Problem Set 5 - Stony Brook Mathematics

... Problem 1. Show that if X is a finite simplicial complex whose underlying topological space is a homology n-manifold, then (a) X consists entirely of n-simplices and their faces, (b) Every (n − 1)-simplex is a face of precisely two n-simplices. Problem 2. Suppose that X is a compact triangulable hom ...

... Problem 1. Show that if X is a finite simplicial complex whose underlying topological space is a homology n-manifold, then (a) X consists entirely of n-simplices and their faces, (b) Every (n − 1)-simplex is a face of precisely two n-simplices. Problem 2. Suppose that X is a compact triangulable hom ...

Complex Bordism (Lecture 5)

... the cohomology theory MU of complex bordism. In fact, we will show that MU is universal among complexoriented cohomology theories. We begin with a general discussion of orientations. Let X be a topological space and let ζ be a vector bundle of rank n on X. We may assume without loss of generality th ...

... the cohomology theory MU of complex bordism. In fact, we will show that MU is universal among complexoriented cohomology theories. We begin with a general discussion of orientations. Let X be a topological space and let ζ be a vector bundle of rank n on X. We may assume without loss of generality th ...

LECTURE NOTES 4: CECH COHOMOLOGY 1

... If G is any old abelian group, then we can consider it as a discrete space. In that case, the sheaf U 7→ G(U ) = T (U, G) is called the constant sheaf with values in G. The very constant presheaf with values in G is the presheaf G! whose value on an open set U is G!(U ) = G. (Check that this is not ...

... If G is any old abelian group, then we can consider it as a discrete space. In that case, the sheaf U 7→ G(U ) = T (U, G) is called the constant sheaf with values in G. The very constant presheaf with values in G is the presheaf G! whose value on an open set U is G!(U ) = G. (Check that this is not ...

Topology Group

... (invariants) of topological spaces specifically we will be dealing with cubical sets • “…Allows one to draw conclusions about global properties of spaces and maps from local computations.” (Mischaikow) ...

... (invariants) of topological spaces specifically we will be dealing with cubical sets • “…Allows one to draw conclusions about global properties of spaces and maps from local computations.” (Mischaikow) ...

Section 07

... where the product ranges over all (n + 1)-tuples i0 , . . . , in for which Ui0 ...in is non-empty. This is a larger complex with much redundant information (the group A(Ui0 ...in ) now occurs (n + 1)! times), hence less convenient for calculations, but it computes the same cohomology as we will now ...

... where the product ranges over all (n + 1)-tuples i0 , . . . , in for which Ui0 ...in is non-empty. This is a larger complex with much redundant information (the group A(Ui0 ...in ) now occurs (n + 1)! times), hence less convenient for calculations, but it computes the same cohomology as we will now ...

Algebraic Topology Introduction

... Frequently quotient spaces arise as equivalence relations, as any surjective map of sets p : X → Y defines an equivalence relation on X by x1 ∼ x2 if and only if p(x1 ) = p(x2 ). In general, if R is an equivalence relation on X, then the quotient space X/R is the set of all equivalence classes of X ...

... Frequently quotient spaces arise as equivalence relations, as any surjective map of sets p : X → Y defines an equivalence relation on X by x1 ∼ x2 if and only if p(x1 ) = p(x2 ). In general, if R is an equivalence relation on X, then the quotient space X/R is the set of all equivalence classes of X ...

Sheaf Cohomology 1. Computing by acyclic resolutions

... sheaf, Čech cohomology (with respect to that cover) and right-derived functor cohomology are the same. Note that the first proposition here holds without any assumption that the sheaf in question be flasque. The acyclicity of the cover gives us a long exact sequence of the Čech groups, allowing an ...

... sheaf, Čech cohomology (with respect to that cover) and right-derived functor cohomology are the same. Note that the first proposition here holds without any assumption that the sheaf in question be flasque. The acyclicity of the cover gives us a long exact sequence of the Čech groups, allowing an ...

AAG, LECTURE 13 If 0 → F 1 → F2 → F3 → 0 is a short exact

... (1) in the category of vector spaces over a field k all objects are injective; (2) Q and Q/Z are injective in the category Ab of abelian groups; (3) if R is a commutative ring then HomAb (R, Q/Z) is an injective R-module. Note that the abelian groups Q and Q/Z are not finitely generated, and the R-m ...

... (1) in the category of vector spaces over a field k all objects are injective; (2) Q and Q/Z are injective in the category Ab of abelian groups; (3) if R is a commutative ring then HomAb (R, Q/Z) is an injective R-module. Note that the abelian groups Q and Q/Z are not finitely generated, and the R-m ...

MANIFOLDS, COHOMOLOGY, AND SHEAVES

... To understand these lectures, it is essential to know some point-set topology, as in [3, Appendix A], and to have a passing acquaintance with the exterior calculus of differential forms on a Euclidean space, as in [3, Sections 1–4]. To be consistent with Eduardo Cattani’s lectures at this summer sch ...

... To understand these lectures, it is essential to know some point-set topology, as in [3, Appendix A], and to have a passing acquaintance with the exterior calculus of differential forms on a Euclidean space, as in [3, Sections 1–4]. To be consistent with Eduardo Cattani’s lectures at this summer sch ...

IV.2 Homology

... Brouwer’s Fixed Point Theorem. A continuous map f : Bd+1 → Bd+1 has at least one fixed point x = f (x). Proof. Let A, B : Sd → Sd be maps defined by A(x) = (x − f (x))/kx − f (x)k and B(x) = x. B is the identity and therefore has degree 1. If f has no fixed point then A is well defined and has degre ...

... Brouwer’s Fixed Point Theorem. A continuous map f : Bd+1 → Bd+1 has at least one fixed point x = f (x). Proof. Let A, B : Sd → Sd be maps defined by A(x) = (x − f (x))/kx − f (x)k and B(x) = x. B is the identity and therefore has degree 1. If f has no fixed point then A is well defined and has degre ...