Exercise Sheet no. 5 of “Topology” - Mathematik@TU
... On the compact space X := [0; 1] ⊆ R, we consider the equivalence relation defined by x ∼ y if either x = y or (x = 0 and y = 1) or (x = 1 and y = 0). Show that: (a) The quotient space [X] := X/ ∼ is Hausdorff and compact. (b) [X] is homeomorphic to S1 := {z ∈ C : |z| = 1}. Hint: Consider f : [X] → ...
... On the compact space X := [0; 1] ⊆ R, we consider the equivalence relation defined by x ∼ y if either x = y or (x = 0 and y = 1) or (x = 1 and y = 0). Show that: (a) The quotient space [X] := X/ ∼ is Hausdorff and compact. (b) [X] is homeomorphic to S1 := {z ∈ C : |z| = 1}. Hint: Consider f : [X] → ...
Problems for Category theory and homological algebra
... (b) An (additively written) abelian group A is called divisible if for each a ∈ A and each n ∈ Z>0 there exists an element b ∈ A such that nb = a. The category of divisible abelian groups (with group homomorphisms as morphisms) is denoted Div. Show that the quotient map Q → Q/Z is a monomorphism in ...
... (b) An (additively written) abelian group A is called divisible if for each a ∈ A and each n ∈ Z>0 there exists an element b ∈ A such that nb = a. The category of divisible abelian groups (with group homomorphisms as morphisms) is denoted Div. Show that the quotient map Q → Q/Z is a monomorphism in ...
problem set #7
... (a) Prove that I( X ) is an ideal of R. (b) If p ∈ [0, 1] and I p = I({ p}), then prove that I p is a maximal ideal of R and R ∼ I p = R. (c) For J ⊂ R, set V( J ) := { x ∈ [0, 1] : f ( x ) = 0 for all f ∈ J }. Prove that V( J ) is a closed subset of [0, 1]. Hint: A function is continuous if and onl ...
... (a) Prove that I( X ) is an ideal of R. (b) If p ∈ [0, 1] and I p = I({ p}), then prove that I p is a maximal ideal of R and R ∼ I p = R. (c) For J ⊂ R, set V( J ) := { x ∈ [0, 1] : f ( x ) = 0 for all f ∈ J }. Prove that V( J ) is a closed subset of [0, 1]. Hint: A function is continuous if and onl ...
The Arithmetic Square (Lecture 32)
... annihilated by the order |A|; in particular, the rational homology groups vanish. ...
... annihilated by the order |A|; in particular, the rational homology groups vanish. ...
Math 210B. Homework 4 1. (i) If X is a topological space and a
... of X admits a finite subcover) and that any subspace Y ⊂ X is noetherian. (iii) Conversely to (ii), if every subspace of a topological space X is quasi-compact then prove X is noetherian. 2. Over a field k = k with char(k) 6= 2, decompose Z(y 4 − x2 , y 4 − x2 y 2 + xy 2 − x3 ) ⊂ k 2 and Z(u2 + v 2 ...
... of X admits a finite subcover) and that any subspace Y ⊂ X is noetherian. (iii) Conversely to (ii), if every subspace of a topological space X is quasi-compact then prove X is noetherian. 2. Over a field k = k with char(k) 6= 2, decompose Z(y 4 − x2 , y 4 − x2 y 2 + xy 2 − x3 ) ⊂ k 2 and Z(u2 + v 2 ...
Solve EACH of the exercises 1-3
... Ex. 2. Show that a continuous image of a separable space is separable, that is, if there exists a continuous function from a separable topological space X onto a topological space Y , then Y is separable. Include the definition of a separable topological space. Ex. 3. Let f be a continuous function ...
... Ex. 2. Show that a continuous image of a separable space is separable, that is, if there exists a continuous function from a separable topological space X onto a topological space Y , then Y is separable. Include the definition of a separable topological space. Ex. 3. Let f be a continuous function ...
Covering Maps and the Monodromy Theorem
... Proof The map p: X̃ → X is a covering map; therefore there exists an open cover U of X such that each open set U belonging to X is evenly covered by the map p. Now the collection consisting of the preimages γ −1 (U ) of the open sets U belonging to U is an open cover of the interval [0, 1]. But [0, ...
... Proof The map p: X̃ → X is a covering map; therefore there exists an open cover U of X such that each open set U belonging to X is evenly covered by the map p. Now the collection consisting of the preimages γ −1 (U ) of the open sets U belonging to U is an open cover of the interval [0, 1]. But [0, ...
MT 3803 - Loyola College
... 01.(a)(i) Let X be a non–empty set, and let d be a real function of ordered pairs of elements of X which satisfies the following two conditions: d(x,y) = 0 x = y, and d(x,y) d(x,z) + d(y,z) Show that d is a metric on X. (OR) (ii) Let X and Y be metric spaces and f be a mapping of X into Y. Prove ...
... 01.(a)(i) Let X be a non–empty set, and let d be a real function of ordered pairs of elements of X which satisfies the following two conditions: d(x,y) = 0 x = y, and d(x,y) d(x,z) + d(y,z) Show that d is a metric on X. (OR) (ii) Let X and Y be metric spaces and f be a mapping of X into Y. Prove ...
Covering space
In mathematics, more specifically algebraic topology, a covering map (also covering projection) is a continuous function p from a topological space, C, to a topological space, X, such that each point in X has an open neighbourhood evenly covered by p (as shown in the image); the precise definition is given below. In this case, C is called a covering space and X the base space of the covering projection. The definition implies that every covering map is a local homeomorphism.Covering spaces play an important role in homotopy theory, harmonic analysis, Riemannian geometry and differential topology. In Riemannian geometry for example, ramification is a generalization of the notion of covering maps. Covering spaces are also deeply intertwined with the study of homotopy groups and, in particular, the fundamental group. An important application comes from the result that, if X is a ""sufficiently good"" topological space, there is a bijection between the collection of all isomorphism classes of connected coverings of X and the conjugacy classes of subgroups of the fundamental group of X.