Homology Theory - Section de mathématiques
... categories of topological spaces, pairs of spaces (called “pairs” for short), and triples of spaces respectively. I.e. the objects of Top(2) are pairs (X, A), where X ∈ Ob(Top) is a topological space and A ⊂ X and a morphism f : (X, A) → (Y, B) is a continuous map f : X → Y with f A ⊂ B. Analogously ...
... categories of topological spaces, pairs of spaces (called “pairs” for short), and triples of spaces respectively. I.e. the objects of Top(2) are pairs (X, A), where X ∈ Ob(Top) is a topological space and A ⊂ X and a morphism f : (X, A) → (Y, B) is a continuous map f : X → Y with f A ⊂ B. Analogously ...
1 Comparing cartesian closed categories of (core) compactly
... A topology on C(X, Y ) is said to be exponential if continuity of a function f : A×X → Y is equivalent to that of its transpose f : A → C(X, Y ). When an exponential topology exists, it is unique, and we denote the resulting space by [X ⇒ Y ]. This is elaborated in Section 2 below. In this case, tra ...
... A topology on C(X, Y ) is said to be exponential if continuity of a function f : A×X → Y is equivalent to that of its transpose f : A → C(X, Y ). When an exponential topology exists, it is unique, and we denote the resulting space by [X ⇒ Y ]. This is elaborated in Section 2 below. In this case, tra ...
Section 16. The Subspace Topology - Faculty
... Example 1. Let X = R with the order topology (which for R is the same as the standard topology) and let Y = [0, 1] have the subspace topology. A basis for the order topology on R is B = {(a, b) | a, b ∈ R, a < b} (by the definition of “order topology,” since R has neither a least or greatest element ...
... Example 1. Let X = R with the order topology (which for R is the same as the standard topology) and let Y = [0, 1] have the subspace topology. A basis for the order topology on R is B = {(a, b) | a, b ∈ R, a < b} (by the definition of “order topology,” since R has neither a least or greatest element ...
![arXiv:0706.3441v1 [math.AG] 25 Jun 2007](http://s1.studyres.com/store/data/017784481_1-dd1160ad974ad834977ad61a663d12c3-300x300.png)
EXISTENCE AND PROPERTIES OF GEOMETRIC QUOTIENTS
... In most applications, group schemes are flat, separated and of finite presentation over the base and many of our results will require these hypotheses. We will not make any general assumptions though. Groups that are finite, flat and locally of finite presentation, or equivalently groups that are fi ...
... In most applications, group schemes are flat, separated and of finite presentation over the base and many of our results will require these hypotheses. We will not make any general assumptions though. Groups that are finite, flat and locally of finite presentation, or equivalently groups that are fi ...
MAT327H1: Introduction to Topology
... A basis B generates a topology T whose elements are all possible unions of elements of B . That is, the topology generated by B is the collection of arbitrary unions of the subsets of B . Proof: Have to prove that if u 1∪⋯∪u n ∈ T then u 1∩⋯∩u n ∈ T . By induction, it suffices to prove that if u1∪ u ...
... A basis B generates a topology T whose elements are all possible unions of elements of B . That is, the topology generated by B is the collection of arbitrary unions of the subsets of B . Proof: Have to prove that if u 1∪⋯∪u n ∈ T then u 1∩⋯∩u n ∈ T . By induction, it suffices to prove that if u1∪ u ...
DISJOINT UNIONS OF TOPOLOGICAL SPACES AND CHOICE Paul
... (iii)→MC. Fix A = {Ai : i ∈ k} a family of infinite pairwise disjoint sets and let A∗i and X be as in the proof of (i) → MC. Clearly, A∗i is a PFCS space. Thus, by (iii), X is also a PFCS space. Let V be a shrinking of the p.f. open cover U = {u : u = Ai or u = A∗i , i ∈ k} of X. Clearly Fi = v, v ∈ ...
... (iii)→MC. Fix A = {Ai : i ∈ k} a family of infinite pairwise disjoint sets and let A∗i and X be as in the proof of (i) → MC. Clearly, A∗i is a PFCS space. Thus, by (iii), X is also a PFCS space. Let V be a shrinking of the p.f. open cover U = {u : u = Ai or u = A∗i , i ∈ k} of X. Clearly Fi = v, v ∈ ...
