Details in the computation of $H_k ( \mathbb{R}^n, \mathbb{R}^n - \{0\} )$ for $k=1,0$ During the proof of Theorem 2.26 (Brouwer's  theorem on invariance of dimension) in his book, Hatcher claims that "$H_k ( \mathbb{R}^n , \mathbb{R}^n - \{ 0 \} )$ is $\mathbb{Z}$ for $k=n$ and $0$ otherwise". I'm perfectly fine with the argument that $ \mathbb{R}^n - \{ 0 \} $ has the same homology as $ S^{n-1} $, and I am also aware of the homology groups of spheres.
I then started to develop the relevant computations on my on, but stumbled upon some details that left me puzzled. They all came up in the following segment of the long exact sequence of relative homologies:
$$
H_1 ( \mathbb{R}^n - \{0\} ) \xrightarrow{ (i_1)_{\ast} } H_1 ( \mathbb{R}^n )  \xrightarrow{ (j_1)_{\ast} } H_1 ( \mathbb{R}^n, \mathbb{R}^n - \{0\} ) \xrightarrow{ \partial_1 } H_0 ( \mathbb{R}^n - \{0\} ) \xrightarrow{ (i_0)_{\ast} } H_0 ( \mathbb{R}^n )  \xrightarrow{ (j_0)_{\ast} } H_0 ( \mathbb{R}^n, \mathbb{R}^n - \{0\} ) \xrightarrow{ \partial_0 } 0
$$
which reads, for every $n$, using that $\mathbb{R}^n$ is path-connected and contractible:
$$
H_1 ( \mathbb{R}^n - \{0\} ) \xrightarrow{ (i_1)_{\ast} } 0  \xrightarrow{ (j_1)_{\ast} } H_1 ( \mathbb{R}^n, \mathbb{R}^n - \{0\} ) \xrightarrow{ \partial_1 } H_0 ( \mathbb{R}^n - \{0\} ) \xrightarrow{ (i_0)_{\ast} } \mathbb{Z} \xrightarrow{ (j_0)_{\ast} } H_0 ( \mathbb{R}^n, \mathbb{R}^n - \{0\} ) \xrightarrow{ \partial_0 } 0
$$
Now, the way things are written in Hatcher suggests his claim follows easily and solely from the properties of exact sequences, for every possible case and every $k \neq n$. But this last segment is always a little thorny for me.
1) First, when $n \geq 2$, $S^{n-1}$ is path-connected, and we can further simplify:
$$
H_1 ( \mathbb{R}^n - \{0\} ) \xrightarrow{ (i_1)_{\ast} } 0  \xrightarrow{ (j_1)_{\ast} } H_1 ( \mathbb{R}^n, \mathbb{R}^n - \{0\} ) \xrightarrow{ \partial_1 } \mathbb{Z} \xrightarrow{ (i_0)_{\ast} } \mathbb{Z} \xrightarrow{ (j_0)_{\ast} } H_0 ( \mathbb{R}^n, \mathbb{R}^n - \{0\} ) \xrightarrow{ \partial_0 } 0
$$
but this didn't take me very far using just algebra (but I'm very new to this). Instead, I tried to explicitly understand the morfism $(i_0)_{\ast}$, induced by the inclusion $i_0 : C_0 ( \mathbb{R}^n - \{0\} ) \to C_0 (  \mathbb{R}^n )$:

Let $\sigma$ be a 0-cycle (i.e. a formal sum of points) in
$\mathbb{R}^n - \{0\}$ such that $\sigma$ is the boundary of a 1-chain
$\sum_{i} n_i \alpha_i$ in $\mathbb{R}^n$ (in other words, $\sigma$
represents a class in $\ker (i_0)_{\ast}$). Then, as long as the
dimension of the ambient space is $n \geq 2$, any $\alpha_i$ that
meets $0$ can be replaced by a 1-chain $\tilde{\alpha}_i$ such that
$\partial_1 \tilde{\alpha}_i = \partial_1 \alpha_i$, but the image of
$\tilde{\alpha}_i$ avoids $0$. Thus, $\sigma$ is also a boundary in $\mathbb{R}^n
 - \{0\}$, and therefore trivial in $H_0 ( \mathbb{R}^n - \{0\} )$.

The above reasoning, if correct, tells me that $(i_0)_{\ast}$ is injective. In this case, yes, I can conclude using exactness + the isomorphism theorem that $H_1 ( \mathbb{R}^n, \mathbb{R}^n - \{0\} ) \approx 0$, but I wonder: is there any way to conclude this in general, solely from the structure of the diagram?
2) When $n=1$, my rudimentary argument above obviously breaks down. Also, $S^0$ is no longer path-connected, so we have:
$$
0  \xrightarrow{ (j_1)_{\ast} } H_1 ( \mathbb{R}, \mathbb{R} - \{0\} ) \xrightarrow{ \partial_1 } \mathbb{Z} \oplus \mathbb{Z} \xrightarrow{ (i_0)_{\ast} } \mathbb{Z}  \xrightarrow{ (j_0)_{\ast} } H_0 ( \mathbb{R}, \mathbb{R} - \{0\} ) \xrightarrow{ \partial_0 } 0
$$
We now wish to prove that $H_1 ( \mathbb{R}, \mathbb{R} - \{0\} ) \approx \mathbb{Z}$, so it definitely can't be the case that $(i_0)_{\ast}$ is injective again (or this homology group would be trivial). This left me wondering: was it indeed injective at 1), or is that reasoning flawed, and both conclusions follow from some purely algebraic argument? Otherwise, how do we conclude this case?
3) I know from exactness that $(j_0)_{\ast}$ must be surjective, but I have no intuition whatsoever on why should this imply $H_0 ( \mathbb{R}^n, \mathbb{R}^n - \{0\} ) \approx 0$. Any hint in this direction would be appreciated, but I feel that this part of the reasoning is related to understanding 1) and 2), so that I can also use information from $(i_0)_{\ast}$.
To give some context: I started studying homology some days ago with the intention to write some paragraphs in my thesis about orientability in a topological setting, so I was basically following the shortest possible path towards it in Hatcher's book, although I have (not very wisely) skipped reduced homology.
Thanks in advance
 A: Let us first look at $ H_0 ( \mathbb{R}^n - \{0\} ) \xrightarrow{(i_0)_{\ast} } H_0 ( \mathbb{R}^n )  \xrightarrow{ (j_0)_{\ast} } H_0 ( \mathbb{R}^n, \mathbb{R}^n - \{0\} ) \xrightarrow{ \partial_0 } 0$, where $n  \ge 1$. Take any map $\phi : \mathbb R^n \to  \mathbb{R}^n - \{0\}$. Since $\mathbb R^n$ is contractible, we have $i_0 \circ \phi \simeq id$ and therefore $(i_0)_{\ast}  \circ \phi_* = id$. This implies that $(i_0)_{\ast}$ is a surjection. Therefore $\ker (j_0)_* = \operatorname{im} (i_0)_* = H_0 ( \mathbb{R}^n )$, i.e. $(j_0)_*$ is the zero map. Thus $\ker \partial_0 = \operatorname{im} (j_0)_* = 0$. Since $\partial_0$ is the zero map, we get $$ H_0 ( \mathbb{R}^n, \mathbb{R}^n - \{0\} ) = \ker \partial_0 = 0 .$$
Therefore we have the short exact sequence
$$
 0  \xrightarrow{ (j_1)_{\ast} } H_1 ( \mathbb{R}^n, \mathbb{R}^n - \{0\} ) \xrightarrow{ \partial_1 } H_0 ( \mathbb{R}^n - \{0\} ) \xrightarrow{ (i_0)_{\ast} } \mathbb{Z} \xrightarrow{ (j_0)_{\ast} }  0
$$
We have $\ker \partial_1 = \operatorname{im} (j_1)_* = 0$, i.e. $\partial_1$ is injective.
1) $n \ge 2$. We get
$$
 0  \xrightarrow{ (j_1)_{\ast} } H_1 ( \mathbb{R}^n, \mathbb{R}^n - \{0\} ) \xrightarrow{ \partial_1 } \mathbb Z \xrightarrow{ (i_0)_{\ast} } \mathbb{Z} \xrightarrow{ (j_0)_{\ast} }  0 .
$$
We know that $(i_0)_*$ is a surjection. This is only possible if $(i_0)_*(1) = \pm 1$, thus $(i_0)_*$ is an isomorphism. We conclude that $ \operatorname{im}\partial_1 = \ker (i_0)_* = 0$. Since $\partial_1$ is injective, this implies
$$ H_1 ( \mathbb{R}^n, \mathbb{R}^n - \{0\} ) = 0 .$$
2) $n = 1$.  We get
$$
 0  \xrightarrow{ (j_1)_{\ast} } H_1 ( \mathbb{R}, \mathbb{R} - \{0\} ) \xrightarrow{ \partial_1 } \mathbb Z \oplus \mathbb Z \xrightarrow{ (i_0)_{\ast} } \mathbb{Z}  \xrightarrow{ (j_0)_{\ast} }  0 .
$$
Since $\partial_1$ is injective,  $H_1 ( \mathbb{R}, \mathbb{R} - \{0\} )$ is isomorphic to the subgroup $G = \ker (i_0)_*$ of $\mathbb Z \oplus \mathbb Z$. The latter is a free abelian group with two generators, thus also $G$ is a free abelian group whose number $g$ of generators is $0, 1$ or $2$.
The above short exact sequence splits because there exists a homomorphism $h : \mathbb Z \to \mathbb Z \oplus \mathbb Z$ such that $(i_0)_* \circ h = id$. This can either be deduced form the fact that $\mathbb Z$ is a free abelian group or by taking the above $\phi_*$. It is well-known that this implies $\mathbb Z \oplus \mathbb Z \approx H_1 ( \mathbb{R}, \mathbb{R} - \{0\} ) \oplus \mathbb Z \approx G \oplus \mathbb Z$. An explicit ismorphism is given by $I : H_1 ( \mathbb{R}, \mathbb{R} - \{0\} ) \oplus \mathbb Z \to \mathbb Z \oplus \mathbb Z, I(\xi,x) = \partial_1(\xi) +  h(x)$.
$G \oplus \mathbb Z$ is a free abelian  group with $g+1$ generators and we must have $g + 1 = 2$ since $\mathbb Z \oplus \mathbb Z$ has two generators. Hence $g = 1$ and
$$H_1 ( \mathbb{R}, \mathbb{R} - \{0\} ) \approx G \approx \mathbb Z .$$
Remark:
The above proof for 2) was purely algebraic / group theoretical. One can alternatively use singular $0$-simplices to show that $(i_0)_* : \mathbb  Z \oplus \mathbb Z = H_0((-\infty,0)) \oplus H_0(0,\infty) = H_0(\mathbb{R}- \{0\}) \to H_0(\mathbb R) = \mathbb Z$ has the form $(i_0)_*(a,b) = a + b$. Then $G = \ker (i_0)_* =\{(a,-a) \mid a \in \mathbb  Z\}$. This group is isomorphic to $\mathbb Z$.
