Hi,
I've question about homology group of S.
How can I show that:
for ?
Thanks for any help.
There are several ways to solve this problem. I mean, you can use a simplicial, singular or cellular homolgy theory to solve this problem, resulting the same homolgy groups in this case.
The simplest one might be a simplicial homology using -complexes.
A circle S is described as one vertex v (0-simplex) and one edge (1-simplex) in the -complex structure. Since the boundary map of 1-simplex e is , we have , where since we have a single 0-simplex (vertex v) whose boundary map is trivially zero.
In a similar manner, we have since there is no 2-simplex in our -complex structure and we have a single 1-simplex (edge e) whose boundary map is zero.
It is trivial to show that all the remaining higher homology groups of S are zero.
This is kind of a standard textbook question and you can find this in most algebraic topology books like Hatcher's Algebraic topology p 106.
In a simplicial homology theory, we concern about the triangulations of the given space. In a singular homology, n-simplexes are replaced by their continuous images in the given topological space, where the continuous images do not have to look like simplexs.
A singular n-simplex is thus a continuous map rather than itself.
The n-chains, denoted as are finite sums for and .
If you define a homomorphism by , a little bit of computation shows that induces an isomorphism and , where X is nonempty path-connected space (verify that .
More generally,
Lemma: If X is a space and are the path components of X, then .
To compute , you might think of an intuitive way. For instance, a circle can be thought of three intervals (or more) gluing together at their endpoints and their boundary map corresponds to zero and we know that there is no nontrivial singular 2-simplex is available to a topological space of a circle.
Or you can use an exact sequence and the above lemma and find its homology group (see here).
Note that .