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.