Homology of $n$-manifold sans reduced homology A number of algebraic topology books, for example Hatcher on page 231, show that for an $n$-manifold $M$, the local homology groups $H_i(M, M \setminus \{ x \}; \mathbb{Z})$ are only nonzero for $i = n$, using the calculation:
\begin{align*}
H_i(M, M \setminus \{ x \}; \mathbb{Z}) &= H_i(\mathbb{R}^n, \mathbb{R}^n \setminus \{ 0 \}; \mathbb{Z}) \\
&= \widetilde{H}_{i-1}(\mathbb{R}^n \setminus \{ 0 \}; \mathbb{Z}) \\
&= \widetilde{H}_{i-1}(S^{n-1}; \mathbb{Z})
\end{align*}
I think the only problem I have with it is the moving from unreduced to reduced homology, and whilst I appreciate that it's not 'deep', this transition from unreduced to reduced homology because of contractibility is profoundly and disturbingly opaque to me. 
To be perhaps more clear, I understand how it arises in the long exact sequence except in the lower dimensions, which I guess is really where the reduced homology is important. With unreduced homology it becomes fuzzier for me because the zero-dimensional absolute homology groups are nonzero, and I don't really understand how the reduced homology groups fit in.
The question I have is, what is the unreduced analogue of this calculation? 
 A: You are correct to suspect that this only depends on contractiblity of $\mathbb{R}^{n}$. 
A suprisingly clean proof of this fact is given by the so called long exact sequence of a triple. If $X$ is a space, let me denote by $C _{\bullet}(X)$ the singular chain complex of $X$, so that $H_{i}(C _{\bullet}(X)) = H_{i}(X, \mathbb{Z})$. If $A \subseteq B$ is a pair of spaces, then one defines $H_{i}(B, A, \mathbb{Z}) = H_{i}(C_{\bullet}(B) / C_{\bullet}(A))$. 
If $(X, x_{0})$ is a pointed space, then one calls the groups $H_{i}(X, \{ x_{0} \})$ the reduced homology groups of $X$ and denotes them by $\widetilde{H}_{i}(X)$ (note I dropped the coefficients from my notation). I strongly suspect that Hatcher uses a slightly different definition of reduced groups that involves adjoining a $\ldots \rightarrow \mathbb{Z} \rightarrow 0$ factor to $C_{\bullet}(X)$, but you can convince yourself that at least in the case of path-connected spaces, the groups are naturally isomorphic for all choices of $x_{0}$. I believe the definiton I gave here is much more prevalent, it also makes it clear that the reduced homology groups are no more complicated then the relative ones.  
Going back to the topic, if $A \subseteq B \subseteq C$ be a triple of spaces, then one sees that we have a short exact sequence of chain complexes

$0 \rightarrow C_{\bullet}(B)/C_{\bullet}(A) \rightarrow C_{\bullet}(C)/C_{\bullet}(A) \rightarrow C_{\bullet}(C)/C_{\bullet}(B) \rightarrow 0$

that gives rise to the long exact sequence 

$\ldots \rightarrow H_{i}(B, A) \rightarrow H_{i}(C, A) \rightarrow H_{i}(C, B) \rightarrow H_{i-1}(B,A) \rightarrow \ldots$

called the homology exact sequence of a triple.
Suppose now that you have a pointed space $(X, x_{0})$ and some contractible space $\widetilde{X}$ that contains $X$ as a subspace. In your case $X = \mathbb{R}^{n} \setminus 0$ and $\widetilde{X} = \mathbb{R}^{n}$ and the choice of point is irrelevant. The long exact sequence of the triple takes the form 

$\ldots \rightarrow \widetilde{H}_{i}(X) \rightarrow \widetilde{H}_{i}(\widetilde{X}) \rightarrow H_{i}(\widetilde{X}, X) \rightarrow \widetilde{H}_{i-1}(X) \rightarrow \ldots$

but since $\widetilde{X}$ is conctractible, $\widetilde{H}_{i}(\widetilde{X}) = 0$ for all $i$, so this proves that $H_{i}(\widetilde{X}, X) \simeq \widetilde{H}_{i-1}(X)$.
