What is analogous statement for lebesgue covering theorem

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Reference no: EM131110319

Honors Examination in Topology 2008

1. Suppose X is a set. The finite-complement topology, Tf is given by Tf = {U ⊂ X|X - U is finite or U = ∅}.

(a) Show that Tf is a topology on X.

(b) When is (X, Tf) a Hausdorff space?

(c) If (X, d) is a metric space with the associated metric topology Td, show that Tf is coarser than Td, that is, Tf ⊂ Td. (d) State the separation axiom T1 and show that if (X, T) is any topology on X, then Tf ⊂ T if and only if (X, T ) is T1.

2. The property of compactness is a generalization of the notion of a finite subset of a space. Take three properties of compact subsets of a topological space X and show how they generalize the same statement for a finite subset of X. (For example, the definition, or perhaps the Extreme Value theorem.) What is the analogous statement for the Lebesgue Covering Theorem when speaking of compact metric spaces?

3. A connected space X may contain points x ∈ X called cut points of order n, defined by the property that X - {x} has exactly n components. (a) Show that the property "x is a cut point of order n in X" is a topologically invariant property. (b) Construct a space containing a sequence of points {p2, p3, p4, . . .} such that pn is a cut point of order n. (c) Classify the alphabet up to homeomorphism (as shown) by analyzing cut points:

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

4. Quotients of the real line can be of all sorts. (a) Let x ∼ y be the relation on R where x ∼ y if x - y ∈ Q. This is a group based relation with quotient given by π: R → R/Q. What is the quotient topology on R/Q in this case? (b) Recall that if A ⊂ X, then the space X/A is the quotient of X by the relation x ≈ y if either x, y ∈ A, or x /∈ A and x = y. Show that the quotient R/(R - [0, 1]) is compact but not Hausdorff.

5. The so-called Dunce Hat is the quotient space of a 2-simplex with the edges glued according to the word aa-1a. Use the Seifert-van Kampen theorem to compute the fundamental group of the Dunce Hat. What is the first homology H1 of the Dunce Hat?

291_Figure.png

6. Show that a continuous mapping f: S2 → S1 × S1 is null-homotopic. (Hint: Try carefully using the lifting property of the universal covering space of S1 × S1.)

7. Let fk: S1 → S1 denote the map of degree k given by z |→ zk. Form the adjunction space Xk = S1fk e2 where we have attached a 2-cell to S1 by attaching the boundary identifying z and fk(z). Use the Seifert-van Kampen theorem to compute π1(Xk). Classify all of the covering spaces of Xk.

8. A graph Γ on S2 is a 1-dimensional simplicial complex with edges given by curves on S2. Suppose that Γ is connected. Show that S2 -Γ is connected if and only if V = E + 1, where V is the number of vertices of Γ and E the number of edges.

9. From the pages of Poincar´e: in 1885, he introduced the notion of the index of a flow on an orientable surface. Suppose that a surface S is compact, orientable, and without boundary, and there is a flow on S, that is, each point in S lies on a unique curve that satisfies some differential equation on the coordinates for the surface. The curves are all disjoint from one another, indexed by initial conditions. Assume that S is triangulated so that no edge in the triangulation is part of the flow. If ? is a cycle on S (a closed path in the edges of the triangulation), then Poincar´e defined the index of ? to be

J(?) = e(?) - i(?) - 2/2

where e(?) is the number of exterior points on the edges of ? where the flow is tangent to the edge and outside ?, i(?) is the number of interior points on the edges of ? where the flow is tangent inside ?. (a) Show that if ?1 and ?2 are cycles sharing an edge, then J(?1) + J(?2) = J(?1 + ?2), where the sum is the oriented sum of geometric cycles (the shared edge becomes interior). (b) Show that

Σ?iJ(?i) = -χ(S),

where the sum is over all triangles of the triangulation of S. This is the 1885 version of the Poincar´e index theorem.

701_Figure1.png

10. Suppose K is a finite polyhedron and F: K × [0, 1] → K is a homotopy between the identity mapping on K and a continuous function f: K → K. Show that χ(K) ≠ 0 implies that f has a fixed point. To which spheres does this result apply?

11. From the pages of Borsuk: in 1933 he proved the following statements:

i. For every continuous mapping f: Sn → Rn, there exists a point x ∈ Sn with f(x) = f(-x).

ii. For every antipodal mapping f: Sn → Rn (that is, f is continuous and f(-x) = -f(x) for all x ∈ Sn) there exists a point x ∈ Sn with f(x) = 0.

iii. There is no antipodal mapping f: Sn → Sn-1.

Show that these statements are equivalent.

12. Among the important constructions possible with simplicial complexes there is the suspension which is defined briefly for a simplicial complex K as

SK = K ∗ S0,

that is, SK is the join of K and the simplicial complex S0 = {a, b}, the zero-sphere or boundary of the standard 1-simplex, ?1. More explicitly, SK has p-simplices given by (x0, . . . , xp-1, a) and (x0, . . . , xp-1, b), where (x0, . . . , xp-1) is a (p - 1)-simplex in K. Consider the algebraic mapping s: Cp-1(K) → Cp(SK), given by

s(i=1Σnλi(v0,i, . . . , vp-1,i) = i=1Σnλi[(v0,i, . . . , vp-1,i, a) - (v0,i, . . . , vp-1,i, b)].

Show that this mapping is a chain map, that is, ∂ ? s = s ? ∂. Show that this induces a homomorphism s∗ : Hp-1(K) → Hp(SK). We can compute the homology of SK by considering the subcomplexes K ∗ {a} and K ∗ {b} with union SK. Recall the MayerVietoris sequence: If A and B are subcomplexes of X and A ∪ B = X, then the inclusion mappings

i1: A ∩ B → A, i2: A ∩ B → B, j1 : A → X, j2: B → X,

determine a long exact sequence

···→ Hp(A ∩ B) →i1∗⊕i2 Hp(A) ⊕ Hp(B) →j1-j2 Hp(A ∪ B) → Hp-1(A ∩ B) → ···

Use this result to determine H(SK) in terms of H(K).

Reference no: EM131110319

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