How to calculate $H_n(\ast)$? Let $X=\{\ast\}$ be the space with just one point.
I want to calculate $H_n(\ast)$. 
The n-th homology.
First I have to give $C_n(X)$.
In my lecture it was stated, that $C_n(X)=\mathbb{Z}$. I want to understand this.
For example $C_0(X)$. $C_0(X)$ is the free abelian group generated by the $0$-simplices.
$n$-simplices (or $C_n(X)$?) are given by formal linear combinations 
$$\sum_{\sigma:\Delta^n\to X} n_\sigma \sigma$$
where $n_\sigma\in\mathbb{Z}$ and just finite many $n_\sigma\neq 0$.
We have $\Delta^0=\{1\}\subseteq\mathbb{R}$. Hence just one $0$-simplice $\sigma:\Delta^0\to X$.
And the sum above simplifies to 
$$\sum_{\sigma:\Delta^0\to X} n_\sigma\sigma=n_\sigma\sigma?$$
What does that mean? $n_\sigma\in\mathbb{Z}$ and gets multiplied with $\sigma$ or even the evaluation at some point, which would have to be $\ast$ here, since $X=\{\ast\}$. But this does not seem correct. Why should we be able to multiply with such a random object. In general it could be $\ast\notin\mathbb{R}$.
What do I misunderstand here?
Can you explain it to me?
Thanks in advance.
 A: Your paragraph

What does that mean? $n_\sigma\in\mathbb{Z}$ and gets multiplied with $\sigma$ or even the evaluation at some point, which would have to be $\ast$ here, since $X=\{\ast\}$. But this does not seem correct. Why should we be able to multiply with such a random object. In general it could be $\ast\notin\mathbb{R}$.

shows that you are confused about the definition of the free abelian group generated by the simplices. In those cases, it may be best to turn back to the formal definition.
Given a set $S$, the free abelian group $F_S$ generated by $S$ is given by
$$F_S:=\{f:S \to \mathbb{Z} \mid f(s) \neq 0 \text{ for finitely many } s\}. $$
The notation $\sum\limits_{s}n_{s}\cdot s=n_{s_1}\cdot s_1+\cdots+n_{s_k}\cdot s_k$ (where the $n_{s_i}$'s are the non-zeros) is intended to mean the function that maps $s \mapsto n_s$. A priori this can be seen just as notation.$^{(1)}$ 
As an example, consider $S=\mathbb{R}$. You must understand why, in $F_{\mathbb{R}}$,
$$2\cdot(\frac{\pi}{2}) \neq 1\cdot \pi. $$
Once you do, your problem is almost solved. Now try to reconcile these things with your intuition of what is supposed to be the free abelian group generated by something.
For your actual question, for any $n\geq 0$, there is only one $n$-simplex to $\{*\}$. Thus, if $\Delta_n(X)$ denotes the set of such simplexes, we have that $\Delta_n(X)$ is a singleton for every $n \geq 0$. It is straightforward to prove that whenever $S$ is a singleton, $F_S \simeq \mathbb{Z}$ (take $f$ to $f(x)$, where $x$ is the element of $S$). 
$^{(1)}$This can be made into something algebraically meaningful if we interpret $s$ to mean the function $s \mapsto 1;$ $s' \mapsto 0$ for all $s' \neq s$. But then you must understand that this algebraic meaning is inherent to the formal algebraic structure of the free abelian group, not to some external structure like the real numbers or something similar.
