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Let us compute these cohomology groups when \( X = \mathbb{R} \) with the \( \Delta \) -complex structure having vertices at the integer points. For a simplicial 0-cochain to be a cocycle it must take the same value on all vertices, but then if the cochain lies in \( {\Delta }_{c}^{0}\left( X\right) \) it must be ident... | Namely, consider the map \( \sum : {\Delta }_{c}^{1}\left( {\mathbb{R};G}\right) \rightarrow G \) sending each cochain to the sum of its values on all the 1-simplices. Note that \( \sum \) is not defined on all of \( {\Delta }^{1}\left( X\right) \), just on \( {\Delta }_{c}^{1}\left( X\right) \) . The map \( \sum \) va... | No |
Proposition 3.33. If a space \( X \) is the union of a directed set of subspaces \( {X}_{\alpha } \) with the property that each compact set in \( X \) is contained in some \( {X}_{\alpha } \), then the natural map \( \mathop{\lim }\limits_{ \rightarrow }{H}_{i}\left( {{X}_{\alpha };G}\right) \rightarrow {H}_{i}\left( ... | Proof: For surjectivity, represent a cycle in \( X \) by a finite sum of singular simplices. The union of the images of these singular simplices is compact in \( X \), hence lies in some \( {X}_{\alpha } \), so the map \( \mathop{\lim }\limits_{ \rightarrow }{H}_{i}\left( {{X}_{\alpha };G}\right) \rightarrow {H}_{i}\le... | Yes |
Theorem 3.35. The duality map \( {D}_{M} : {H}_{c}^{k}\left( {M;R}\right) \rightarrow {H}_{n - k}\left( {M;R}\right) \) is an isomorphism \( \parallel \) for all \( k \) whenever \( M \) is an \( R \) -oriented \( n \) -manifold. | The proof will not be difficult once we establish a technical result stated in the next lemma, concerning the commutativity of a certain diagram. Commutativity statements of this sort are usually routine to prove, but this one seems to be an exception. The reader who consults other books for alternative expositions wil... | No |
Lemma 3.36. If \( M \) is the union of two open sets \( U \) and \( V \), then there is a diagram of Mayer-Vietoris sequences, commutative up to sign: | Proof: Compact sets \( K \subset U \) and \( L \subset V \) give rise to the Mayer-Vietoris sequence in the upper row of the following diagram, whose lower row is also a Mayer-Vietoris sequence. | No |
Proposition 3.38. The cup product pairing is nonsingular for closed \( R \) -orientable manifolds when \( R = \mathbb{Z} \) and torsion in \( {H}^{ * }\left( {M;\mathbb{Z}}\right) \) is factored out. | Proof: Consider the composition\n\n\[ \n{H}^{n - k}\left( {M;R}\right) \overset{h}{ \rightarrow }{\operatorname{Hom}}_{R}\left( {{H}_{n - k}\left( {M;R}\right), R}\right) \overset{{D}^{ * }}{ \rightarrow }{\operatorname{Hom}}_{R}\left( {{H}^{k}\left( {M;R}\right), R}\right) \]\n\nwhere \( h \) is the map appearing in t... | Yes |
Corollary 3.39. If \( M \) is a closed connected orientable \( n \) -manifold, then for each element \( \alpha \in {H}^{k}\left( {M;\mathbb{Z}}\right) \) of infinite order that is not a proper multiple of another element, there exists an element \( \beta \in {H}^{n - k}\left( {M;\mathbb{Z}}\right) \) such that \( \alph... | Proof: The hypotheses on \( \alpha \) mean that it generates a \( \mathbb{Z} \) summand of \( {H}^{k}\left( {M;\mathbb{Z}}\right) \) . There is then a homomorphism \( \varphi : {H}^{k}\left( {M;\mathbb{Z}}\right) \rightarrow \mathbb{Z} \) with \( \varphi \left( \alpha \right) = 1 \) . By the nonsingularity of the cup p... | Yes |
Proposition 3.42. If \( M \) is a compact manifold with boundary, then \( \partial M \) has a collar neighborhood. | Proof: Let \( {M}^{\prime } \) be \( M \) with an external collar attached, the quotient of the disjoint union of \( M \) and \( \partial M \times \left\lbrack {0,1}\right\rbrack \) in which \( x \in \partial M \) is identified with \( \left( {x,0}\right) \in \partial M \times \left\lbrack {0,1}\right\rbrack \) . It wi... | Yes |
Theorem 3.43. Suppose \( M \) is a compact \( R \) -orientable \( n \) -manifold whose boundary \( \partial M \) is decomposed as the union of two compact \( \left( {n - 1}\right) \) -dimensional manifolds \( A \) and \( B \) with a common boundary \( \partial A = \partial B = A \cap B \) . Then cap product with a fund... | Proof: The cap product map \( {D}_{M} : {H}^{k}\left( {M, A;R}\right) \rightarrow {H}_{n - k}\left( {M, B;R}\right) \) is defined since the existence of collar neighborhoods of \( A \cap B \) in \( A \) and \( B \) and \( \partial M \) in \( M \) implies that \( A \) and \( B \) are deformation retracts of open neighbo... | No |
Theorem 3.44. If \( K \) is a compact, locally contractible, nonempty, proper subspace of \( {S}^{n} \), then \( {\widetilde{H}}_{i}\left( {{S}^{n} - K;\mathbb{Z}}\right) \approx {\widetilde{H}}^{n - i - 1}\left( {K;\mathbb{Z}}\right) \) for all \( i \) . | Proof: We will obtain the desired isomorphism when \( i \neq 0 \) as the composition of five isomorphisms\n\n\[ \n{H}_{i}\left( {{S}^{n} - K}\right) \approx {H}_{c}^{n - i}\left( {{S}^{n} - K}\right) \]\n\n\[ \n\approx \mathop{\lim }\limits_{ \rightarrow }{H}^{n - i}\left( {{S}^{n} - K, U - K}\right) \]\n\n\[ \n\approx... | Yes |
Proposition 4.1. A covering space projection \( p : \left( {\widetilde{X},{\widetilde{x}}_{0}}\right) \rightarrow \left( {X,{x}_{0}}\right) \) induces isomor- \( \parallel \) phisms \( {p}_{ * } : {\pi }_{n}\left( {\widetilde{X},{\widetilde{x}}_{0}}\right) \rightarrow {\pi }_{n}\left( {X,{x}_{0}}\right) \) for all \( n... | Proof: For surjectivity of \( {p}_{ * } \) we apply the lifting criterion in Proposition 1.33, which implies that every map \( \left( {{S}^{n},{s}_{0}}\right) \rightarrow \left( {X,{x}_{0}}\right) \) lifts to \( \left( {\widetilde{X},{\widetilde{x}}_{0}}\right) \) provided that \( n \geq 2 \) so that \( {S}^{n} \) is s... | Yes |
For a product \( {\Pi }_{\alpha }{X}_{\alpha } \) of an arbitrary collection of path-connected \( \parallel \) spaces \( {X}_{\alpha } \) there are isomorphisms \( {\pi }_{n}\left( {{\Pi }_{\alpha }{X}_{\alpha }}\right) \approx {\Pi }_{\alpha }{\pi }_{n}\left( {X}_{\alpha }\right) \) for all \( n \) . | A map \( f : Y \rightarrow \mathop{\prod }\limits_{\alpha }{X}_{\alpha } \) is the same thing as a collection of maps \( {f}_{\alpha } : Y \rightarrow {X}_{\alpha } \) . Taking \( Y \) to be \( {S}^{n} \) and \( {S}^{n} \times I \) gives the result. | No |
Lemma 4.6. Let \( \left( {X, A}\right) \) be a \( {CW} \) pair and let \( \left( {Y, B}\right) \) be any pair with \( B \neq \varnothing \) . For each \( n \) such that \( X - A \) has cells of dimension \( n \), assume that \( {\pi }_{n}\left( {Y, B,{y}_{0}}\right) = 0 \) for all \( {y}_{0} \in B \) . Then every map \... | Proof: Assume inductively that \( f \) has already been homotoped to take the skeleton \( {X}^{k - 1} \) to \( B \) . If \( \Phi \) is the characteristic map of a cell \( {e}^{k} \) of \( X - A \), the composition \( {f\Phi } : \left( {{D}^{k},\partial {D}^{k}}\right) \rightarrow \left( {Y, B}\right) \) can be homotope... | Yes |
Theorem 4.8. Every map \( f : X \rightarrow Y \) of CW complexes is homotopic to a cellular map. If \( f \) is already cellular on a subcomplex \( A \subset X \), the homotopy may be taken to be stationary on \( A \) . | Proof of 4.8: Suppose inductively that \( f : X \rightarrow Y \) is already cellular on the skeleton \( {X}^{n - 1} \), and let \( {e}^{n} \) be an \( n \) -cell of \( X \) . The closure of \( {e}^{n} \) in \( X \) is compact, being the image of a characteristic map for \( {e}^{n} \), so \( f \) takes the closure of \(... | Yes |
Lemma 4.10. Let \( f : {I}^{n} \rightarrow Z \) be a map, where \( Z \) is obtained from a subspace \( W \) by attaching a cell \( {e}^{k} \) . Then \( f \) is homotopic \( \operatorname{rel}{f}^{-1}\left( W\right) \) to a map \( {f}_{1} \) for which there is a simplex \( {\Delta }^{k} \subset {e}^{k} \) with \( {f}_{1... | Proof of 4.10: Identifying \( {e}^{k} \) with \( {\mathbb{R}}^{k} \), let \( {B}_{1},{B}_{2} \subset {e}^{k} \) be the closed balls of radius 1 and 2 centered at the origin. Since \( {f}^{-1}\left( {B}_{2}\right) \) is closed and therefore compact in \( {I}^{n} \), it follows that \( f \) is uniformly continuous on \( ... | Yes |
Corollary 4.12. A CW pair \( \left( {X, A}\right) \) is \( n \) -connected if all the cells in \( X - A \) have dimension greater than \( n \) . In particular the pair \( \left( {X,{X}^{n}}\right) \) is \( n \) -connected, hence the inclusion \( {X}^{n} \hookrightarrow X \) induces isomorphisms on \( {\pi }_{i} \) for ... | Proof: Applying cellular approximation to maps \( \left( {{D}^{i},\partial {D}^{i}}\right) \rightarrow \left( {X, A}\right) \) with \( i \leq n \) gives the first statement. The last statement comes from the long exact sequence of the pair \( \left( {X,{X}^{n}}\right) \) . | No |
When \( X \) is path-connected and \( A \) is a point, the construction of a 0 -connected CW model for \( \left( {X, A}\right) \) gives a CW approximation to \( X \) with a single 0 -cell and all higher cells attached by basepoint-preserving maps. | In particular, any connected CW complex is homotopy equivalent to a CW complex with these properties. | No |
Corollary 4.16. If \(\left( {X, A}\right)\) is an \(n\)-connected CW pair, then there exists a CW pair \(\left( {Z, A}\right) \simeq \left( {X, A}\right)\) rel \(A\) such that all cells of \(Z - A\) have dimension greater than \(n\). | Proof: An \( n \) -connected CW approximation \( f : \left( {Z, A}\right) \rightarrow \left( {X, A}\right) \) given by the preceding proposition will do the trick. First we check that \( f \) induces isomorphisms \( {\pi }_{i}\left( Z\right) \approx \) \( {\pi }_{i}\left( X\right) \) for all \( i \) . This is true for ... | Yes |
Proposition 4.18. Suppose we are given:\n\n(i) an \( n \) -connected \( {CW} \) model \( f : \left( {Z, A}\right) \rightarrow \left( {X, A}\right) \) ,\n\n(ii) an \( {n}^{\prime } \) -connected CW model \( {f}^{\prime } : \left( {{Z}^{\prime },{A}^{\prime }}\right) \rightarrow \left( {{X}^{\prime },{A}^{\prime }}\right... | Proof: By Corollary 4.16 we may assume all cells of \( Z - A \) have dimension greater than \( n \) . Let \( W \) be the quotient space of the mapping cylinder of \( {f}^{\prime } \) obtained by collapsing each line segment \( \left\{ {a}^{\prime }\right\} \times I \) to a point, for \( {a}^{\prime } \in {A}^{\prime } ... | Yes |
Corollary 4.19. An \( n \) -connected \( {CW} \) model for \( \left( {X, A}\right) \) is unique up to homotopy equivalence \( \operatorname{rel}A \) . In particular, CW approximations to spaces are unique up to homotopy equivalence. | Proof: Given two \( n \) -connected CW models \( \left( {Z, A}\right) \) and \( \left( {{Z}^{\prime }, A}\right) \) for \( \left( {X, A}\right) \), we apply the proposition twice with \( g \) the identity map to obtain maps \( h : Z \rightarrow {Z}^{\prime } \) and \( {h}^{\prime } : {Z}^{\prime } \rightarrow Z \) . Th... | Yes |
Proposition 4.21. A weak homotopy equivalence \( f : X \rightarrow Y \) induces isomorphisms \( {f}_{ * } : {H}_{n}\left( {X;G}\right) \rightarrow {H}_{n}\left( {Y;G}\right) \) and \( {f}^{ * } : {H}^{n}\left( {Y;G}\right) \rightarrow {H}^{n}\left( {X;G}\right) \) for all \( n \) and all coefficient groups \( G \) . | Proof: Replacing \( Y \) by the mapping cylinder \( {M}_{f} \) and looking at the long exact sequences of homotopy, homology, and cohomology groups for \( \left( {{M}_{f}, X}\right) \), we see that it suffices to show:\n\n- If \( \left( {Z, X}\right) \) is an \( n \) -connected pair of path-connected spaces, then \( {H... | Yes |
Corollary 4.24. The suspension map \( {\pi }_{i}\left( {S}^{n}\right) \rightarrow {\pi }_{i + 1}\left( {S}^{n + 1}\right) \) is an isomorphism for \( i < {2n} - 1 \) and a surjection for \( i = {2n} - 1 \) . More generally this holds for the suspension \( {\pi }_{i}\left( X\right) \rightarrow {\pi }_{i + 1}\left( {SX}\... | Proof: Decompose the suspension \( {SX} \) as the union of two cones \( {C}_{ + }X \) and \( {C}_{ - }X \) intersecting in a copy of \( X \) . The suspension map is the same as the map\n\n\[ \n{\pi }_{i}\left( X\right) \approx {\pi }_{i + 1}\left( {{C}_{ + }X, X}\right) \rightarrow {\pi }_{i + 1}\left( {{SX},{C}_{ - }X... | Yes |
Corollary 4.25. \( {\pi }_{n}\left( {S}^{n}\right) \approx \mathbb{Z} \), generated by the identity map, for all \( n \geq 1 \) . In \( \parallel \) particular, the degree map \( {\pi }_{n}\left( {S}^{n}\right) \rightarrow \mathbb{Z} \) is an isomorphism. | Proof: From the preceding corollary we know that in the suspension sequence\n\n\[ \n{\pi }_{1}\left( {S}^{1}\right) \rightarrow {\pi }_{2}\left( {S}^{2}\right) \rightarrow {\pi }_{3}\left( {S}^{3}\right) \rightarrow \cdots \n\]\n\nthe first map is surjective and all the subsequent maps are isomorphisms. Since \( {\pi }... | Yes |
Let us show that \( {\pi }_{n}\left( {{S}^{1} \vee {S}^{n}}\right) \) for \( n \geq 2 \) is free abelian on a countably infinite number of generators. | By Proposition 4.1 we may compute \( {\pi }_{i}\left( {{S}^{1} \vee {S}^{n}}\right) \) for \( i \geq 2 \) by passing to the universal cover. This consists of a copy of \( \mathbb{R} \) with a sphere \( {S}_{k}^{n} \) attached at each integer point \( k \in \mathbb{R} \), so it is homotopy equivalent to \( { \vee }_{k}{... | Yes |
Proposition 4.28. If a CW pair \( \left( {X, A}\right) \) is \( r \) -connected and \( A \) is \( s \) -connected, with \( \parallel r, s\parallel \geq 0 \), then the map \( {\pi }_{i}\left( {X, A}\right) \rightarrow {\pi }_{i}\left( {X/A}\right) \) induced by the quotient map \( X \rightarrow X/A \) . I is an isomorph... | Proof: Consider \( X \cup {CA} \), the complex obtained from \( X \) by attaching a cone \( {CA} \) along \( A \subset X \) . Since \( {CA} \) is a contractible subcomplex of \( X \cup {CA} \), the quotient map \( X \cup {CA} \rightarrow \left( {X \cup {CA}}\right) /{CA} = X/A \) is a homotopy equivalence by Propositio... | Yes |
Suppose \( X \) is obtained from a wedge of spheres \( { \vee }_{\alpha }{S}_{\alpha }^{n} \) by attaching cells \( {e}_{\beta }^{n + 1} \) via basepoint-preserving maps \( {\varphi }_{\beta } : {S}^{n} \rightarrow { \vee }_{\alpha }{S}_{\alpha }^{n} \), with \( n \geq 2 \). By cellular approximation we know that \( {\... | To see that \( {\pi }_{n}\left( X\right) \) is as claimed, consider the following portion of the long exact sequence of the pair \( \left( {X,\mathop{\bigvee }\limits_{\alpha }{S}_{\alpha }^{n}}\right) \):\n\n\[ \n{\pi }_{n + 1}\left( {X,\mathop{\bigvee }\limits_{\alpha }{S}_{\alpha }^{n}}\right) \overset{\partial }{ \... | Yes |
Theorem 4.32. If a space \( X \) is \( \left( {n - 1}\right) \) -connected, \( n \geq 2 \), then \( {\widetilde{H}}_{i}\left( X\right) = 0 \) for \( i < n \) and \( {\pi }_{n}\left( X\right) \approx {H}_{n}\left( X\right) \) . If a pair \( \left( {X, A}\right) \) is \( \left( {n - 1}\right) \) -connected, \( n \geq 2 \... | Proof: We may assume \( X \) is a CW complex and \( \left( {X, A}\right) \) is a CW pair by taking CW approximations to \( X \) and \( \left( {X, A}\right) \) . For CW pairs the relative case then reduces to the absolute case since \( {\pi }_{i}\left( {X, A}\right) \approx {\pi }_{i}\left( {X/A}\right) \) for \( i \leq... | Yes |
Corollary 4.33. A map \( f : X \rightarrow Y \) between simply-connected CW complexes is a ho- \( \parallel \) motopy equivalence if \( {f}_{ * } : {H}_{n}\left( X\right) \rightarrow {H}_{n}\left( Y\right) \) is an isomorphism for each \( n \) . | Proof: After replacing \( Y \) by the mapping cylinder \( {M}_{f} \) we may take \( f \) to be an inclusion \( X \hookrightarrow Y \) . Since \( X \) and \( Y \) are simply-connected, we have \( {\pi }_{1}\left( {Y, X}\right) = 0 \) . The relative Hurewicz theorem then says that the first nonzero \( {\pi }_{n}\left( {Y... | Yes |
We construct a space \( X = \left( {{S}^{1} \vee {S}^{n}}\right) \cup {e}^{n + 1} \), for arbitrary \( n > 1 \), such that the inclusion \( {S}^{1} \hookrightarrow X \) induces an isomorphism on all homology groups and on \( {\pi }_{i} \) for \( i < n \), but not on \( {\pi }_{n} \) . | From Example 4.27 we have \( {\pi }_{n}\left( {{S}^{1} \vee {S}^{n}}\right) \approx \mathbb{Z}\left\lbrack {t,{t}^{-1}}\right\rbrack \) . Let \( X \) be obtained from \( {S}^{1} \vee {S}^{n} \) by attaching a cell \( {e}^{n + 1} \) via a map \( {S}^{n} \rightarrow {S}^{1} \vee {S}^{n} \) corresponding to \( {2t} - 1 \i... | Yes |
Proposition 4.36. The Hurewicz map \( h : {\pi }_{n}\left( {X, A,{x}_{0}}\right) \rightarrow {H}_{n}\left( {X, A}\right) \) is a homomorphism, assuming \( n > 1 \) so that \( {\pi }_{n}\left( {X, A,{x}_{0}}\right) \) is a group. | Proof: It suffices to show that for maps \( f, g : \left( {{D}^{n},\partial {D}^{n}}\right) \rightarrow \left( {X, A}\right) \), the induced maps on homology satisfy \( {\left( f + g\right) }_{ * } = {f}_{ * } + {g}_{ * } \), for if this is the case then \( h\left( \left\lbrack {f + g}\right\rbrack \right) = \) \( {\le... | Yes |
Lemma 4.38. If \( X \) is a connected \( {CW} \) complex to which cells \( {e}_{\alpha }^{n} \) are attached for a fixed \( n \geq 2 \), forming a \( {CW} \) complex \( W = X\mathop{\bigcup }\limits_{\alpha }{e}_{\alpha }^{n} \), then \( {\pi }_{n}\left( {W, X}\right) \) is a free \( {\pi }_{1}\left( X\right) \) -modul... | Proof: Since \( W/X = \mathop{\bigvee }\limits_{\alpha }{S}_{\alpha }^{n} \), we have \( {\pi }_{n}\left( {W, X}\right) \approx {\pi }_{n}\left( {\mathop{\bigvee }\limits_{\alpha }{S}_{\alpha }^{n}}\right) \) when \( X \) is simply-connected, by Proposition 4.28. The conclusion of the lemma in this case is then immedia... | Yes |
Lemma 4.39. For any \( \left( {X, A,{x}_{0}}\right) \), the formula \( a + b - a = \left( {\partial a}\right) b \) holds for all \( \parallel a, b\parallel \in {\pi }_{2}\left( {X, A,{x}_{0}}\right) \), where \( \partial : {\pi }_{2}\left( {X, A,{x}_{0}}\right) \rightarrow {\pi }_{1}\left( {A,{x}_{0}}\right) \) is the ... | Proof: The formula is obtained by constructing a homotopy from \( a + b - a \) to \( \left( {\partial a}\right) b \) as indicated in the pictures below. | No |
Proposition 4.40. Let \( X \) be a connected \( {CW} \) complex with \( {H}_{1}\left( X\right) = 0 \) . Then there is a simply-connected \( {CW} \) complex \( {X}^{ + } \) and a map \( X \rightarrow {X}^{ + } \) inducing isomorphisms on all homology groups. | Proof: Choose loops \( {\varphi }_{\alpha } : {S}^{1} \rightarrow {X}^{1} \) generating \( {\pi }_{1}\left( X\right) \) and use these to attach cells \( {e}_{\alpha }^{2} \) to \( X \) to form a simply-connected CW complex \( {X}^{\prime } \) . The homology exact sequence\n\n\[ 0 \rightarrow {H}_{2}\left( X\right) \rig... | Yes |
Theorem 4.41. Suppose \( p : E \rightarrow B \) has the homotopy lifting property with respect to disks \( {D}^{k} \) for all \( k \geq 0 \) . Choose basepoints \( {b}_{0} \in B \) and \( {x}_{0} \in F = {p}^{-1}\left( {b}_{0}\right) \) . Then the map \( {p}_{ * } : {\pi }_{n}\left( {E, F,{x}_{0}}\right) \rightarrow {\... | Proof: First we show that \( {p}_{ * } \) is onto. Represent an element of \( {\pi }_{n}\left( {B,{b}_{0}}\right) \) by a map \( f : \left( {{I}^{n},\partial {I}^{n}}\right) \rightarrow \left( {B,{b}_{0}}\right) \) . The constant map to \( {x}_{0} \) provides a lift of \( f \) to \( E \) over the subspace \( {J}^{n - 1... | Yes |
One of the simplest nontrivial fiber bundles is the Möbius band, which is a bundle over \( {S}^{1} \) with fiber an interval. Specifically, take \( E \) to be the quotient of \( I \times \left\lbrack {-1,1}\right\rbrack \) under the identifications \( \left( {0, v}\right) \sim \left( {1, - v}\right) \), with \( p : E \... | Glueing two copies of \( E \) together by the identity map between their boundary circles produces a Klein bottle, a bundle over \( {S}^{1} \) with fiber \( {S}^{1} \) . | No |
Projective spaces yield interesting fiber bundles. In the real case we have the familiar covering spaces \( {S}^{n} \rightarrow {\mathbb{{RP}}}^{n} \) with fiber \( {S}^{0} \) . Over the complex numbers the analog of this is a fiber bundle \( {S}^{1} \rightarrow {S}^{{2n} + 1} \rightarrow {\mathbb{{CP}}}^{n} \) . Here ... | To see that the local triviality condition for fiber bundles is satisfied, let \( {U}_{i} \subset {\mathbb{{CP}}}^{n} \) be the open set of equivalence classes \( \left\lbrack {{z}_{0},\cdots ,{z}_{n}}\right\rbrack \) with \( {z}_{i} \neq 0 \) . Define \( {h}_{i} : {p}^{-1}\left( {U}_{i}\right) \rightarrow {U}_{i} \tim... | Yes |
The case \( n = 1 \) is particularly interesting since \( {\mathbb{{CP}}}^{1} = {S}^{2} \) and the bundle becomes \( {S}^{1} \rightarrow {S}^{3} \rightarrow {S}^{2} \) with fiber, total space, and base all spheres. This is known as the Hopf bundle, and is of low enough dimension to be seen explicitly. The projection \(... | In polar coordinates we have \( p\left( {{r}_{0}{e}^{i{\theta }_{0}},{r}_{1}{e}^{i{\theta }_{1}}}\right) = \left( {{r}_{0}/{r}_{1}}\right) {e}^{i\left( {{\theta }_{0} - {\theta }_{1}}\right) } \) where \( {r}_{0}^{2} + {r}_{1}^{2} = 1 \) . For a fixed ratio \( \rho = {r}_{0}/{r}_{1} \in \left( {0,\infty }\right) \) the... | Yes |
Another Hopf bundle \( {S}^{7} \rightarrow {S}^{15} \rightarrow {S}^{8} \) can be defined using the octonion algebra \( \mathbb{O} \). Elements of \( \mathbb{O} \) are pairs of quaternions \( \left( {{a}_{1},{a}_{2}}\right) \) with multiplication given by \( \left( {{a}_{1},{a}_{2}}\right) \left( {{b}_{1},{b}_{2}}\righ... | Let \( {U}_{0} \) and \( {U}_{1} \) be the complements of \( \infty \) and 0 in the base space \( \mathbb{O} \cup \{ \infty \} \). Define \( {h}_{i} : {p}^{-1}\left( {U}_{i}\right) \rightarrow {U}_{i} \times {S}^{7} \) and \( {g}_{i} : {U}_{i} \times {S}^{7} \rightarrow {p}^{-1}\left( {U}_{i}\right) \) by\n\n\[ \n{h}_{... | Yes |
Proposition 4.48. A fiber bundle \( p : E \rightarrow B \) has the homotopy lifting property with \( \parallel \) respect to all \( {CW} \) pairs \( \left( {X, A}\right) \) . | Proof: As noted earlier, the homotopy lifting property for CW pairs is equivalent to the homotopy lifting property for disks, or equivalently, cubes. Let \( G : {I}^{n} \times I \rightarrow B \) , \( G\left( {x, t}\right) = {g}_{t}\left( x\right) \), be a homotopy we wish to lift, starting with a given lift \( {\wideti... | Yes |
Applying this theorem to a covering space \( p : E \rightarrow B \) with \( E \) and \( B \) path-connected, and discrete fiber \( F \), the resulting long exact sequence of homotopy groups yields Proposition 4.1 that \( {p}_{ * } : {\pi }_{n}\left( E\right) \rightarrow {\pi }_{n}\left( B\right) \) is an isomorphism fo... | We also obtain a short exact sequence \( 0 \rightarrow {\pi }_{1}\left( E\right) \rightarrow {\pi }_{1}\left( B\right) \rightarrow {\pi }_{0}\left( F\right) \rightarrow 0 \), consistent with the covering space theory facts that \( {p}_{ * } : {\pi }_{1}\left( E\right) \rightarrow {\pi }_{1}\left( B\right) \) is injecti... | No |
For \( m < n \leq k \) there are fiber bundles\n\n\[ \n{V}_{n - m}\left( {\mathbb{R}}^{k - m}\right) \rightarrow {V}_{n}\left( {\mathbb{R}}^{k}\right) \overset{p}{ \rightarrow }{V}_{m}\left( {\mathbb{R}}^{k}\right) \n\]\n\nwhere the projection \( p \) sends an \( n \) -frame onto the \( m \) -frame formed by its first ... | Local trivializations can be constructed as follows. For an \( m \) -frame \( F \) , choose an orthonormal basis for the \( \left( {k - m}\right) \) -plane orthogonal to \( F \) . This determines orthonormal bases for the \( \left( {k - m}\right) \) -planes orthogonal to all nearby \( m \) -frames by orthogonal project... | Yes |
Theorem 4.57. There are natural bijections \( T : \langle X, K\left( {G, n}\right) \rangle \rightarrow {H}^{n}\left( {X;G}\right) \) for all \( {CW} \) complexes \( X \) and all \( n > 0 \), with \( G \) any abelian group. Such a \( T \) has the form \( T\left( \left\lbrack f\right\rbrack \right) = {f}^{ * }\left( \alp... | In the course of the proof we will define a natural group structure on \( \langle X, K\left( {G, n}\right) \rangle \) such that the transformation \( T \) is an isomorphism. | No |
Theorem 4.58. If \( \left\{ {K}_{n}\right\} \) is an \( \Omega \) -spectrum, then the functors \( X \mapsto {h}^{n}\left( X\right) = \left\langle {X,{K}_{n}}\right\rangle \) , \( \parallel n \in \mathbb{Z} \), define a reduced cohomology theory on the category of basepointed CW complexes and basepoint-preserving maps. | Proof: Two of the three axioms for a cohomology theory, the homotopy axiom and the wedge sum axiom, are quite easy to check. For the homotopy axiom, a basepoint-preserving map \( f : X \rightarrow Y \) induces \( {f}^{ * } : \left\langle {Y,{K}_{n}}\right\rangle \rightarrow \left\langle {X,{K}_{n}}\right\rangle \) by c... | No |
Theorem 4.59. If \( {h}^{ * } \) is an unreduced cohomology theory on the category of \( {CW} \) pairs and \( {h}^{n}\left( \text{point}\right) = 0 \) for \( n \neq 0 \), then there are natural isomorphisms \( {h}^{n}\left( {X, A}\right) \approx \) \( {H}^{n}\left( {X, A;{h}^{0}\left( \text{point}\right) }\right) \) fo... | Proof: The case of homology is slightly simpler, so let us consider this first. For CW complexes, relative homology groups reduce to absolute groups, so it suffices to deal with the latter. For a CW complex \( X \) the long exact sequences of \( {h}_{ * } \) homology groups for the pairs \( \left( {{X}^{n},{X}^{n - 1}}... | No |
For a fibration \( p : E \rightarrow B \), the fibers \( {F}_{b} = {p}^{-1}\left( b\right) \) over each path component of \( B \) are all homotopy equivalent. | Proof: A path \( y : I \rightarrow B \) gives rise to a homotopy \( {g}_{t} : {F}_{y\left( 0\right) } \rightarrow B \) with \( {g}_{t}\left( {F}_{y\left( 0\right) }\right) = y\left( t\right) \) . The inclusion \( {F}_{y\left( 0\right) } \hookrightarrow E \) provides a lift \( {\widetilde{g}}_{0} \), so by the homotopy ... | Yes |
Corollary 4.63. A fibration \( E \rightarrow B \) over a contractible base \( B \) is fiber homotopy equiv- alent to a product fibration \( B \times F \rightarrow B \) . | Proof: The pullback of \( E \) by the identity map \( B \rightarrow B \) is \( E \) itself, while the pullback by a constant map \( B \rightarrow B \) is a product \( B \times F \) . | No |
Proposition 4.64. The map \( p : {E}_{f} \rightarrow B, p\left( {a,\gamma }\right) = \gamma \left( 1\right) \), is a fibration. | Proof: Continuity of \( p \) follows from (a) of Proposition A. 14 in the Appendix which says that the evaluation map \( {B}^{I} \times I \rightarrow B,\left( {y, s}\right) \mapsto y\left( s\right) \), is continuous.\n\nTo verify the fibration property, let a homotopy \( {g}_{t} : X \rightarrow B \) and a lift \( {\wid... | Yes |
If \( p : E \rightarrow B \) is a fibration, then the inclusion \( E \hookrightarrow {E}_{p} \) is a fiber homotopy equivalence. In particular, the homotopy fibers of \( p \) are homotopy equivalent to the actual fibers. | Proof: We apply the homotopy lifting property to the homotopy \( {g}_{t} : {E}_{p} \rightarrow B,{g}_{t}\left( {e, y}\right) = \) \( y\left( t\right) \), with initial lift \( {\widetilde{g}}_{0} : {E}_{p} \rightarrow E,{\widetilde{g}}_{0}\left( {e, y}\right) = e \) . The lifting \( {\widetilde{g}}_{t} : {E}_{p} \righta... | Yes |
Proposition 4.66. If \( F \rightarrow E \rightarrow B \) is a fibration or fiber bundle with \( E \) contractible, then there is a weak homotopy equivalence \( F \rightarrow {\Omega B} \) . | Proof: If we compose a contraction of \( E \) with the projection \( p : E \rightarrow B \) then we have for each point \( x \in E \) a path \( {y}_{x} \) in \( B \) from \( p\left( x\right) \) to a basepoint \( {b}_{0} = p\left( {x}_{0}\right) \), where \( {x}_{0} \) is the point to which \( E \) contracts. This yield... | Yes |
For the Postnikov tower of a connected CW complex \( X \) the natural map \( X \rightarrow \lim {X}_{n} \) is a weak homotopy equivalence, so \( X \) is a CW approximation to \( \mathop{\lim }\limits_{ \leftarrow }{X}_{n} \) | The composition \( {\pi }_{i}\left( X\right) \rightarrow {\pi }_{i}\left( {\mathop{\lim }\limits_{ \rightarrow }{X}_{n}}\right) \overset{\lambda }{ \rightarrow }\mathop{\lim }\limits_{ \leftarrow }{\pi }_{i}\left( {X}_{n}\right) \) is an isomorphism since \( {\pi }_{i}\left( X\right) \rightarrow {\pi }_{i}\left( {X}_{n... | Yes |
Lemma 4.70. Let \( \left( {X, A}\right) \) be a CW pair with both \( X \) and \( A \) connected, such that the homotopy fiber of the inclusion \( A \hookrightarrow X \) is a \( K\left( {\pi, n}\right), n \geq 1 \) . Then there exists a fibration \( F \rightarrow E \rightarrow B \) and a map \( \left( {X, A}\right) \rig... | Proof: It remains only to prove the ’if’ implication. As we noted just before the statement of the lemma, the groups \( {\pi }_{i}\left( {X, A}\right) \) are zero except for \( {\pi }_{n + 1}\left( {X, A}\right) \approx \pi \) . If the action of \( {\pi }_{1}\left( A\right) \) on \( {\pi }_{n + 1}\left( {X, A}\right) \... | Yes |
Theorem 4.71. Every map \( f : X \rightarrow Y \) between connected CW complexes has a Moore-Postnikov tower, which is unique up to homotopy equivalence. A Moore-Postnikov tower of principal fibrations exists iff \( {\pi }_{1}\left( X\right) \) acts trivially on \( {\pi }_{n}\left( {{M}_{f}, X}\right) \) for all \( n >... | Proof: The existence and uniqueness of a diagram satisfying (1) and (2) and commutative at least up to homotopy follows from Propositions 4.13 and 4.18 applied to the pair \( \left( {{M}_{f}, X}\right) \) with \( {M}_{f} \) the mapping cylinder of \( f \) . Having such a diagram, we proceed as in the earlier case of Po... | Yes |
Consider the experiment of rolling a pair of 4-sided dice (cf. Fig. 1.4). We assume the dice are fair, and we interpret this assumption to mean that each of the sixteen possible outcomes [ordered pairs \( \left( {i, j}\right) \), with \( i, j = 1,2,3,4 \) ], has the same probability of \( 1/{16} \) . To calculate the p... | \[ \mathbf{P}\left( {\{ \text{the sum of the rolls is even}\} }\right) = 8/{16} = 1/2\text{,} \] \[ \mathbf{P}\left( {\{ \text{the sum of the rolls is odd}\} }\right) = 8/{16} = 1/2\text{,} \] \( \mathbf{P}\left( {\{ \text{the first roll is equal to the second}\} }\right) = 4/{16} = 1/4 \) \( \mathbf{P}\left( {\{ \text... | Yes |
A wheel of fortune is continuously calibrated from 0 to 1, so the possible outcomes of an experiment consisting of a single spin are the numbers in the interval \( \Omega = \left\lbrack {0,1}\right\rbrack \) . Assuming a fair wheel, it is appropriate to consider all outcomes equally likely, but what is the probability ... | Therefore, the probability of any event that consists of a single element must be 0. | Yes |
Romeo and Juliet have a date at a given time, and each will arrive at the meeting place with a delay between 0 and 1 hour, with all pairs of delays being equally likely. The first to arrive will wait for 15 minutes and will leave if the other has not yet arrived. What is the probability that they will meet? | Let us use as sample space the square \( \Omega = \left\lbrack {0,1}\right\rbrack \times \left\lbrack {0,1}\right\rbrack \), whose elements are the possible pairs of delays for the two of them. Our interpretation of \ | No |
We toss a fair coin three successive times. We wish to find the conditional probability \( \mathbf{P}\left( {A \mid B}\right) \) when \( A \) and \( B \) are the events\n\n\[ A = \{ \text{more heads than tails come up}\} ,\;B = \{ \text{1st toss is a head}\} \text{.} \]\n\nThe sample space consists of eight sequences,\... | Thus, the conditional probability \( \mathbf{P}\left( {A \mid B}\right) \) is\n\n\[ \mathbf{P}\left( {A \mid B}\right) = \frac{\mathbf{P}\left( {A \cap B}\right) }{\mathbf{P}\left( B\right) } = \frac{3/8}{4/8} = \frac{3}{4}. \] | Yes |
Example 1.7. A fair 4-sided die is rolled twice and we assume that all sixteen possible outcomes are equally likely. Let \( X \) and \( Y \) be the result of the 1st and the 2nd roll, respectively. We wish to determine the conditional probability \( \mathbf{P}\left( {A \mid B}\right) \) where\n\n\[ A = \{ \max \left( {... | As in the preceding example, we can first determine the probabilities \( \mathbf{P}\left( {A \cap B}\right) \) and \( \mathbf{P}\left( B\right) \) by counting the number of elements of \( A \cap B \) and \( B \), respectively, and dividing by 16. Alternatively, we can directly divide the number of elements of \( A \cap... | No |
If both teams are successful, the design of team N is adopted. Assuming that exactly one successful design is produced, what is the probability that it was designed by team \( \mathrm{N} \) ? | There are four possible outcomes here, corresponding to the four combinations of success and failure of the two teams:\n\n\( {SS} \) : both succeed, \( {FF} \) : both fail,\n\n\( {SF} \) : C succeeds, \( \mathrm{N} \) fails, \( {FS} \) : C fails, \( \mathrm{N} \) succeeds.\n\nWe are given that the probabilities of thes... | Yes |
If an aircraft is present in a certain area, a radar correctly registers its presence with probability 0.99 . If it is not present, the radar falsely registers an aircraft presence with probability 0.10 . We assume that an aircraft is present with probability 0.05 . What is the probability of false alarm (a false indic... | A sequential representation of the sample space is appropriate here, as shown in Fig. 1.8. Let \( A \) and \( B \) be the events\n\n\[ A = \{ \\text{ an aircraft is present }\} \]\n\n\[ B = \{ \\text{the radar registers an aircraft presence}\} \\text{,}\]\n\nand consider also their complements\n\n\[ {A}^{c} = \{ \\text... | Yes |
Three cards are drawn from an ordinary 52-card deck without replacement (drawn cards are not placed back in the deck). We wish to find the probability that none of the three cards is a heart. | Define the events\n\n\[ \n{A}_{i} = \{ \text{the}i\text{th card is not a heart}\} ,\;i = 1,2,3\text{.} \n\]\n\nWe will calculate \( \mathbf{P}\left( {{A}_{1} \cap {A}_{2} \cap {A}_{3}}\right) \), the probability that none of the three cards is a heart, using the multiplication rule,\n\n\[ \n\mathbf{P}\left( {{A}_{1} \c... | Yes |
A class consisting of 4 graduate and 12 undergraduate students is randomly divided into 4 groups of 4 . What is the probability that each group includes a graduate student? | We interpret randomly to mean that given the assignment of some students to certain slots, any of the remaining students is equally likely to be assigned to any of the remaining slots. We then calculate the desired probability using the multiplication rule, based on the sequential description shown in Fig. 1.11. Let us... | Yes |
You enter a chess tournament where your probability of winning a game is 0.3 against half the players (call them type 1), 0.4 against a quarter of the players (call them type 2), and 0.5 against the remaining quarter of the players (call them type 3). You play a game against a randomly chosen opponent. What is the prob... | Let \( {A}_{i} \) be the event of playing with an opponent of type \( i \) . We have\n\n\[ \mathbf{P}\left( {A}_{1}\right) = {0.5},\;\mathbf{P}\left( {A}_{2}\right) = {0.25},\;\mathbf{P}\left( {A}_{3}\right) = {0.25}. \]\n\nLet also \( B \) be the event of winning. We have\n\n\[ \mathbf{P}\left( {B \mid {A}_{1}}\right)... | Yes |
We roll a fair four-sided die. If the result is 1 or 2, we roll once more but otherwise, we stop. What is the probability that the sum total of our rolls is at least 4 ? | Let \( {A}_{i} \) be the event that the result of first roll is \( i \), and note that \( \mathbf{P}\left( {A}_{i}\right) = 1/4 \) for each \( i \) . Let \( B \) be the event that the sum total is at least 4 . Given the event \( {A}_{1} \) , the sum total will be at least 4 if the second roll results in 3 or 4 , which ... | Yes |
Let us return to the radar detection problem of Example 1.9 and Fig. 1.8. Let\n\n\[ A = \{ \\text{ an aircraft is present }\} \]\n\n\\( B = \{ \\text{ the radar registers an aircraft presence } \} \\) .\n\nWe are given that\n\n\[ \\mathbf{P}\\left( A\\right) = {0.05},\\;\\mathbf{P}\\left( {B \\mid A}\\right) = {0.99},\... | \n\[ = \\frac{\\mathbf{P}\\left( A\\right) \\mathbf{P}\\left( {B \\mid A}\\right) }{\\mathbf{P}\\left( B\\right) } \]\n\n\[ = \\frac{\\mathbf{P}\\left( A\\right) \\mathbf{P}\\left( {B \\mid A}\\right) }{\\mathbf{P}\\left( A\\right) \\mathbf{P}\\left( {B \\mid A}\\right) + \\mathbf{P}\\left( {A}^{c}\\right) \\mathbf{P}\... | Yes |
Suppose that you win. What is the probability \( \mathbf{P}\left( {{A}_{1} \mid B}\right) \) that you had an opponent of type 1 ? | Using Bayes' rule, we have\n\n\[ \mathbf{P}\left( {{A}_{1} \mid B}\right) = \frac{\mathbf{P}\left( {A}_{1}\right) \mathbf{P}\left( {B \mid {A}_{1}}\right) }{\mathbf{P}\left( {A}_{1}\right) \mathbf{P}\left( {B \mid {A}_{1}}\right) + \mathbf{P}\left( {A}_{2}\right) \mathbf{P}\left( {B \mid {A}_{2}}\right) + \mathbf{P}\le... | Yes |
Consider an experiment involving two successive rolls of a 4-sided die in which all 16 possible outcomes are equally likely and have probability \( 1/{16} \). Are the events \( {A}_{i} = \{ 1\text{st roll results in}i\} \) and \( {B}_{j} = \{ 2\text{nd roll results in}j\} \) independent? | We have \( \mathbf{P}\left( {A \cap B}\right) = \mathbf{P}\left( {\text{ the result of the two rolls is }\left( {i, j}\right) }\right) = \frac{1}{16} \), \( \mathbf{P}\left( {A}_{i}\right) = \frac{\text{ number of elements of }{A}_{i}}{\text{ total number of possible outcomes }} = \frac{4}{16} \), and \( \mathbf{P}\lef... | Yes |
Consider two independent fair coin tosses, in which all four possible outcomes are equally likely. Let\n\n\[ \n{H}_{1} = \{ 1\text{st toss is a head}\} \text{,}\n\]\n\n\[ \n{H}_{2} = \{ 2\mathrm{{nd}}\text{toss is a head}\} \n\]\n\n\[ \nD = \{ \text{the two tosses have different results}\} \text{.} \n\]\n\nThe events \... | \[ \n\mathbf{P}\left( {{H}_{1} \mid D}\right) = \frac{1}{2},\;\mathbf{P}\left( {{H}_{2} \mid D}\right) = \frac{1}{2},\;\mathbf{P}\left( {{H}_{1} \cap {H}_{2} \mid D}\right) = 0, \n\]\n\nso that \( \mathbf{P}\left( {{H}_{1} \cap {H}_{2} \mid D}\right) \neq \mathbf{P}\left( {{H}_{1} \mid D}\right) \mathbf{P}\left( {{H}_{... | Yes |
There are two coins, a blue and a red one. We choose one of the two at random, each being chosen with probability \( 1/2 \), and proceed with two independent tosses. The coins are biased: with the blue coin, the probability of heads in any given toss is 0.99 , whereas for the red coin it is 0.01 . Let \( B \) be the ev... | \[ \mathbf{P}\left( {{H}_{1} \cap {H}_{2} \mid B}\right) = \mathbf{P}\left( {{H}_{1} \mid B}\right) \mathbf{P}\left( {{H}_{2} \mid B}\right) = {0.99} \cdot {0.99}. \] On the other hand, the events \( {H}_{1} \) and \( {H}_{2} \) are not independent. Intuitively, if we are told that the first toss resulted in heads, thi... | Yes |
Pairwise independence does not imply independence. Consider two independent fair coin tosses, and the following events:\n\n\[ \n{H}_{1} = \{ 1\text{st toss is a head}\} \n\]\n\n\[ \n{H}_{2} = \{ 2\mathrm{{nd}}\text{toss is a head}\} \n\]\n\n\[ \nD = \{ \text{the two tosses have different results}\} \text{.} \n\] | The events \( {H}_{1} \) and \( {H}_{2} \) are independent, by definition. To see that \( {H}_{1} \) and \( D \) are independent, we note that\n\n\[ \n\mathbf{P}\left( {D \mid {H}_{1}}\right) = \frac{\mathbf{P}\left( {{H}_{1} \cap D}\right) }{\mathbf{P}\left( {H}_{1}\right) } = \frac{1/4}{1/2} = \frac{1}{2} = \mathbf{P... | Yes |
The equality \( \mathbf{P}\left( {{A}_{1} \cap {A}_{2} \cap {A}_{3}}\right) = \mathbf{P}\left( {A}_{1}\right) \mathbf{P}\left( {A}_{2}\right) \mathbf{P}\left( {A}_{3}\right) \) is not enough for independence. Consider two independent rolls of a fair die, and the following events:\n\n\[ A = \{ 1\text{st roll is}1,2\text... | The intuition behind the independence of a collection of events is analogous to the case of two events. Independence means that the occurrence or non-occurrence of any number of the events from that collection carries no information on the remaining events or their complements. For example, if the events \( {A}_{1},{A}... | No |
What is the probability that there is a path connecting A and B in which all links are up? | Returning now to the network of Fig. 1.14(a), we can calculate the probability of success (a path from A to B is available) sequentially, using the preceding formulas, and starting from the end. Let us use the notation \( X \rightarrow Y \) to denote the event that there is a (possibly indirect) connection from node \(... | Yes |
Example 1.23. Grade of service. An internet service provider has installed \( c \) modems to serve the needs of a population of \( n \) customers. It is estimated that at a given time, each customer will need a connection with probability \( p \), independently of the others. What is the probability that there are more... | Here we are interested in the probability that more than \( c \) customers simultaneously need a connection. It is equal to\n\n\[ \mathop{\sum }\limits_{{k = c + 1}}^{n}p\left( k\right) \]\n\nwhere\n\n\[ p\left( k\right) = \left( \begin{array}{l} n \\ k \end{array}\right) {p}^{k}{\left( 1 - p\right) }^{n - k} \]\n\nare... | Yes |
Example 1.24. The number of telephone numbers. A telephone number is a 7-digit sequence, but the first digit has to be different from 0 or 1 . How many distinct telephone numbers are there? | We can visualize the choice of a sequence as a sequential process, where we select one digit at a time. We have a total of 7 stages, and a choice of one out of 10 elements at each stage, except for the first stage where we only have 8 choices. Therefore, the answer is\n\n\[ \n8 \cdot \underset{6\text{ times }}{\underbr... | Yes |
Example 1.25. The number of subsets of an \( n \) -element set. Consider an \( n \) -element set \( \left\{ {{s}_{1},{s}_{2},\ldots ,{s}_{n}}\right\} \) . How many subsets does it have (including itself and the empty set)? | We can visualize the choice of a subset as a sequential process where we examine one element at a time and decide whether to include it in the set or not. We have a total of \( n \) stages, and a binary choice at each stage. Therefore the number of subsets is\n\n\[ \underset{n\text{ times }}{\underbrace{2 \cdot 2\cdots... | Yes |
Let us count the number of words that consist of four distinct letters. This is the problem of counting the number of 4-permutations of the 26 letters in the alphabet. | The desired number is\n\n\[ \frac{n!}{\left( {n - k}\right) !} = \frac{{26}!}{{22}!} = {26} \cdot {25} \cdot {24} \cdot {23} = {358},{800}. \] | Yes |
You have \( {n}_{1} \) classical music CDs, \( {n}_{2} \) rock music CDs, and \( {n}_{3} \) country music CDs. In how many different ways can you arrange them so that the CDs of the same type are contiguous? | We break down the problem in two stages, where we first select the order of the CD types, and then the order of the CDs of each type. There are 3 ! ordered sequences of the types of CDs (such as classical/rock/country, rock/country/classical, etc), and there are \( {n}_{1} \) ! (or \( {n}_{2} \) !, or \( {n}_{3} \) !) ... | Yes |
The number of combinations of two out of the four letters A, B, C, and D is found by letting \( n = 4 \) and \( k = 2 \) . | \[ \left( \begin{array}{l} 4 \\ 2 \end{array}\right) = \frac{4!}{2!2!} = 6 \] | Yes |
How many different letter sequences can be obtained by rearranging the letters in the word TATTOO? | There are six positions to be filled by the available letters. Each rearrangement corresponds to a partition of the set of the six positions into a group of size 3 (the positions that get the letter T), a group of size 1 (the position that gets the letter A), and a group of size 2 (the positions that get the letter O).... | Yes |
A class consisting of 4 graduate and 12 undergraduate students is randomly divided into four groups of 4 . What is the probability that each group includes a graduate student? | We first determine the nature of the sample space. A typical outcome is a particular way of partitioning the 16 students into four groups of 4 . We take the term \ | No |
Let \( Y = \left| X\right| \) and let us apply the preceding formula for the PMF \( {p}_{Y} \) to the case where\n\n\[ \n{p}_{X}\left( x\right) = \left\{ \begin{array}{ll} 1/9 & \text{ if }x\text{ is an integer in the range }\left\lbrack {-4,4}\right\rbrack \\ 0 & \text{ otherwise. } \end{array}\right.\n\]\n\nThe possi... | To compute \( {p}_{Y}\left( y\right) \) for some given value \( y \) from this range, we must add \( {p}_{X}\left( x\right) \) over all values \( x \) such that \( \left| x\right| = y \) . In particular, there is only one value of \( X \) that corresponds to \( y = 0 \), namely \( x = 0 \) . Thus,\n\n\[ \n{p}_{Y}\left(... | Yes |
Consider two independent coin tosses, each with a \( 3/4 \) probability of a head, and let \( X \) be the number of heads obtained. This is a binomial random variable with parameters \( n = 2 \) and \( p = 3/4 \) . Its PMF is\n\n\[ \n{p}_{X}\left( k\right) = \left\{ \begin{array}{ll} {\left( 1/4\right) }^{2} & \text{ i... | so the mean is\n\n\[ \n\mathbf{E}\left\lbrack X\right\rbrack = 0 \cdot {\left( \frac{1}{4}\right) }^{2} + 1 \cdot \left( {2 \cdot \frac{1}{4} \cdot \frac{3}{4}}\right) + 2 \cdot {\left( \frac{3}{4}\right) }^{2} = \frac{24}{16} = \frac{3}{2}. \n\] | Yes |
Consider the random variable \( X \) of Example 2.1, which has the PMF\n\n\[ \n{p}_{X}\left( x\right) = \left\{ \begin{array}{ll} 1/9 & \text{ if }x\text{ is an integer in the range }\left\lbrack {-4,4}\right\rbrack , \\ 0 & \text{ otherwise. } \end{array}\right. \]\n\nThe mean \( \mathbf{E}\left\lbrack X\right\rbrack ... | This can be seen from the symmetry of the PMF of \( X \) around 0, and can also be verified from the definition:\n\n\[ \n\mathbf{E}\left\lbrack X\right\rbrack = \mathop{\sum }\limits_{x}x{p}_{X}\left( x\right) = \frac{1}{9}\mathop{\sum }\limits_{{x = - 4}}^{4}x = 0. \]\n | Yes |
For the random variable \( X \) with PMF\n\n\[ \n{p}_{X}\left( x\right) = \left\{ \begin{array}{ll} 1/9 & \text{ if }x\text{ is an integer in the range }\left\lbrack {-4,4}\right\rbrack , \\ 0 & \text{ otherwise,} \end{array}\right.\n\]\n\nwe have\n\n\[ \n\operatorname{var}\left( X\right) = \mathbf{E}\left\lbrack {\lef... | \[ \n= \mathop{\sum }\limits_{x}{\left( x - \mathbf{E}\left\lbrack X\right\rbrack \right) }^{2}{p}_{X}\left( x\right) \n\]\n\n\[ \n= \frac{1}{9}\mathop{\sum }\limits_{{x = - 4}}^{4}{x}^{2}\;\text{ since }\mathbf{E}\left\lbrack X\right\rbrack = 0 \n\]\n\n\[ \n= \frac{1}{9}\left( {{16} + 9 + 4 + 1 + 0 + 1 + 4 + 9 + {16}}... | Yes |
Consider the experiment of tossing a biased coin, which comes up a head with probability \( p \) and a tail with probability \( 1 - p \), and the Bernoulli random variable \( X \) with PMF\n\n\[ \n{p}_{X}\left( k\right) = \left\{ \begin{array}{ll} p & \text{ if }k = 1 \\ 1 - p & \text{ if }k = 0 \end{array}\right.\n\] | Its mean, second moment, and variance are given by the following calculations:\n\n\[ \n\mathbf{E}\left\lbrack X\right\rbrack = 1 \cdot p + 0 \cdot \left( {1 - p}\right) = p \n\]\n\n\[ \n\mathbf{E}\left\lbrack {X}^{2}\right\rbrack = {1}^{2} \cdot p + 0 \cdot \left( {1 - p}\right) = p, \n\]\n\n\[ \n\operatorname{var}\lef... | Yes |
What is the mean and variance of the roll of a fair six-sided die? | If we view the result of the roll as a random variable \( X \), its PMF is\n\n\[ \n{p}_{X}\left( k\right) = \left\{ \begin{array}{ll} 1/6 & \text{ if }k = 1,2,3,4,5,6 \\ 0 & \text{ otherwise. } \end{array}\right.\n\]\n\nSince the PMF is symmetric around 3.5, we conclude that \( \mathbf{E}\left\lbrack X\right\rbrack = {... | Yes |
The Mean of the Poisson. The mean of the Poisson PMF\n\n\[ \n{p}_{X}\left( k\right) = {e}^{-\lambda }\frac{{\lambda }^{k}}{k!},\;k = 0,1,2,\ldots ,\n\] | can be calculated is follows:\n\n\[ \nE\left\lbrack X\right\rbrack = \mathop{\sum }\limits_{{k = 0}}^{\infty }k{e}^{-\lambda }\frac{{\lambda }^{k}}{k!}\n\]\n\n\[ \n= \mathop{\sum }\limits_{{k = 1}}^{\infty }k{e}^{-\lambda }\frac{{\lambda }^{k}}{k!}\;\text{ the }k = 0\text{ term is zero }\n\]\n\n\[ \n= \lambda \mathop{\... | Yes |
Consider a quiz game where a person is given two questions and must decide which question to answer first. Question 1 will be answered correctly with probability 0.8, and the person will then receive as prize $100, while question 2 will be answered correctly with probability 0.5, and the person will then receive as pri... | (a) Answer question 1 first: Then the PMF of \( X \) is (cf. the left side of Fig. 2.10)\n\n\[ \n{p}_{X}\left( 0\right) = {0.2},\;{p}_{X}\left( {100}\right) = {0.8} \cdot {0.5},\;{p}_{X}\left( {300}\right) = {0.8} \cdot {0.5}, \n\]\n\nand we have\n\n\[ \n\mathbf{E}\left\lbrack X\right\rbrack = {0.8} \cdot {0.5} \cdot {... | Yes |
If the weather is good (which happens with probability 0.6 ), Alice walks the 2 miles to class at a speed of \( V = 5 \) miles per hour, and otherwise drives her motorcycle at a speed of \( V = {30} \) miles per hour. What is the mean of the time \( T \) to get to class? | The correct way to solve the problem is to first derive the PMF of \( T \) ,\n\n\[ \n{p}_{T}\left( t\right) = \left\{ \begin{array}{ll} {0.6} & \text{ if }t = 2/5\text{ hours,} \\ {0.4} & \text{ if }t = 2/{30}\text{ hours,} \end{array}\right. \n\]\n\nand then calculate its mean by\n\n\[ \n\mathbf{E}\left\lbrack T\right... | Yes |
Your probability class has 300 students and each student has probability \( 1/3 \) of getting an A, independently of any other student. What is the mean of \( X \), the number of students that get an A? | Let\n\n\[ \n{X}_{i} = \left\{ \begin{array}{ll} 1 & \text{ if the }i\text{ th student gets an }\mathrm{A}, \\ 0 & \text{ otherwise. } \end{array}\right.\n\]\n\nThus \( {X}_{1},{X}_{2},\ldots ,{X}_{n} \) are Bernoulli random variables with common mean \( p = 1/3 \) and variance \( p\left( {1 - p}\right) = \left( {1/3}\r... | No |
Example 2.10. The Hat Problem. Suppose that \( n \) people throw their hats in a box and then each picks up one hat at random. What is the expected value of \( X \), the number of people that get back their own hat? | For the \( i \) th person, we introduce a random variable \( {X}_{i} \) that takes the value 1 if the person selects his/her own hat, and takes the value 0 otherwise. Since \( \mathbf{P}\left( {{X}_{i} = 1}\right) = 1/n \) and \( \mathbf{P}\left( {{X}_{i} = 0}\right) = 1 - 1/n \), the mean of \( {X}_{i} \) is\n\n\[ \ma... | Yes |
Example 2.11. Professor May B. Right often has her facts wrong, and answers each of her students’ questions incorrectly with probability \( 1/4 \), independently of other questions. In each lecture May is asked 0,1 , or 2 questions with equal probability \( 1/3 \) . Let \( X \) and \( Y \) be the number of questions Ma... | This can be done by using a sequential description of the experiment and the multiplication rule \( {p}_{X, Y}\left( {x, y}\right) = {p}_{Y}\left( y\right) {p}_{X \mid Y}\left( {x \mid y}\right) \) , as shown in Fig. 2.14. For example, for the case where one question is asked and is answered wrong, we have\n\n\[ \n{p}_... | Yes |
Consider four independent rolls of a 6-sided die. Let \( X \) be the number of 1’s and let \( Y \) be the number of 2’s obtained. What is the joint PMF of \( X \) and \( Y \) ? | The marginal PMF \( {p}_{Y} \) is given by the binomial formula\n\n\[ \n{p}_{Y}\left( y\right) = \left( \begin{array}{l} 4 \\ y \end{array}\right) {\left( \frac{1}{6}\right) }^{y}{\left( \frac{5}{6}\right) }^{4 - y},\;y = 0,1,\ldots ,4. \]\n\nTo compute the conditional PMF \( {p}_{X \mid Y} \), note that given that \( ... | Yes |
Consider a transmitter that is sending messages over a computer network. Let us define the following two random variables:\n\n\\( X \\) : the travel time of a given message, \\( Y \\) : the length of the given message.\n\nWe know the PMF of the travel time of a message that has a given length, and we know the PMF of th... | We assume that the length of a message can take two possible values: \\( y = {10}^{2} \\) bytes with probability \\( 5/6 \\), and \\( y = {10}^{4} \\) bytes with probability \\( 1/6 \\), so that\n\n\\[ \n{p}_{Y}\\left( y\\right) = \\left\\{ \\begin{array}{ll} 5/6 & \\text{ if }y = {10}^{2} \\\\ 1/6 & \\text{ if }y = {1... | Yes |
Messages transmitted by a computer in Boston through a data network are destined for New York with probability 0.5, for Chicago with probability 0.3, and for San Francisco with probability 0.2. The transit time \( X \) of a message is random. Its mean is 0.05 secs if it is destined for New York, 0.1 secs if it is desti... | \[ \mathbf{E}\left\lbrack X\right\rbrack = {0.5} \cdot {0.05} + {0.3} \cdot {0.1} + {0.2} \cdot {0.3} = {0.115}\text{secs.} \] | Yes |
What is the mean and variance of \( X \), the number of tries until the program works correctly? | We recognize \( X \) as a geometric random variable with PMF\n\n\[ \n{p}_{X}\left( k\right) = {\left( 1 - p\right) }^{k - 1}p,\;k = 1,2,\ldots \n\]\n\nThe mean and variance of \( X \) are given by\n\n\[ \n\mathbf{E}\left\lbrack X\right\rbrack = \mathop{\sum }\limits_{{k = 1}}^{\infty }k{\left( 1 - p\right) }^{k - 1}p,\... | Yes |
Consider two independent tosses of a fair coin. Let \( X \) be the number of heads and let \( A \) be the event that the number of heads is even. The (unconditional) PMF of \( X \) is\n\n\[ \n{p}_{X}\left( x\right) = \left\{ \begin{array}{ll} 1/4 & \text{ if }x = 0 \\ 1/2 & \text{ if }x = 1 \\ 1/4 & \text{ if }x = 2 \e... | The conditional PMF is obtained from the definition \( {p}_{X \mid A}\left( x\right) = \) \( \mathbf{P}\left( {X = x\text{ and }A}\right) /\mathbf{P}\left( A\right) \) :\n\n\[ \n{p}_{X \mid A}\left( x\right) = \left\{ \begin{array}{ll} 1/2 & \text{ if }x = 0 \\ 0 & \text{ if }x = 1 \\ 1/2 & \text{ if }x = 2 \end{array}... | Yes |
Example 2.17. Variance of the Binomial. We consider \( n \) independent coin tosses, with each toss having probability \( p \) of coming up a head. For each \( i \), we let \( {X}_{i} \) be the Bernoulli random variable which is equal to 1 if the \( i \) th toss comes up a head, and is 0 otherwise. Then, \( X = {X}_{1}... | \[ \operatorname{var}\left( X\right) = \mathop{\sum }\limits_{{i = 1}}^{n}\operatorname{var}\left( {X}_{i}\right) = {np}\left( {1 - p}\right) \] | Yes |
Example 2.19. Estimating Probabilities by Simulation. In many practical situations, the analytical calculation of the probability of some event of interest is very difficult. However, if we have a physical or computer model that can generate outcomes of a given experiment in accordance with their true probabilities, we... | To see how accurate this process is, consider \( n \) independent Bernoulli random variables \( {X}_{1},\ldots ,{X}_{n} \), each with PMF\n\n\[ \n{p}_{{X}_{i}}\left( {x}_{i}\right) = \left\{ \begin{array}{ll} \mathbf{P}\left( A\right) & \text{ if }{x}_{i} = 1 \\ 0 & \text{ if }{x}_{i} = 0 \end{array}\right.\n\]\n\nIn a... | Yes |
A gambler spins a wheel of fortune, continuously calibrated between 0 and 1 , and observes the resulting number. Assuming that all subintervals of \( \\left\\lbrack {0,1}\\right\\rbrack \) of the same length are equally likely, this experiment can be modeled in terms a random variable \( X \) with PDF\n\n\[ \n{f}_{X}\\... | This constant can be determined by using the normalization property\n\n\[ \n1 = {\\int }_{-\\infty }^{\\infty }{f}_{X}\\left( x\\right) {dx} = {\\int }_{0}^{1}{cdx} = c{\\int }_{0}^{1}{dx} = c \n\]\n\nso that \( c = 1 \) . | Yes |
Alvin's driving time to work is between 15 and 20 minutes if the day is sunny, and between 20 and 25 minutes if the day is rainy, with all times being equally likely in each case. Assume that a day is sunny with probability \( 2/3 \) and rainy with probability \( 1/3 \) . What is the PDF of the driving time, viewed as ... | We interpret the statement that \ | No |
A PDF can be arbitrarily large. Consider a random variable \( X \) with PDF \[ {f}_{X}\left( x\right) = \left\{ \begin{array}{ll} \frac{1}{2\sqrt{x}} & \text{ if }0 < x \leq 1 \\ 0 & \text{ otherwise. } \end{array}\right. \] | Even though \( {f}_{X}\left( x\right) \) becomes infinitely large as \( x \) approaches zero, this is still a valid PDF, because \[ {\int }_{-\infty }^{\infty }{f}_{X}\left( x\right) {dx} = {\int }_{0}^{1}\frac{1}{2\sqrt{x}}{dx} = {\left. \sqrt{x}\right| }_{0}^{1} = 1. \] | Yes |
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