Following Qualifying exam question (Algebra, Representation theory, Category Theory and Homological Algebra), this post will discuss questions in Algebraic Topology.
Suppose is a finite CW complex such that is finite and nontrivial. Show that the universal covering of cannot be contractible. (Note that , in particular it cannot be 1.)
Let denotes the closed orientable surface of genus . There is a degree 1 map from to iff . (If , we can construct a degree one map by writing and contracting to a point; If and is a degree 1 map, then is not injective for dimensional reason. Let be in the kernel, and by Poincare duality let such that generates . Then . However, by the (cohomological) definition of degree, induces an isomorphism of , contradiction. This in fact shows that any continuous map from to has degree 0.)
Remark: See this question and this paper for what degree could arise when .
- What are the homology of ,
with coefficients in ?
with coefficients in ?
with coefficients in ?
(For (a) the fastest way to do this is via cellular homology, the cellular chain complex for is corresponding to one cell in each dimension; the gluing map is via ; the computation of degree of this map can be done locally, and it is the sum of the degree of the identity map and the antipodal map, i.e. . Therefore the map in the cellular chain complex is 0 and multiplication by 2 alternately. The answer is in ascending degrees. Then we can compute the integral cohomology of using Kunneth’s formula, and we get ; For (b) and (c) we use universal coefficient theorem and get and .)
Remark: Don’t forget the term when using Kunneth and Universal Coefficient.
Compute the cohomology for all and . (Use the universal coefficient theorem for Ext, the degree cohomology with coefficients in surjects onto with kernel . The answer is for all and 0 otherwise.)
Show that every closed, connected, compact (but not necessarily orientable) smooth manifold of odd dimension has Euler characteristic 0. (Suppose first the manifold is orientable, then we just use Poincaré duality and note that . For nonoroentable manifolds we can either consider its orientable double cover, or apply Poincaré duality with coefficients.)
Let be a closed orientable surface of genus 2.
Prove the homotopy group . (Its universal cover can be taken to be the hyperbolic plane, which is contractible, so has all homotopy groups zero. Note that any map of into lifts to its universal cover.)
Show that is nonabelian. (The fundamental group admits a surjection onto the free group in two generators.)
Find the fundamental group of minus two points. (Use van-Kampen theorem.)
Let be a path connected, locally path connected, semi-locally simply connected space, and assume that
Classify all isomorphism types of path connected covering spaces of . (By Galois correspondence of covering spaces, there is a bijection between subgroups of and path-connected covering spaces of .)
For each path connected covering space, describe the group of deck trans- formations. (The group of deck transformation is the automorphism group of the cover. It is equal to if is associated to the subgroup .)
Compute the singular homology of the Klein bottle with -coefficient. (The answer is in degree and 0 otherwise, which is consistent with Poincare duality.)
Prove the theorem on invariance of dimension, i.e. two manifolds of different dimensions cannot be homeomorphic, by considering the relative homology (Using excision, we can compute the relative chomology where is a compact neighborhood of homeomorphic to the n-dimensional closed disk.)
Universal coefficient theorem proof: Use the exact sequence . Take Hom (note that it remains exact since is free abelian) and then take the LES in homology; On the other hand, we have a SES that is actually a free resolution of . The proof essentially identifies the two LES in homology. Note that cohomology is determined by homology, but it is not the other way around. See this question for a reference. The intuition is that cohomology is contravariant, so from the diagonal map we get a map . From Eilenberg-Zilber (right to left) and Alexander-Whitney (left to right) we know that . However, for the (co)homology of tensor product of complexes, we only have the following SES: this sequence splits but not naturally. In any case the composite gives a ring operation on . Note that this doesn’t work to give a coalgebra structure on homology in general precisely because the Kunneth formula for chain complexes goes the wrong way. This doesn work over a field since then the Kunneth formula is an isomorphism. See this question for more details.
The relative cohomology SES is also more geometric than relative homology one. Essentially consists of functions on simplicial complexes that have support in , and it also suggests generalizing to sheaves other than the constant sheaf .
To show that the cup product is graded commutative on cohomology, consider given by mapping to . We want to show that this map is chain homotopic to the identity, which will imply the graded commutativity since chain homotopic maps induce the same maps on homology. In other words, given a singular -complex, we want to produce a singular chain from to , essentially the idea is to build up the swapping in a step-by-step fashion: can be triangulated by , say one of them is for some , then we apply to the ’s, and keep the v’s, and take the alternating sum over . All this is really geometry and combinatorics of simplices, in the sense that plays no role. To be more precise, on the level of chains , the difference naturally. The proof indeed says on the level of chains, the operation is commutative up to a specified homotopy that is part of the data.