Math 8301, Manifolds and Topology Homework 3
... 2. Using the previous exercise, show that if K : [0, 1] × [0, 1] → X is a continuous map, and we define α(t) = K(t, 0), γ(t) = K(0, t), ...
... 2. Using the previous exercise, show that if K : [0, 1] × [0, 1] → X is a continuous map, and we define α(t) = K(t, 0), γ(t) = K(0, t), ...
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... based spaces. Note that any cover can be made into a based cover by choosing a basepoint from the pre-images of x. The universal covering space has the following universal property: If π : (X̃, x0 ) → (X, x) is a based universal cover, then for any connected based cover π 0 : (X 0 , x0 ) → (X, x), t ...
... based spaces. Note that any cover can be made into a based cover by choosing a basepoint from the pre-images of x. The universal covering space has the following universal property: If π : (X̃, x0 ) → (X, x) is a based universal cover, then for any connected based cover π 0 : (X 0 , x0 ) → (X, x), t ...
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... 1. For example, any topological space is trivially an ordered space, with the partial order defined by a ≤ b iff a = b. But this is not so interesting. A more interesting example is to take a T0 space X, and define a ≤ b iff a ∈ {b}. The relation so defined turns out to be a partial order on X, call ...
... 1. For example, any topological space is trivially an ordered space, with the partial order defined by a ≤ b iff a = b. But this is not so interesting. A more interesting example is to take a T0 space X, and define a ≤ b iff a ∈ {b}. The relation so defined turns out to be a partial order on X, call ...
ON ULTRACONNECTED SPACES
... Any compact, maximal F-connected topology on a set X is of the form {,X} II where II is an ultrafilter on X{a}, for some a e X. PROOF. Let (X, z) be compact and maximal F-connected. Since the family of all the nonempty closed sets has finite intersection property and (x,z) is compact, it has is clos ...
... Any compact, maximal F-connected topology on a set X is of the form {,X} II where II is an ultrafilter on X{a}, for some a e X. PROOF. Let (X, z) be compact and maximal F-connected. Since the family of all the nonempty closed sets has finite intersection property and (x,z) is compact, it has is clos ...
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... The collection of sets can be arbitrary, that is, I can be finite, countable, or uncountable. The cover is correspondingly called a finite cover, countable cover, or uncountable cover. A subcover of U is a subset U 0 ⊂ U such that U 0 is also a cover of X. A refinement V of U is a cover of X such th ...
... The collection of sets can be arbitrary, that is, I can be finite, countable, or uncountable. The cover is correspondingly called a finite cover, countable cover, or uncountable cover. A subcover of U is a subset U 0 ⊂ U such that U 0 is also a cover of X. A refinement V of U is a cover of X such th ...
On finite $ T_0 $
... that throws (|, £) onto p'; just retract [0, -J) onto [0, J ] . Now define a function / : [0, 1] -> X b y / | [0, -J) = / i , / ( [ i , 1]) = b. Now the minimal open set containing b in X is Ut say. / - 1 ( U i ) = f~x(b) uf~1(U1 — b) which is open. So / is a map. Verification of the inductive hypot ...
... that throws (|, £) onto p'; just retract [0, -J) onto [0, J ] . Now define a function / : [0, 1] -> X b y / | [0, -J) = / i , / ( [ i , 1]) = b. Now the minimal open set containing b in X is Ut say. / - 1 ( U i ) = f~x(b) uf~1(U1 — b) which is open. So / is a map. Verification of the inductive hypot ...
Homework Set 1
... Remark. A morphism of schemes f : Y → X is an open immersion if it factors as iu ◦ g, for some U as above and an isomorphism of schemes g : Y → U . Problem 2. S Let f : Y → X be a morphism of schemes. Show that if there is an open cover X = i Ui such that the induced morphisms f −1 (Ui ) → Ui are is ...
... Remark. A morphism of schemes f : Y → X is an open immersion if it factors as iu ◦ g, for some U as above and an isomorphism of schemes g : Y → U . Problem 2. S Let f : Y → X be a morphism of schemes. Show that if there is an open cover X = i Ui such that the induced morphisms f −1 (Ui ) → Ui are is ...
REVIEW OF POINT-SET TOPOLOGY I WOMP 2006 The
... Proposition 4.5. All closed subsets of a compact space are compact. The converse holds if the space is also Hausdorff. Proposition 4.6. Let f : X → Y be a continuous bijection, and suppose X is compact and Y is Hausdorff. Then f is a homeomorphism; i.e., f −1 is also continuous. Definition 4.7. In t ...
... Proposition 4.5. All closed subsets of a compact space are compact. The converse holds if the space is also Hausdorff. Proposition 4.6. Let f : X → Y be a continuous bijection, and suppose X is compact and Y is Hausdorff. Then f is a homeomorphism; i.e., f −1 is also continuous. Definition 4.7. In t ...
Garrett 02-15-2012 1 Harmonic analysis, on R, R/Z, Q , A, and A
... and the quotient (A/k)b/k is still discrete. These maps are continuous group homs, so the image of (A/k)b/k in A/k is a discrete subgroup of a compact group, so is finite. Since (A/k)b is a k-vectorspace, (A/k)b/k is a singleton. Thus, (A/k)b ≈ k, if b is the usual diagonal copy. the image of k · ψ ...
... and the quotient (A/k)b/k is still discrete. These maps are continuous group homs, so the image of (A/k)b/k in A/k is a discrete subgroup of a compact group, so is finite. Since (A/k)b is a k-vectorspace, (A/k)b/k is a singleton. Thus, (A/k)b ≈ k, if b is the usual diagonal copy. the image of k · ψ ...
Solutions to Problem Set 3: Limits and closures
... V ) = π −1 (U ) ∩ π −1 (V ) is open in X. Hence, U ∩ V is open in X/∼. It is similar for unions. Hence, this defines a topology. Preimages of opens are open by definition, hence π is continuous. b. No. For instance, let X = [0, 1] and let x ∼ y if and only if x = y or x, y ∈ {0, 1}, i.e. we are iden ...
... V ) = π −1 (U ) ∩ π −1 (V ) is open in X. Hence, U ∩ V is open in X/∼. It is similar for unions. Hence, this defines a topology. Preimages of opens are open by definition, hence π is continuous. b. No. For instance, let X = [0, 1] and let x ∼ y if and only if x = y or x, y ∈ {0, 1}, i.e. we are iden ...
Topology Ph.D. Qualifying Exam Gerard Thompson Mao-Pei Tsui April 2009
... 1. (i) Define what is meant by a Möbius band. Identify the space obtained by identifying the boundary of a Möbius band to a point. Give a brief explanation. (ii) Let X be a topological space and let x0 ∈ X. Define the product of homotopy classes of loops [α] x0 based at x0 and verify in detail tha ...
... 1. (i) Define what is meant by a Möbius band. Identify the space obtained by identifying the boundary of a Möbius band to a point. Give a brief explanation. (ii) Let X be a topological space and let x0 ∈ X. Define the product of homotopy classes of loops [α] x0 based at x0 and verify in detail tha ...
