \( {S}_{k} \) is the set of all \( \omega \left( {{w}_{1},\ldots ,{w}_{n}}\right) \) in which \( \omega \in T \) has ar... | By 2.1, every element of \( \langle X\rangle \) can be calculated in finitely many steps from elements of \( X \) and operations on \( A \) (using \( k \) operations when \( x \in {S}_{k} \) ). In general, this calculation can be performed in several different ways. The simplest way to construct an algebra of type \( T... | No |
Proposition 2.2. If \( w \in {W}_{X}^{T} \), then \( w \in {W}_{Y}^{T} \) for some finite subset \( Y \) of \( X \) . | Proof. We have \( w \in {W}_{k} \) for some \( k \) and prove the result by induction on \( k \) . If \( w \in {W}_{0} \), then \( w \in X \) and \( Y = \{ w\} \) serves. If \( k > 0 \) and \( w \in {W}_{k} \), then \( w = \left( {\omega ,{w}_{1},\ldots ,{w}_{n}}\right) \), where \( \omega \in T \) has arity \( n \) an... | Yes |
Proposition 2.3. The word algebra \( {W}_{X}^{T} \) of type \( T \) on a set \( X \) is generated by \( X \) . Moreover, every mapping of \( X \) into a universal algebra of type \( T \) extends uniquely to a homomorphism of \( {W}_{X}^{T} \) into \( A \) . | Proof. \( W = {W}_{X}^{T} \) is generated by \( X \), by 2.1. Let \( f \) be a mapping of \( X \) into a universal algebra \( A \) of type \( T \) . If \( \varphi : W \rightarrow A \) is a homomorphism that extends \( f \), then necessarily \( \varphi \left( x\right) = f\left( x\right) \) for all \( x \in X \) and \( \... | Yes |
Proposition 3.1. Every variety is closed under subalgebras, homomorphic images, direct products, and directed unions. | Proof. Let \( \mathcal{V} = \mathrm{V}\left( \mathcal{J}\right) \) be the variety of all universal algebras \( A \) of type \( T \) that satisfy a set \( \mathcal{J} \subseteq {W}_{X}^{T} \times {W}_{X}^{T} \) of identities. An algebra \( A \) of type \( T \) belongs to \( \mathcal{V} \) if and only if \( \varphi \left... | Yes |
Theorem 3.3. Let \( \mathcal{C} \) be a class of universal algebras of the same type, that is closed under isomorphisms, direct products, and subalgebras (for instance, a variety). For every set \( X \) there exists a universal algebra that is free on \( X \) in the class C. | Proof. We give a direct proof; a better proof will be found in Section XVI.10. Given a set \( X \), let \( {\left( {\mathcal{E}}_{i}\right) }_{i \in I} \) be the set of all congruences \( {\mathcal{E}}_{i} \) on \( {W}_{X}^{T} \) such that \( {W}_{X}^{T}/{\mathcal{E}}_{i} \in \mathcal{C} \) ; let \( {C}_{i} = {W}_{X}^{... | Yes |
Theorem 3.4 (Birkhoff [1935]). A nonempty class of universal algebras of the same type is a variety if and only if it is closed under direct products, subalgebras, and homomorphic images. | Proof. First we prove the following: when \( F \) is free in a class \( \mathcal{C} \) on an infinite set, relations that hold in \( F \) yield identities that hold in every \( C \in \mathcal{C} \) : | No |
(1) There exist homomorphisms \( \sigma : {W}_{Y}^{T} \rightarrow {W}_{X}^{T} \) and \( \mu : {W}_{X}^{T} \rightarrow {W}_{Y}^{T} \) such that \( \sigma \circ \mu \) is the identity on \( {W}_{X}^{T} \) and \( \mu \left( {\sigma \left( p\right) }\right) = p,\mu \left( {\sigma \left( q\right) }\right) = q \) . | Proof. (1). By 2.2, \( p, q \in {W}_{Z}^{T} \) for some finite subset \( Z \) of \( Y \) . There is an injection \( h : X \rightarrow Y \) such that \( h\left( X\right) \) contains \( Z \) . The inverse bijection \( h\left( X\right) \rightarrow X \) can be extended to a surjection \( g : Y \rightarrow X \) ; then \( g ... | Yes |
Lemma 3.6. Let \( \mathcal{C} \) be a class of universal algebras of the same type \( T \), that is closed under isomorphisms, direct products, and subalgebras. Let A be a nonempty universal algebra of type \( T \) such that every identity that holds in every \( C \in \mathcal{C} \) also holds in \( A \) . Then \( A \)... | Proof. There is an infinite set \( Y \) and a mapping \( f \) of \( Y \) into \( A \) such that \( A \) is generated by \( f\left( Y\right) \) : indeed, \( A \) is generated by some subset \( S \) ; if \( S \) is infinite, then \( Y = S \) serves; otherwise, construct \( Y \) by adding new elements to \( S \), which \(... | Yes |
Proposition 3.7. Let \( \mathcal{C} \) be a class of universal algebras of type \( T \) . The variety generated by \( \mathcal{C} \) consists of all homomorphic images of subalgebras of direct products of members of \( \mathcal{C} \) . | Proof. For any class \( \mathcal{C} \) of universal algebras of type \( T \) :\n\n(1) a homomorphic image of a homomorphic image of a member of \( \mathcal{C} \) is a homomorphic image of a member of \( \mathcal{C} \) ; symbolically, \( \mathrm{{HH}}\mathcal{C} \subseteq \mathrm{H}\mathcal{C} \) ;\n\n(2) a subalgebra o... | Yes |
Proposition 4.1. Let \( {\left( {A}_{i}\right) }_{i \in I} \) be universal algebras of type \( T \) . A universal algebra \( A \) of type \( T \) is isomorphic to a subdirect product of \( {\left( {A}_{i}\right) }_{i \in I} \) if and only if there exist surjective homomorphisms \( {\varphi }_{i} : A \rightarrow {A}_{i}... | Proof. Let \( P \) be a subdirect product of \( {\left( {A}_{i}\right) }_{i \in I} \) . The inclusion homomorphism \( \iota : P \rightarrow \mathop{\prod }\limits_{{i \in I}}{A}_{i} \) and projections \( {\pi }_{j} : \mathop{\prod }\limits_{{i \in I}}{A}_{i} \rightarrow {A}_{j} \) yield surjective homomorphisms \( {\rh... | Yes |
Proposition 4.3. A universal algebra \( A \) is subdirectly irreducible if and only if \( A \) has more than one element and the equality on \( A \) is not the intersection of congruences on \( A \) that are different from the equality. | The proof is an exercise in further deduction from Proposition 4.1. | No |
Theorem 4.4 (Birkhoff [1944]). Every nonempty universal algebra is isomorphic to a subdirect product of subdirectly irreducible universal algebras. In any variety \( \mathcal{V} \), every nonempty universal algebra \( A \in \mathcal{V} \) is isomorphic to a subdirect product of subdirectly irreducible universal algebra... | Proof. Let \( A \) be a nonempty algebra of type \( T \) . By 1.5, the union of a chain of congruences on \( A \) is a congruence on \( A \) . Let \( a, b \in A, a \neq b \) of \( A \) . If \( {\left( {\mathcal{C}}_{i}\right) }_{i \in I} \) is a chain of congruences on \( A \), none of which contains the pair \( \left(... | Yes |
An abelian group is subdirectly irreducible if and only if it is isomorphic to \( {\mathbb{Z}}_{{p}^{\infty }} \) or to \( {\mathbb{Z}}_{{p}^{n}} \) for some \( n > 0 \) . | Proof. Readers will verify that \( {\mathbb{Z}}_{{p}^{\infty }} \) and \( {\mathbb{Z}}_{{p}^{n}} \) (where \( n > 0 \) ) are subdirectly irreducible. Conversely, every abelian group \( A \) can, by X.4.9 and X.4.10, be embedded into a direct product of copies of \( \mathbb{Q} \) and \( {\mathbb{Z}}_{{p}^{\infty }} \) f... | No |
Every distributive lattice is isomorphic to a subdirect product of two-element lattices. A distributive lattice is subdirectly irreducible if and only if it has just two elements. | Proof. To each prime ideal \( P \neq \varnothing, L \) of a distributive lattice \( L \) there corresponds a lattice homomorphism \( {\varphi }_{P} \) of \( L \) onto \( {L}_{2} \), defined by \( {\varphi }_{P}\left( x\right) = 0 \) if \( x \in P,{\varphi }_{P}\left( x\right) = 1 \) if \( x \notin P \) . The homomorphi... | Yes |
Lemma 4.7. Let \( F \) be the free commutative semigroup on \( {X}_{1},\ldots ,{X}_{n} \) . Every congruence \( \mathcal{E} \) on \( F \) is induced by an ideal of \( \mathbb{Z}\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) . Every commutative semigroup with \( n \) generators is determined by an ideal of \( \m... | Proof. Let \( \mathfrak{E} \) be the ideal of \( \mathbb{Z}\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) generated by all binomials \( {X}^{a} - {X}^{b} \) such that \( {X}^{a}\mathcal{E}{X}^{b} \) in \( F \) . The ideal \( \mathfrak{E} \) induces a congruence \( \overline{\mathcal{E}} \) on \( F \), in which ... | Yes |
Every commutative semigroup with \( n \) generators has a sub-direct decomposition into finitely many commutative semigroups determined by primary ideals of \( \mathbb{Z}\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) . | Let \( S \) be a commutative semigroup with \( n \) generators \( {x}_{1},\ldots ,{x}_{n} \). By 4.7, \( S \cong F/\mathcal{E} \), where \( \mathcal{E} \) is induced by an ideal \( \mathfrak{E} \) of \( \mathbb{Z}\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \). In the Noetherian ring \( \mathbb{Z}\left\lbrack {{... | Yes |
Proposition 1.2. In the category Grps of groups and homomorphisms, a morphism is a monomorphism if and only if it is injective. | Proof. Let \( \mu : A \rightarrow B \) be a monomorphism in Grps. Assume that \( \mu \left( x\right) = \) \( \mu \left( y\right) \) . Since \( \mathbb{Z} \) is free on \( \{ 1\} \) there exist homomorphisms \( \alpha ,\beta : \mathbb{Z} \rightarrow A \) such that \( \alpha \left( 1\right) = x \) and \( \beta \left( 1\r... | Yes |
Proposition 1.3. In the category Grps of groups and homomorphisms, a morphism is an epimorphism if and only if it is surjective. | Proof. Surjective homomorphisms are epimorphisms. Conversely, let \( \varphi \) : \( G \rightarrow H \) be a homomorphism of groups that is not surjective. Construct isomorphic copies \( {H}_{1},{H}_{2} \) of \( H \) that contain \( \operatorname{Im}\varphi \) as a subgroup and satisfy \( {H}_{1} \cap {H}_{2} = \operat... | Yes |
Proposition 3.2. Let \( \mathcal{C} \) be a category and let \( \mathcal{G} \) be a [small] graph. Every diagram \( D : \mathcal{G} \rightarrow \mathcal{C} \) extends uniquely to a functor \( \widehat{D} : \widehat{\mathcal{G}} \rightarrow \mathcal{C} \) ; every morphism \( D \rightarrow E \) of diagrams over \( \mathc... | Proof. The objects of \( \widehat{\mathcal{G}} \) are the vertices of \( \mathcal{G} \), and the morphisms of \( \widehat{\mathcal{G}} \) are the empty paths \( \left( {i, i}\right) \) and all nonempty paths \( \left( {i,{a}_{1},\ldots ,{a}_{n}, j}\right) \) . The latter are compositions in \( \widehat{\mathcal{G}} : \... | Yes |
Proposition 4.1. The category Sets is complete; in fact, in Sets, a limit can be assigned to every diagram. | Proof. Let \( D \) be a diagram in Sets over a graph \( \mathcal{G} \) . Let \( P = \mathop{\prod }\limits_{{i \in \mathcal{G}}}{D}_{i} \) be the Cartesian product (where \ | No |
Proposition 4.4. A category that has products and equalizers is complete. If in a category \( \mathcal{C} \) a product can be assigned to every family of objects of \( \mathcal{C} \), and an equalizer can be assigned to every pair of coterminal morphisms of \( \mathcal{C} \), then a limit can be assigned to every diagr... | Proof. Let \( \mathcal{G} \) be a graph, let \( E \) be the set of its edges, and let \( o, d \) be the origin and destination mappings of \( \mathcal{G} \), so that \( a : o\left( a\right) \rightarrow d\left( a\right) \) for every edge \( a \) . Let \( D \) be a diagram over \( \mathcal{G} \) . Let \( P = \mathop{\pro... | Yes |
Proposition 4.6. The category \( {}_{R} \) Mods is cocomplete; in fact, in \( {}_{R} \) Mods, a colimit can be assigned to every diagram. | Proof. Let \( D \) be a diagram in \( {}_{R} \) Mods over a graph \( \mathcal{G} \) . Let \( S = {\bigoplus }_{i \in \mathcal{G}}{D}_{i} \) be the direct sum and let \( {\iota }_{i} : {D}_{i} \rightarrow S \) be the injections. Let \( K \) be the submodule of \( S \) generated by all \( {\iota }_{j}\left( {{D}_{a}\left... | Yes |
Proposition 4.7. Let \( \mathcal{G} \) be a graph and let \( \mathcal{C} \) be a category in which a limit can be assigned to every diagram over \( \mathcal{G} \) . There is a limit functor from Diag \( \left( {\mathcal{G},\mathcal{C}}\right) \) to \( \mathcal{C} \) that assigns to each diagram \( D \) over \( \mathcal... | Dually, if a colimit can be assigned to every diagram over \( \mathcal{G} \), then there is a colimit functor Diag \( \left( {\mathcal{G},\mathcal{C}}\right) \rightarrow \mathcal{C} \) . | No |
Proposition 5.1. For objects \( A, B, P \) of an additive category the following conditions are equivalent:\n\n(1) There exist morphisms \( \pi : P \rightarrow A \) and \( \rho : P \rightarrow A \) such that \( P \) is a product of \( A \) and \( B \) with projections \( \pi \) and \( \rho \) .\n\n(2) There exist morph... | Proof. (1) implies (3). The universal property of products yields morphisms \( \iota : A \rightarrow P,\kappa : B \rightarrow P \) such that \( {\pi \iota } = {1}_{A},{\rho \iota } = 0,{\pi \kappa } = 0 \), and \( {\rho \kappa } = {1}_{B} \) . Then \( \pi \left( {{\iota \pi } + {\kappa \rho }}\right) = \pi ,\rho \left(... | Yes |
Proposition 5.2. If \( \alpha ,\beta : A \rightarrow B \) are morphisms of an additive category, and the biproducts \( A \oplus A \) and \( B \oplus B \) exist, then \( \alpha + \beta = {\nabla }_{B}\left( {\alpha \oplus \beta }\right) {\Delta }_{A} \) . | Proof. Let \( \pi ,\rho : A \oplus A \rightarrow A \) and \( {\pi }^{\prime },{\rho }^{\prime } : B \oplus B \rightarrow B \) be the projections, and let \( {\iota }^{\prime },{\kappa }^{\prime } : B \rightarrow B \oplus B \) be the injections. By 5.1, \( {\iota }^{\prime }{\pi }^{\prime } + {\kappa }^{\prime }{\rho }^... | Yes |
Proposition 5.3. An abelian category has finite limits and colimits. | Proof. In an abelian category, every nonempty finite family \( {A}_{1},\ldots ,{A}_{n} \) has a product \( \left( {\ldots \left( {\left( {{A}_{1} \times {A}_{2}}\right) \times {A}_{3}}\right) \times \ldots }\right) \times {A}_{n} \) . The empty sequence also has a product, the zero object. Hence an abelian category has... | Yes |
Lemma 5.4. Let \( \mu \) be a monomorphism and let \( \sigma \) be an epimorphism. In an abelian category, \( \mu \) is a kernel of \( \sigma \) if and only if \( \sigma \) is a cokernel of \( \mu \) . | Proof. First, \( \sigma \) is a cokernel of some \( \alpha \) . Assume that \( \mu \) is a kernel of \( \sigma \) . Then \( {\sigma \mu } = 0 \), and \( \alpha \) factors through \( \mu : \alpha = {\mu \xi } \), since \( {\sigma \alpha } = 0 \) . If \( {\varphi \mu } = 0 \) , then \( {\varphi \alpha } = 0 \) and \( \va... | Yes |
Proposition 5.5. A morphism of an abelian category is a monomorphism if and only if it has a kernel that is a zero morphism; an epimorphism if and only if it has a cokernel that is a zero morphism; an isomorphism if and only if it is both a monomorphism and an epimorphism. | Proof. We prove the last part and leave the first two parts as exercises. In any category, an isomorphism is both a monomorphism and an epimorphism. Conversely, let \( \alpha : A \rightarrow B \) be both a monomorphism and an epimorphism of an abelian category. Let \( \gamma \) be a cokernel of \( \alpha \) . Then \( {... | No |
Proposition 5.6 (Homomorphism Theorem). Let \( \alpha \) be a morphism of an abelian category; let \( \iota \) be an image of \( \alpha \), and let \( \rho \) be a coimage of \( \alpha \) . There exists a unique isomorphism \( \theta \) such that \( \alpha = {\iota \theta \rho } \) . | Proof. Construct the following diagram:\n\n\n\nLet \( \kappa : K \rightarrow A \) and \( \gamma : B \rightarrow C \) be a kernel and cokernel of \( \alpha : A \rightarrow B \) . Let \( \iota : I \rightarrow B \) be a... | Yes |
Proposition 6.1. For two functors \( F : \mathcal{A} \rightarrow \mathcal{C} \) and \( G : \mathcal{C} \rightarrow \mathcal{A} \) the following conditions are equivalent:\n\n(1) there exists a natural transformation \( \eta : {1}_{\mathcal{A}} \rightarrow G \circ F \) such that for every morphism \( \alpha : A \rightar... | Proof. (1) implies (2). Applying (1) to \( {1}_{G\left( C\right) } : G\left( C\right) \rightarrow G\left( C\right) \) yields a morphism \( {\varepsilon }_{C} : F\left( {G\left( C\right) }\right) \rightarrow C \), unique such that\n\n\[ G\left( {\varepsilon }_{C}\right) {\eta }_{G\left( C\right) } = {1}_{G\left( C\right... | Yes |
Proposition 6.2. Any two left adjoints of the same functor are naturally isomorphic. Any two right adjoints of the same functor are naturally isomorphic. | This follows from the universal properties. | No |
Proposition 6.6. Every right adjoint functor preserves limits. Every left adjoint functor preserves colimits. | Proof. Let \( F : \mathcal{A} \rightarrow \mathcal{C} \) be a left adjoint of \( G : \mathcal{C} \rightarrow \mathcal{A} \), and let \( \lambda : L \rightarrow D \) be a limit cone of a diagram \( D \) in \( \mathcal{C} \) over some graph \( \mathcal{G} \). We want to show that \( G\left( \lambda \right) : G\left( L\ri... | Yes |
Proposition 7.1. In every category, any two initial objects are isomorphic, and any two terminal objects are isomorphic. | Proof. Let \( A \) and \( B \) be initial objects. There exists a morphism \( \alpha : A \rightarrow B \) and a morphism \( \beta : B \rightarrow A \) . Then \( {1}_{A} \) and \( {\beta \alpha } \) are morphisms from \( A \) to \( A \) ; since \( A \) is an initial object, \( {\beta \alpha } = {1}_{A} \) . Similarly, \... | Yes |
Proposition 7.2. A locally small, complete category \( \mathcal{C} \) has an initial object if and only if it satisfies the solution set condition:\n\nthere exists a set \( \mathcal{S} \) of objects of \( \mathcal{C} \) such that there is for every object \( A \) of \( \mathcal{C} \) at least one morphism from some \( ... | Proof. If \( \mathcal{C} \) has an initial object \( C \), then \( \{ C\} \) is a solution set. Conversely, let \( \mathcal{C} \) have a solution set \( \mathcal{S} \) . Since \( \mathcal{C} \) is complete, \( \mathcal{S} \) has a product \( P \) in \( \mathcal{C} \) . Then \( \{ P\} \) is a solution set: for every obj... | Yes |
Lemma 7.4. Let \( \mathcal{C} \) be a locally small category in which a limit can be assigned to every diagram, let \( G : \mathcal{C} \rightarrow \mathcal{A} \) be a functor, and let \( A \) be an object of \( \mathcal{A} \) . If \( G \) preserves limits, then a limit can be assigned to every diagram in \( \left( {A \... | Proof. Let \( \Delta \) be a diagram in \( \left( {A \downarrow G}\right) \) over a graph \( \mathcal{G} \) . For every vertex \( i \) of \( \mathcal{G},{\Delta }_{i} = \left( {{\delta }_{i},{D}_{i}}\right) \), where \( {\delta }_{i} : A \rightarrow G\left( {D}_{i}\right) \) ; for every edge \( a : i \rightarrow j \) ,... | Yes |
Proposition 8.1 Let \( \\left( {F, G,\\eta ,\\varepsilon }\\right) \) be an adjunction from \( \\mathcal{A} \) to \( \\mathcal{C} \) . Let \( T = {GF} \) and \( \\mu = {G\\varepsilon F} : {TT} \\rightarrow T \) . The following diagrams commute:  is a natural transformation, the square below left commutes  \n\nfor every objec... | Yes |
Proposition 8.2. Let \( \\left( {F, G,\\eta ,\\varepsilon }\\right) \) be an adjunction from \( \\mathcal{A} \) to \( \\mathcal{C} \) and let \( \\left( {T,\\eta ,\\mu }\\right) = \\left( {{GF},\\eta ,{G\\varepsilon F}}\\right) \) be the triple it induces on \( \\mathcal{A} \). For every object \( C \) of \( \\mathcal{... | Proof. The diagrams below commute, the first since \( \\varepsilon \) is a natural transformation, the second by 6.5; hence \( \\left( {{GC}, G{\\varepsilon }_{C}}\\right) \) is a \( T \) -algebra. \n\nIf \( \\gamma ... | Yes |
If \( \left( {T,\eta ,\mu }\right) \) is a triple on a category \( \mathcal{A} \), then \( T \) -algebras and their homomorphisms are the objects and morphisms of a category \( {\mathcal{A}}^{T} \) ; moreover, there is an adjunction \( \left( {{F}^{T},{G}^{T},{\eta }^{T},{\varepsilon }^{T}}\right) \) from \( \mathcal{A... | Proof. It is immediate that \( T \) -algebras and their homomorphisms constitute a category \( {\mathcal{A}}^{T} \) with a forgetful functor \( {G}^{T} : {\mathcal{A}}^{T} \rightarrow \mathcal{A} \) . The definition of triples shows that \( {F}^{T}A = \left( {{TA},{\mu }_{A}}\right) \) is a \( T \) -algebra. If \( \alp... | Yes |
Proposition 8.4. Let \( \left( {F, G,\eta ,\varepsilon }\right) \) be an adjunction from \( \mathcal{A} \) to \( \mathcal{C} \) and let \( \left( {T,\eta ,\mu }\right) = \left( {{GF},\eta ,{G\varepsilon F}}\right) \) be the triple it induces on \( \mathcal{A} \) . There is a unique functor \( Q : \mathcal{C} \rightarro... | Proof. By 8.2, \( \left( {{GC}, G{\varepsilon }_{C}}\right) \) is a \( T \) -algebra for every object \( C \) of \( \mathcal{C} \), and, if \( \gamma : C \rightarrow D \) is a morphism of \( \mathcal{C} \), then \( {G\gamma } : \left( {{GC}, G{\varepsilon }_{C}}\right) \rightarrow \left( {{GD}, G{\varepsilon }_{D}}\rig... | Yes |
Proposition 9.1. (1) If \( \left( {A,\varphi }\right) \) is a T-algebra, then \( \varphi \) is a split coequalizer of \( {\mu }_{A} \) and \( {T\varphi } \). | Proof. (1). We saw that \( \varphi \) is a split coequalizer of \( \alpha = {\mu }_{A} \) and \( \beta = {T\varphi } \), with \( v = {\eta }_{A} \) and \( \kappa = {\eta }_{TA} \) . \( ▱ \) | Yes |
Proposition 9.2. The functor \( {G}^{T} \) creates coequalizers of pairs \( \alpha ,\beta \) with a split coequalizer in \( \mathcal{A} \) . | Proof. Let \( \alpha ,\beta : \left( {A,\varphi }\right) \rightarrow \left( {B,\psi }\right) \) be homomorphisms of \( T \) -algebras that have a split coequalizer \( \sigma : B \rightarrow C \) in \( \mathcal{A} \) . We need to show that there is a unique \( T \) -algebra \( \left( {C,\chi }\right) \) such that \( \si... | Yes |
Theorem 9.3 (Beck). A functor \( G : \mathcal{C} \rightarrow \mathcal{A} \) with a left adjoint is tripleable if and only if it creates coequalizers of pairs \( \alpha ,\beta \) such that \( {G\alpha },{G\beta } \) have a split coequalizer in \( \mathcal{A} \) . | Theorem 9.3 follows from Proposition 9.2 and from a result that is similar to Proposition 8.4: | No |
Proposition 9.5. The forgetful functor from Grps to Sets is tripleable. | Proof. We show that this functor creates coequalizers of pairs \( \alpha ,\beta : G \rightarrow H \) of group homomorphisms that have a split coequalizer \( \sigma : H \rightarrow K \) in Sets .\n\nLet \( {m}_{G} : G \times G \rightarrow G \) and \( {m}_{H} : H \times H \rightarrow H \) be the group operations on \( G ... | Yes |
Proposition 10.1. If \( \mathcal{C} \) is a class of universal algebras of type \( T \) that is closed under products and subalgebras (for instance, a variety), then \( \mathcal{C} \) is complete; in fact, a limit can be assigned to every diagram in \( \mathcal{C} \), and the forgetful functor from C to Sets preserves ... | Proof. Let \( D \) be a diagram in \( \mathcal{C} \) over a graph \( \mathcal{G} \) . Let \( P = \mathop{\prod }\limits_{{i \in \mathcal{G}}}{D}_{i} \) be the direct product, with projections \( {\pi }_{i} : P \rightarrow {D}_{i} \) and componentwise operations,\n\n\[ \omega \left( {{\left( {x}_{1i}\right) }_{i \in \ma... | Yes |
Theorem 10.2. Let \( \mathcal{C} \) be a class of universal algebras of the same type, that is closed under isomorphisms, direct products, and subalgebras (for instance, a variety). The forgetful functor from \( \mathcal{C} \) to Sets has a left adjoint. Hence there is for every set \( X \) a universal algebra that is ... | Proof. We show that the forgetful functor \( G : \mathcal{C} \rightarrow \) Sets has a left adjoint \( F \) ; then \( F \) assigns to each set \( X \) an algebra \( {F}_{X} \in \mathcal{C} \) that is free on \( X \) in the class \( \mathcal{C} \) : for every mapping \( f \) of \( X \) into a universal algebra \( A \in ... | Yes |
Theorem 10.4. For every variety \( \mathcal{V} \), the forgetful functor from \( \mathcal{V} \) to Sets is tripleable. | Proof. We invoke Beck’s theorem. Let \( \mathcal{V} \) be a variety of type \( T \) . The forgetful functor from \( \mathcal{V} \) to Sets has a left adjoint by 10.2 ; we show that it creates coequalizers of pairs \( \alpha ,\beta : A \rightarrow B \) of homomorphisms of algebras \( A, B \in \mathcal{V} \) , which have... | Yes |
Proposition 1.1. For a partially ordered set \( X \) the following conditions are equivalent:\n\n(1) every infinite ascending sequence \( {x}_{1} \leqq {x}_{2} \leqq \cdots \leqq {x}_{n} \leqq {x}_{n + 1} \leqq \cdots \) of elements of \( X \) terminates (is eventually stationary): there exists \( N > 0 \) such that \(... | Proof. (1) implies (2), since a strictly ascending infinite sequence cannot terminate.\n\n(2) implies (3). If the nonempty set \( S \) in (c) has no maximal element, then one can choose \( {x}_{1} \in S \) ; since \( {x}_{1} \) is not maximal in \( S \) one can choose \( {x}_{1} < {x}_{2} \in S \) ; since \( {x}_{2} \)... | Yes |
Proposition 1.2. The subgroups of a group \( G \) satisfy the ascending chain condition if and only if every subgroup of \( G \) is finitely generated. | Proof. Assume that every subgroup of \( G \) is finitely generated, and let \( {H}_{1} \subseteq \) \( {H}_{2} \subseteq \cdots \subseteq {H}_{n} \subseteq {H}_{n + 1} \subseteq \cdots \) be an infinite ascending sequence of subgroups of \( G \) . The union \( H = \mathop{\bigcup }\limits_{{n > 0}}{H}_{n} \) is a subgr... | Yes |
Proposition 2.1. The axiom of choice is equivalent to the following statement: when \( I \) is a nonempty set, and \( {\left( {S}_{i}\right) }_{i \in I} \) is a family of nonempty sets, then \( \mathop{\prod }\limits_{{i \in I}}{S}_{i} \) is nonempty. | Proof. Recall that \( \mathop{\prod }\limits_{{i \in I}}{S}_{i} \) is the set of all mappings (usually written as families) that assign to each \( i \in I \) some element of \( {S}_{i} \) . If \( I \neq \varnothing \) and \( {S}_{i} \neq \varnothing \) for all \( i \), and the axiom of choice holds, then \( \mathop{\bi... | Yes |
Corollary 2.3. Every equivalence relation has a cross section. | Proof. Let \( X \) be a set with an equivalence relation. Let \( \mathcal{S} \) be the set of all subsets \( S \) of \( X \) such that every equivalence class contains at most one element of \( S \) . Then \( \mathcal{S} \neq \varnothing \), since \( \varnothing \in \mathcal{S} \) . Partially order \( \mathcal{S} \) by... | Yes |
Proposition 3.2. If \( \alpha \) and \( \beta \) are ordinal numbers, then \( \alpha \in \beta \) if and only \( \alpha \subsetneqq \beta \) . Hence the class Ord of all ordinal numbers is totally ordered, with \( \alpha < \beta \) if and only if \( \alpha \in \beta \) . | Proof. Since \( \beta \) is transitive, \( \alpha \in \beta \) implies \( \alpha \subseteq \beta \) ; moreover, \( \alpha \notin \alpha \) (otherwise, \( \alpha < \alpha \) in \( \beta \) ), whence \( \alpha \subsetneqq \beta \) . Conversely, assume \( \alpha \subsetneqq \beta \) . Then \( \beta \smallsetminus \alpha \... | Yes |
Proposition 3.3. Every nonempty class of ordinal numbers has a least element. | Proof. Let \( \mathcal{C} \) be a nonempty class of ordinals. Let \( \alpha \in \mathcal{C} \) . We may assume that \( \alpha \) is not the least element of \( \mathcal{C} \) . Then \( \mathcal{C} \cap \alpha \neq \varnothing \) and \( \mathcal{C} \cap \alpha \subseteq \alpha \) has a least element \( \gamma \) . In fa... | Yes |
Proposition 3.4. The union of a set of ordinal numbers is an ordinal number. | Proof. Let \( S \) be a set of ordinal numbers. Then \( v = \mathop{\bigcup }\limits_{{\sigma \in S}}\sigma \) is a set. By 3.2, \( S \) is a chain, and any two elements of \( S \) are elements of some \( \sigma \in S \). It follows that \( v \) is transitive, and is totally ordered, with \( x < y \) in \( v \) if and ... | Yes |
Corollary 3.5. The class Ord of all ordinal numbers is not a set. | Proof. Let \( \alpha \) be an ordinal. The successor \( \beta = \alpha \cup \{ \alpha \} \) of \( \alpha \) is also an ordinal, as readers will verify. The result follows from this and 3.4. If \( {Ord} \) were a set, then \( \mathop{\bigcup }\limits_{{\alpha \in \text{ Ord }}}\alpha \) would be an ordinal number, and w... | No |
Lemma 3.6. A subset \( S \) of a well ordered set \( X \) is a lower section of \( X \) if and only if either \( S = X \) or \( S = X\left( a\right) \) for some \( a \in X \) . | Proof. Let \( S \neq X \) be a lower section. Then \( X \smallsetminus S \) has a least element \( a \) . If \( x < a \), then \( x \in S \) : otherwise, \( a \) would not be the least element of \( X \smallsetminus S \) . If \( x \in S \), then \( x < a \) : otherwise, \( a \leqq x \in S \) and \( a \in S \) . Thus \(... | Yes |
Lemma 3.7. Let \( S \) and \( T \) be lower sections of a well ordered set \( X \) . If \( S \cong T \) , then \( S = T \) . | Proof. Let \( S \neq T \) and let \( \theta : S \rightarrow T \) be an isomorphism. By \( {3.6}, S \subseteq T \) or \( T \subseteq S \), and we may exchange \( S \) and \( T \) if necessary and assume that \( S \nsubseteq T \) . Then we cannot have \( \theta \left( x\right) = x \) for all \( x \in S \), and the set \(... | Yes |
Proposition 3.9. If \( \alpha \) is an ordinal number, then so is \( \alpha \cup \{ \alpha \} \) ; in fact, \( \alpha \cup \{ \alpha \} \) is the least ordinal \( \beta > \alpha \) . | The proof is an exercise for our avid readers. | No |
Proposition 3.10. An ordinal number \( \alpha \) is a successor if and only if it has a greatest element; then the greatest element of \( \alpha \) is \( \mathop{\bigcup }\limits_{{\gamma < \alpha }}\gamma < \alpha \) and \( \alpha \) is its successor. Otherwise, \( \mathop{\bigcup }\limits_{{\gamma < \alpha }}\gamma =... | Proof. A successor \( \alpha = \beta \cup \{ \beta \} \) has a greatest element \( \beta \) . Conversely, assume that \( \alpha \) has a greatest element \( \beta \) . Then \( {\mathop{\bigcup }\limits_{{\gamma < \alpha }}}^{\prime } = \beta < \alpha \), and \( \beta < \delta \) implies \( \delta \geqq \alpha \), since... | Yes |
Proposition 4.1. Let \( \mathcal{C} \) be a class of ordinal numbers such that\n\n(1) \( 0 \in \mathcal{C} \) ;\n\n(2) \( \alpha \in \mathcal{C} \) implies \( \alpha + 1 \in \mathcal{C} \) ;\n\n(3) if \( \alpha \) is a limit ordinal and \( \beta \in \mathcal{C} \) for all \( \beta < \alpha \), then \( \alpha \in \mathc... | Proof. If \( \mathcal{C} \neq \) Ord, then Ord \( \smallsetminus \mathcal{C} \) has a least element \( \alpha \), by 3.3. Then \( \mathcal{C} \) contains every \( \beta < \alpha \) . But \( \alpha \neq 0 \), by (1); \( \alpha \) is not a successor ordinal, by (2); and \( \alpha \) is not a limit ordinal, by (3). Theref... | Yes |
Lemma 4.4. No set can contain a transfinite sequence \( {\left( {x}_{\alpha }\right) }_{\alpha \in \text{ Ord }} \) indexed by all ordinals, such that \( {x}_{\alpha } \neq {x}_{\beta } \) whenever \( \alpha \neq \beta \) . | Proof. In the next section we shall see that such a sequence would force the poor set to have entirely too many elements. For now we argue as follows. Let \( X \) be the subset of all \( {x}_{\alpha } \) . Order \( X \) so that \( {x}_{\alpha } < {x}_{\beta } \) if and only if \( \alpha < \beta \) . Then \( X \) is wel... | No |
Proposition 5.1. Let \( {I}_{n} = \{ 1,2,\ldots, n\} \) . If \( m < n \), then \( {I}_{m} \) has fewer elements than \( {I}_{n} \) ; in fact, there is no injection \( {I}_{n} \rightarrow {I}_{m} \) . | Proof. This is not obvious since we have not established that we can count elements as usual. What is obvious is that \( {I}_{m} \subseteq {I}_{n} \) has at most as many elements as \( {I}_{n} \) . We prove by induction on \( m \) that there is no injection \( f : {I}_{n} \rightarrow {I}_{m} \) .\n\nIf \( m = 0 < n \),... | Yes |
Proposition 5.2 (Cantor [1883]). Every set \( X \) has fewer elements than the set \( {2}^{X} \) of all its subsets. | Proof. There is an injection \( x \mapsto \{ x\} \) of \( X \) into \( {2}^{X} \) . To show that \( X \) has fewer elements than \( {2}^{X} \) we prove that there is no bijection of \( X \) onto \( {2}^{X} \) . Let \( f : X \rightarrow {2}^{X} \) be any mapping. Then \( S = \{ x \in X \mid x \notin f\left( x\right) \} ... | Yes |
Theorem 5.3 (Cantor-Bernstein). Let \( X \) and \( Y \) be sets. If there exist an injection of \( X \) into \( Y \) and an injection of \( Y \) into \( X \), then there exists a bijection of \( X \) onto \( Y \) . | Proof. We may assume that \( X \) and \( Y \) are disjoint. Let \( f : X \rightarrow Y \) and \( g : Y \rightarrow X \) be injections. Arrange \( X \cup Y \) into disjoint families in which every element of one set begets (all by itself) one child in the other set. This imagery is due to Halmos. The child of \( x \in X... | Yes |
For every set \( X \) there exists a unique cardinal number \( \left| X\right| \) such that there is a bijection of \( X \) onto \( \left| X\right| \) . | By the axiom of choice, every set \( X \) can be well ordered (Theorem 2.4) and has the same number of elements as an ordinal number, by 3.9. The least ordinal number \( \kappa \) with this property is a cardinal number (since all ordinals \( \alpha < \kappa \) have fewer elements). Moreover, \( \kappa \) is the only c... | Yes |
A direct product of finitely many countable sets is countable. A union of countably many countable sets is countable. | The elements of \( \mathbb{N} \times \mathbb{N} \) can be arranged by increasing sums into a sequence \( \left( {1,1}\right) ;\left( {1,2}\right) ,\left( {2,1}\right) ;\left( {1,3}\right) ,\left( {2,2}\right) ,\left( {3,1}\right) ;\ldots \) Thus \( \mathbb{N} \times \mathbb{N} \) is countable. If now \( X \) and \( Y \... | Yes |
Proposition 5.6 (Cantor [1873]). \( \mathbb{R} \) is not countable. | Proof. Let \( X \) be the set of all real numbers with a decimal expansion \( 0.{d}_{1}{d}_{2}\ldots {d}_{n}\ldots \) in which every digit \( {d}_{n} \) is either 0 or 1 . Every such \( 0.{d}_{1}{d}_{2}\ldots {d}_{n}\ldots \) is determined by a subset \( \left\{ {n \in \mathbb{N} \mid {d}_{n} = 1}\right\} \) of \( \mat... | Yes |
Proposition 5.8. Let \( \kappa \) and \( \lambda \) be cardinal numbers. If \( \kappa \) or \( \lambda \) is infinite, then \( \kappa + \lambda = \max \left( {\kappa ,\lambda }\right) \) . | Proof. We show that \( \kappa + \kappa = \kappa \) when \( \kappa \) is infinite. Then \( \lambda \leqq \kappa \) implies \( \kappa \leqq \kappa + \lambda \leqq \kappa + \kappa = \kappa \) and \( \kappa + \lambda = \kappa \), and 5.8 holds.\n\nFor every set \( X,\left| X\right| + \left| X\right| = \left| {2 \times X}\r... | Yes |
Proposition 5.9. Let \( \kappa \) and \( \lambda \) be nonzero cardinal numbers. If \( \kappa \) or \( \lambda \) is infinite, then \( {\kappa \lambda } = \max \left( {\kappa ,\lambda }\right) \) . | Proof. We show that \( {\kappa \kappa } = \kappa \) when \( \kappa \) is infinite. Then \( 1 \leqq \lambda \leqq \kappa \) implies \( \kappa \leqq {\kappa \lambda } \leqq {\kappa \kappa } = \kappa \) and \( {\kappa \lambda } = \kappa \), and 5.9 holds.\n\nLet \( A \) be an infinite set. As in the proof of 5.8, let \( \... | Yes |
Corollary 5.10. An infinite set \( X \) has \( \left| X\right| \) finite subsets; moreover, there are \( \left| X\right| \) finite sequences of elements of \( X \) . | Proof. The set \( X \) has at least \( \left| X\right| \) finite subsets, since it has \( \left| X\right| \) subsets with one element. On the other hand, \( X \) has \( 1 \leqq \left| X\right| \) empty subset, \( \left| X\right| \) subsets with one element, at most \( \left| X\right| \left| X\right| = \left| X\right| \... | Yes |
Proposition 1.2.1. Let \( q : A \coprod {B}^{n}\left( \mathcal{A}\right) \rightarrow Y \) be the quotient map taking each point to its equivalence class. Then \( q \mid A : A \rightarrow Y \) is a closed embedding, and \( q \mid {B}^{n}\left( \mathcal{A}\right) - {S}^{n - 1}\left( \mathcal{A}\right) \) is an open embed... | Proof. Equivalence classes in \( A\coprod {B}^{n}\left( \mathcal{A}\right) \) have the form \( \{ a\} \cup {f}^{-1}\left( a\right) \) with \( a \in A \), or the form \( \{ z\} \) with \( z \in {B}^{n}\left( \mathcal{A}\right) - {S}^{n - 1}\left( \mathcal{A}\right) \) . Thus \( q \mid A \) and \( q \mid \) \( {B}^{n}\le... | Yes |
Proposition 1.2.2. If \( A \) is Hausdorff, \( Y \) is Hausdorff. Hence \( {e}_{\alpha }^{n} = {\operatorname{cl}}_{Y}{\overset{ \circ }{e}}_{\alpha }^{n} \) . | Proof. Let \( {y}_{1} \neq {y}_{2} \in Y \) . We seek saturated disjoint open subsets \( {U}_{1},{U}_{2} \subset \) \( {B}^{n}\left( \mathcal{A}\right) \coprod A \) whose images contain \( {y}_{1} \) and \( {y}_{2} \) respectively. There are three cases: (i) \( {q}^{-1}\left( \overrightarrow{{y}_{i}}\right) = \left\{ {... | Yes |
Proposition 1.2.3. Let \( f = p \mid {S}^{n - 1}\left( \mathcal{A}\right) : {S}^{n - 1}\left( \mathcal{A}\right) \rightarrow A \) . Let \( X \) be the space obtained by attaching \( {B}^{n}\left( \mathcal{A}\right) \) to \( A \) using \( f \) . Let \( q : A \coprod {B}^{n}\left( \mathcal{A}\right) \rightarrow X \) be t... | Proof. Consider the following commutative diagram:\n\n\n\n\n\n\n\nFig. 1.1.\n\nBy definition of \( p \) and \( q \),... | Yes |
Example 1.2.4. Let \( \left( {Y, A}\right) = \left( {{B}^{2},{S}^{1}}\right) \) . Here \( Y \) is obtained from \( A \) by attaching 2-cells, since we can take \( p : {S}^{1} \coprod {B}^{2} \rightarrow Y \) to be the identity on \( {B}^{2} \) and the inclusion on \( {S}^{1} \) . The corresponding attaching map \( f : ... | For example, define \( h : I \times I \rightarrow I \) by \( \left( {x,0}\right) \mapsto 0,\left( {x,\frac{1}{3}}\right) \mapsto \frac{1}{3} + \frac{x}{6},\left( {x,\frac{2}{3}}\right) \mapsto \frac{2}{3} - \frac{x}{6},\left( {x,1}\right) \mapsto 1 \) , and for each \( x, h \) linear on the segments \( \{ x\} \times \l... | Yes |
Proposition 1.2.5. Let \( A \) be Hausdorff and let \( Y \) be obtained from \( A \) by attaching n-cells. Then the space \( Y \) has the weak topology with respect to \( \left\{ {{e}_{\alpha }^{n} \mid \alpha \in \mathcal{A}}\right\} \cup \{ A\} . \) | Proof. First, we check that \( \left\{ {{e}_{\alpha }^{n} \mid \alpha \in \mathcal{A}}\right\} \cup \{ A\} \) is suitable for defining a weak topology (see Sect. 1.1). By 1.2.1, \( A \) inherits its original topology from \( Y \) . When \( \alpha \neq \beta ,{e}_{\alpha }^{n} \cap {e}_{\beta }^{n} = {e}_{\alpha }^{n} \... | Yes |
Proposition 1.2.6. Let \( A \) be Hausdorff and let \( Y \) be obtained from \( A \) by attaching n-cells. Any compact subset of \( Y \) lies in the union of \( A \) and finitely many cells of \( \left( {Y, A}\right) \) . | Proof. Suppose this were false. Then there would be a compact subset \( C \) of \( Y \) such that \( C \cap {\overset{ \circ }{e}}_{\alpha }^{n} \neq \varnothing \) for infinitely many values of \( \alpha \) . For each such \( \alpha \), pick \( {x}_{\alpha } \in {e}_{\alpha }^{n} \cap C \) . Let \( D \) be the set of ... | Yes |
Proposition 1.2.8. Let \( \left( {Y, A}\right) \) be a Hausdorff pair such that \( \left\{ {{e}_{\alpha }^{ \circ } \mid \alpha \in \mathcal{A}}\right\} \) , the set of path components of \( Y - A \), is finite. Let \( n \in \mathbb{N}.Y \) is obtained from \( A \) by attaching \( n \) -cells if \( A \) is closed in \(... | Proof. \ | No |
Proposition 1.2.9. Let \( \left( {Y, A}\right) ,\left\{ {{e}_{\alpha } \mid \alpha \in \mathcal{A}}\right\} \) and \( n \in \mathbb{N} \) be as in 1.2.7, and let \( {p}_{\alpha } : \left( {{B}^{n},{S}^{n - 1}}\right) \rightarrow \left( {A \cup {e}_{\alpha }, A}\right) \) be a map such that \( {p}_{\alpha } \) maps \( {... | Proof. Since \( {B}^{n} \) is compact, \( {p}_{\alpha } : {B}^{n} \rightarrow {e}_{\alpha }^{n} \) is a quotient map. The restriction \( {p}_{\alpha } \mid : {B}^{n} - {S}^{n - 1} \rightarrow {\overset{ \circ }{e}}_{\alpha }^{n} \) is a bijective quotient map, hence a homeomorphism. | Yes |
Proposition 1.2.12. A CW complex \( X \) has the weak topology with respect to its cells. | Proof. This is proved for \( n \) -dimensional CW complexes by induction on \( n \) , using 1.2.5. For an arbitrary CW complex \( X, U \subset X \) is open iff \( U \cap {X}^{n} \) is open in \( {X}^{n} \) for all \( n \), iff \( U \cap {e}^{i} \) is open in \( {e}^{i} \) for every \( i \) -cell \( {e}^{i} \) of \( X \... | Yes |
Proposition 1.2.13. A compact subset of a CW complex lies in the union of finitely many cells. In particular, a CW complex is finite iff its underlying space is compact. | Proof. Suppose this were false. Then there would be a compact subset \( C \) of \( X \) such that \( C \cap {\overset{ \circ }{e}}_{\alpha } \neq \varnothing \) for infinitely many values of \( \alpha \) (where \( \left\{ {{e}_{\alpha } \mid \alpha \in \mathcal{A}}\right\} \) is the set of cells of \( X \) ). For each ... | Yes |
Proposition 1.2.14. Let \( X \) be a Hausdorff space and let \( \left\{ {{e}_{\alpha } \mid \alpha \in \mathcal{A}}\right\} \) be a family of subspaces with the following properties:\n\n(i) \( X = \mathop{\bigcup }\limits_{\alpha }\left\{ {\overset{ \circ }{e}}_{\alpha }\right\} \) and \( {\overset{ \circ }{e}}_{\alpha... | Proof. Let \( A \subset {X}^{0} \) and let \( {e}_{\alpha } \) be a cell of \( X \) . Then \( A \cap {e}_{\alpha } \) is finite by (iv), hence compact, hence closed in \( {e}_{\alpha } \) . So \( A \) is closed in \( X \), hence also in \( {X}^{0} \) . So \( {X}^{0} \) is discrete.\n\nNext, we show that \( {X}^{n} \) h... | No |
Proposition 1.2.20. Let \( \left( {X,\left\{ {X}^{n}\right\} }\right) \) be a CW complex, let \( \left\{ {{e}_{\alpha } \mid \alpha \in \mathcal{A}}\right\} \) be the set of cells of \( X \), let \( \mathcal{B} \subset \mathcal{A} \), let \( A = \cup \left\{ {{\overset{ \circ }{e}}_{\alpha } \mid \alpha \in \mathcal{B}... | Proof. We verify the axioms (i)-(iv) in the definition of a CW complex. Axioms (i) and (iii) clearly hold. To verify Axiom (ii), we show by induction that \( {A}^{n} \) is obtained from \( {A}^{n - 1} \) by attaching \( n \) -cells and that \( {A}^{n} \) is closed in \( {X}^{n} \) . This is clear when \( n = 0 \) ; ass... | Yes |
Proposition 1.2.21. Each path component of a CW complex \( X \) is a subcomplex, an open subset of \( X \), and a closed subset of \( X \). Hence, a non-empty \( {CW} \) complex is connected iff it is path connected. | Proof. The first part is clear. For the rest, let \( A \) be a path component of the CW complex \( X \). Cells are path connected, being the images of balls under maps. Hence, for each cell \( e \) of \( X \), either \( e \cap A = e \) or \( e \cap A = \varnothing \). This proves \( A \) is both open and closed in \( X... | Yes |
Proposition 1.2.22. If there exist pairwise disjoint open sets \( {U}_{\alpha } \subset X \) such that, for each \( \alpha ,{A}_{\alpha } \subset {U}_{\alpha } \), then \( \left( {X/ \sim ,\left\{ {\left( X/ \sim \right) }^{n}\right\} }\right) \) is a CW complex. In particular, if \( A \) is a subcomplex of \( X \), th... | Proof. Apply 1.2.14. The Hausdorff property is clear under these hypotheses.\n\nWith this CW structure, \( X/ \sim \) is the quotient complex. | No |
Proposition 1.2.23. Let \( \left( {X, A}\right) \) be a CW pair and let the n-cells of \( X \) which are not cells of \( A \) be indexed by \( \mathcal{A} \) . Let \( \left\{ {{h}_{\alpha } : {B}_{\alpha }^{n} \rightarrow {e}_{\alpha }^{n} \mid \alpha \in \mathcal{A}}\right\} \) be a set of characteristic maps for thos... | Proof. By 1.2.7, \( {X}^{n} \cup A \) is obtained from \( {X}^{n - 1} \cup A \) by attaching \( n \) -cells. The result follows from the properties of the quotient topology stated in Sect. 1.1. | No |
Proposition 1.3.1. Homotopy is an equivalence relation on the set of maps from \( X \) to \( Y \) . | Proof. Given \( f : X \rightarrow Y \), define \( F : f \simeq f \) by \( F\left( {x, t}\right) = f\left( x\right) \) for all \( t \in I \) ; \( F = f \circ \) (projection: \( X \times I \rightarrow X \) ), the composition of two maps. So \( F \) is a map. Thus reflexivity. Given \( F : f \simeq g \), define \( {F}^{\p... | Yes |
Proposition 1.3.3. Let \( {f}_{0},{f}_{1} : \left( {X, A}\right) \rightarrow \left( {Y, B}\right) \) be homotopic rel \( {X}^{\prime } \), let \( {g}_{0},{g}_{1} : \left( {Y, B}\right) \rightarrow \left( {Z, C}\right) \) be homotopic rel \( {Y}^{\prime } \), where \( {f}_{0}\left( {X}^{\prime }\right) \subset {Y}^{\pri... | Proof. Let \( F : {f}_{0} \simeq {f}_{1} \) and \( G : {g}_{0} \simeq {g}_{1} \) be homotopies which behave as required on \( A, B, C \) and \( {X}^{\prime } \) . Let \( p : X \times I \rightarrow I \) be projection. Let \( \left( {F, p}\right) \) : \( X \times I \rightarrow Y \times I \) denote the function \( \left( ... | Yes |
The spaces \( \mathbb{R},{\mathbb{R}}_{ + } \), and \( I \) are contractible. | Hence also (by 1.3.6) \( {\mathbb{R}}^{n},{\mathbb{R}}_{ + }^{n} \) and \( {B}^{n} \) are contractible. If \( p \in {S}^{n} \), then \( {S}^{n} - \{ p\} \) is homeomorphic to \( {\mathbb{R}}^{n} \) ; hence \( {S}^{n} - \{ p\} \) is contractible. | No |
Proposition 1.3.8. If \( A \) is a strong deformation retract of \( X \) then \( A \hookrightarrow X \) is a homotopy equivalence. | Proof. In the notation above, \( r \) is a homotopy inverse for inclusion. | No |
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