1. Lecture 4, February 21 1.1. Open immersion. Let (X,O X) be a
... 1.1. Open immersion. Let (X, OX ) be a scheme. If U ⊆ X is an open subset then (U, OX|U ) is a scheme, and we have a natural map of schemes (i, i] ) : (U, OX|U ) −→ (X, OX ). We say that U is an open subscheme of X. A morphism of schemes f : Y −→ X is an open immersion if f induces an isomorphism wi ...
... 1.1. Open immersion. Let (X, OX ) be a scheme. If U ⊆ X is an open subset then (U, OX|U ) is a scheme, and we have a natural map of schemes (i, i] ) : (U, OX|U ) −→ (X, OX ). We say that U is an open subscheme of X. A morphism of schemes f : Y −→ X is an open immersion if f induces an isomorphism wi ...
Chapter 3. Topology of the Real Numbers. - Faculty
... Together, the sets X and T are called a topological space. If U ∈ T , then U is said to be open. Example. A topology on the real line is given by the collection of intervals of the form (a, b) along with arbitrary unions of such intervals. Let I = {(a, b) | a, b ∈ R}. Then the sets X = R and T = {∪α ...
... Together, the sets X and T are called a topological space. If U ∈ T , then U is said to be open. Example. A topology on the real line is given by the collection of intervals of the form (a, b) along with arbitrary unions of such intervals. Let I = {(a, b) | a, b ∈ R}. Then the sets X = R and T = {∪α ...
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... Using the notion of ordered pair, we soon get the very important structure called the product A × B of two sets A and B. Next, we can get such things as equivalence relations and order relations on a set A, for they are subsets of A×A. And we get the critical notion of a function AQ→ B, as a subset ...
... Using the notion of ordered pair, we soon get the very important structure called the product A × B of two sets A and B. Next, we can get such things as equivalence relations and order relations on a set A, for they are subsets of A×A. And we get the critical notion of a function AQ→ B, as a subset ...
Elements of Homotopy Fall 2008 Prof. Kathryn Hess Series 13 Let B
... (b) If B is path connected and E is nonempty, then p is a surjective map. Exercise 4. Let p : E−→B be a fibration, b0 ∈ B, and F := p−1 (b0 ) ⊆ E the fiber over b0 . Assume F is nonempty. Denote by i : F −→E the inclusion map. Prove the following: (a) If B is path connected, then the induced map π0 ...
... (b) If B is path connected and E is nonempty, then p is a surjective map. Exercise 4. Let p : E−→B be a fibration, b0 ∈ B, and F := p−1 (b0 ) ⊆ E the fiber over b0 . Assume F is nonempty. Denote by i : F −→E the inclusion map. Prove the following: (a) If B is path connected, then the induced map π0 ...
June 2010
... Do three of the problems from section A and three questions from section B. If you work more than the required number of problems, make sure that you clearly mark which problems you want to have counted. If you have doubts about the wording of a problem or about what results may be assumed without p ...
... Do three of the problems from section A and three questions from section B. If you work more than the required number of problems, make sure that you clearly mark which problems you want to have counted. If you have doubts about the wording of a problem or about what results may be assumed without p ...
THE UNIVERSITY OF TOLEDO Topology M.A. Comprehensive Examination L. Bentley H. Wolff
... This exam has been checked carefully for errors. If you find what you believe to be an error in a question, report this to the proctor. If the proctor’s interpretation still seems unsatisfactory to you, you may alter the question so that in your view it is correctly stated, but not in such a way tha ...
... This exam has been checked carefully for errors. If you find what you believe to be an error in a question, report this to the proctor. If the proctor’s interpretation still seems unsatisfactory to you, you may alter the question so that in your view it is correctly stated, but not in such a way tha ...
Lecture 4
... a unique bijective map φ : (S 1 × S 1 )/(S 1 ∨ S 1 ) −→ S 2 such that φ ◦ η2 = φ, from which follows that φ is continuous and a closed map since the domain is compact and the codomain is Hausdorff. Hence φ is a homeomorphism between (S 1 × S 1 )/(S 1 ∨ S 1 ) and S 2 . Surfaces: The sphere S 2 , toru ...
... a unique bijective map φ : (S 1 × S 1 )/(S 1 ∨ S 1 ) −→ S 2 such that φ ◦ η2 = φ, from which follows that φ is continuous and a closed map since the domain is compact and the codomain is Hausdorff. Hence φ is a homeomorphism between (S 1 × S 1 )/(S 1 ∨ S 1 ) and S 2 . Surfaces: The sphere S 2 , toru ...
Week 3
... (4) Let’s turn again to R`` . We saw that [0, 1) was already closed. What about (0, 1]? Since [0, 1] is closed in the usual topology, this must be closed in R`` as well. (Recall that the topology on R`` is finer than the standard one). It follows that (0, 1] is either already closed, or its closure ...
... (4) Let’s turn again to R`` . We saw that [0, 1) was already closed. What about (0, 1]? Since [0, 1] is closed in the usual topology, this must be closed in R`` as well. (Recall that the topology on R`` is finer than the standard one). It follows that (0, 1] is either already closed, or its closure ...
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... Definition 0.1 Let X be a topological space. We say X is compact if any open cover of X has a finite subcover. Definition 0.2 Let X and Y be topological spaces. The following is called the product topology on X × Y : The open sets of X × Y are given as arbitrary unions of sets of the form U × V , wh ...
... Definition 0.1 Let X be a topological space. We say X is compact if any open cover of X has a finite subcover. Definition 0.2 Let X and Y be topological spaces. The following is called the product topology on X × Y : The open sets of X × Y are given as arbitrary unions of sets of the form U × V , wh ...