I'm an absolute beginner to the concepts of algebraic topology and we're only now covering the basics of singular homology in class. I'm horribly confused by a few things, particularly by this example where we compute the homology of a single point set, $X = \{x\_0\}$. I'd be incredibly grateful if I could try explain how I see the problem and someone could explain my misapprehensions.

We begin by saying that $C\_n(X) = \mathbb{Z}$; this is a freely generated abelian group whose generators are all the possible singular n-simplices that could exist in the space $X$, and these singular n-simplices are themselves are maps that embed the regular simplices in $X$.

Then since only one singular simplex exists per dimension because they're mapping into a singular point we can define: $\sigma\_n : \Delta^n \rightarrow \{x\_0\}$

Then computing the boundary of each of these:

$\partial\_0(\sigma\_0) = 0$... Does this holds for all 0-d simplices?

$ \partial\_1(\sigma\_1) = \sigma\_0 - \sigma\_0 = 0$ Is this because there is only one permitted 0-simplex in $X$? 

$ \partial\_2(\sigma\_2) = \sigma\_1 - \sigma\_1 + \sigma\_1 = \sigma\_1$ Is this because there is only one permitted 1-simplex in $X$? Also, is it possible for $\sigma\_2$ to be orientated such that this gives the same result but with a negative sign?

So the chain complex looks like:

$C\_2 = \mathbb{Z} \xrightarrow{\partial\_2} C\_1 = \mathbb{Z} \xrightarrow{\partial\_1} C\_0 = \mathbb{Z} \xrightarrow{\partial\_0} 0$ because there is exactly one generator per $C\_n$?

And so, $H\_0(X) = \dfrac{\ker(\partial\_0)}{Im({\partial\_1})} = \mathbb{Z} / \{0\} = \mathbb{Z}$

$H\_1(X) = \dfrac{\ker(\partial\_1)}{Im({\partial\_2})} = \mathbb{Z} / \mathbb{Z} = \{0\} $

Is this right?