# Thread: Homology group of S

1. ## Homology group of S

Hi,

I've question about homology group of S.

How can I show that:

$H_{n}\left(S\right)\cong\mathbb{Z}$ for $n=0,1$ ?

Thanks for any help.

2. Originally Posted by Arczi1984
Hi,

I've question about homology group of S.

How can I show that:

$H_{n}\left(S\right)\cong\mathbb{Z}$ for $n=0,1$ ?

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 $\Delta$-complexes.

A circle S is described as one vertex v (0-simplex) and one edge (1-simplex) in the $\Delta$-complex structure. Since the boundary map of 1-simplex e is $\partial e=v - v = 0$, we have $H_0(S)\cong \frac{\text{Ker } \partial_0}{\text{Im} \partial_1} = \frac{\mathbb{Z}}{\{e\}}=\mathbb{Z}$, where $\text{Ker } \partial_0 = \mathbb{Z}$ since we have a single 0-simplex (vertex v) whose boundary map is trivially zero.

In a similar manner, we have $H_1(S) \cong \frac{\text{Ker } \partial_1}{\text{Im} \partial_2} = \frac{\mathbb{Z}}{\{e\}}=\mathbb{Z}$ since there is no 2-simplex in our $\Delta$-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.

3. Thanks for help.

I've one more question, maybe two.

Is it possible to show this by exact sequences?
Could You show me how can I solve this problem via singular homology?

4. Originally Posted by Arczi1984
Hi,

I've question about homology group of S.

How can I show that:

$H_{n}\left(S\right)\cong\mathbb{Z}$ for $n=0,1$ ?

Thanks for any help.
Or one can note that since $S$ is connected, $H_{0}(S) \cong \mathbb{Z}$.

Also, $H_{1}(S) := \ \text{the number of} \ 1- \text{cycles that are not} \ 1 \ \text{boundaries}$. Now $S$ itself is the only 1-cycle which is not a 1-boundary, since there are no higher dimensional objects. Thus $H_{1}(S) \cong \mathbb{Z}$.

5. Originally Posted by Arczi1984
Thanks for help.

I've one more question, maybe two.

Is it possible to show this by exact sequences?
Could You show me how can I solve this problem via singular homology?
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 $\sigma:\Delta^n \rightarrow X$ rather than $\Delta^n$ itself.

The n-chains, denoted as $C_n$ are finite sums $\sum_in_i\sigma_i$ for $n_i \in \mathbb{Z}$ and $\sigma_i:\Delta^n \rightarrow X$.

If you define a homomorphism $\epsilon:C_0(X) \rightarrow \mathbb{Z}$ by $\epsilon(\sum_in_i\sigma_i)=\sum_in_i$, a little bit of computation shows that $\epsilon$ induces an isomorphism and $H_0(X) \cong \mathbb{Z}$, where X is nonempty path-connected space (verify that $\text{Im }\partial_1=\text{Ker } \epsilon)$.

More generally,

Lemma: If X is a space and $\{X_i:i \in I \}$ are the path components of X, then $H_k(X) \cong \sum_{i \in I}H_k(X_i)$.

To compute $H_1(S^1)$, 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 $H_0(X)=\tilde{H}_0(X) \oplus \mathbb{Z}$.