For a space $X$, the suspension of it is defined by $$\Sigma X:=C_{+}(X)\cup_{X} C_{-}(X),$$ where $C_{+}(X)$ and $C_{-}(X)$ are the upper and lower cone of it.
The reduced homology, roughly speaking, it is the homology obtained by removing $H_{k}(\text{point})$ from $H_{k}(X)$.
For a usual space, this means that
$$\overline{H}_{k}(X)=\left\{
\begin{array}{ll}
H_{k}(X),\ \ \text{for}\ k>0\\
\\
\mathbb{Z}^{\#\ \text{of path components-1}},\ \ \text{for}\ k=0.
\end{array}
\right.
$$
A well-known theorem and corollary are as follows:
Theorem. $\overline{H}_{k+1}(\Sigma X)\cong \overline{H}_{k}(X)$.
Corollary: $\overline{H}_{k}\Big(\Sigma^{\ell}X\Big)\cong \overline{H}_{k-\ell}(X).$
Now, I want to apply these two results to compute the homology group of $\mathbb{S}^{n}$. I computed the case of $k=n$ as follows:
Recall from the first theorem that $\overline{H}_{k+1}(\Sigma X)\cong \overline{H}_{k}(X)$.
For $X=\mathbb{S}^{n}$, we know that $\Sigma X=\mathbb{S}^{n+1}$, so the above identity gives us $$\overline{H}_{k+1}(\mathbb{S}^{n+1})\cong\overline{H}_{k}(\mathbb{S}^{n}).$$
For $k=n$, we firstly consider the case of $n=0$ where $\mathbb{S}^{0}$ is two points, then $$\overline{H}_{0}(\mathbb{S}^{0})=\mathbb{Z}^{\#\ \text{of path components of}\ \mathbb{S}^{0}}=\mathbb{Z},$$ and thus $$H_{1}(\mathbb{S}^{1})=\overline{H}_{1}(\mathbb{S}^{1})=\overline{H}_{0}(\mathbb{S}^{0})=\mathbb{Z}.$$
Then consider $n=1$, we have $$H_{2}(\mathbb{S}^{2})=\overline{H}_{2}(\mathbb{S}^{2})=\overline{H}_{1}(\mathbb{S}^{1})=\mathbb{Z}.$$
Keep iterating, an inductive argument will show that $H_{k}(\mathbb{S}^{n})=\mathbb{Z}$ for $k=n$.
But then I got stuck, when I tried to compute in the case of $k=0$.
Firstly, from the case of $k=n$, we see that $H_{0}(\mathbb{S}^{0})=\mathbb{Z}$, but then for $n=1$, we have we have $$\overline{H}_{1}(\mathbb{S}^{2})=\overline{H}_{0}(\mathbb{S}^{1}),$$ what should I do next without knowing the number of path-components of $\mathbb{S}^{n}$ for $n\geq 1$? I understand that $\mathbb{S}^{n}$ is path-connected, but I don't know how to start from this to compute the number of path-components.
If path-connectedness implies there is one path-components, then okay no problem we have $$H_{0}(\mathbb{S}^{n})=\mathbb{Z}.$$ But then do we have contradiction? since then we have $$H_{1}(\mathbb{S}^{n})=\overline{H}_{1}(\mathbb{S}^{n})=\overline{H}_{0}(\mathbb{S}^{n})=\mathbb{Z}$$ but we know that $H_{k}(\mathbb{S}^{n})=0$ for $0<k<n$.
so path-connected means there does not exist path-component? and thus $$H_{0}(\mathbb{S}^{n})=0$$ but then what should I do to compute $H_{0}(\mathbb{S}^{n})$ for $n>0$?
There seems no other way to iterate…
Similarly problem also arose when I tried to compute for $k<n$. If we know the case of $k=0$, then we can separate the case where $k=0<n$, and for all other $1\leq k<n$, $$\overline{H}_{k}(\mathbb{S}^{n})=H_{k}(\mathbb{S}^{n})$$ but what should I do to compute this homology group if I don't know $H_{k}(\mathbb{S}^{n})=0$ for $0<k<n$??
I am kind of believing that there is no way to compute the homology using only these two result other than the case of $k=n$…
Thank you for any advice and help!
Edit 1:
Okay, I figured it out. This can be done using induction. See my answer.
Best Answer
First note that for all $n\geq 1$, $H_{0}(\mathbb{S}^{n})=\mathbb{Z}$ since $\mathbb{S}^{n}$ is path-connected for all $n\geq 1$.
To compute the higher dimensions, we use induction on $n$. Recall from the first theorem that $\overline{H}_{k+1}(\Sigma X)\cong \overline{H}_{k}(X)$.
Suppose we know the homology group for $S^{n-1}$ for some fixed $n> 1$, which is $$H_{k}(\mathbb{S}^{n-1})=\left\{ \begin{array}{ll} \mathbb{Z},\ \ \ k=n-1\\ 0,\ \ \ 1\leq k\leq n-2\\ \mathbb{Z},\ \ \ k=0. \end{array} \right. $$ This theorem implies that $$\overline{H}_{k}(\mathbb{S}^{n-1})=\left\{ \begin{array}{ll} H_{k}(\mathbb{S}^{n-1}),\ \ \ k>0\\ \\ \mathbb{Z}^{\{\#\ \text{of path components}-1\}},\ \ \ k=0 \end{array} \right. =\left\{ \begin{array}{ll} \mathbb{Z},\ \ \ k=n-1\\ 0,\ \ \ 1\leq k\leq n-2\\ 0,\ \ \ k=0 \end{array} \right. $$
Therefore, $$\overline{H}_{k+1}(\mathbb{S}^{n})=\overline{H}_{k+1}(\Sigma \mathbb{S}^{n-1})=\overline{H}_{k}(\mathbb{S}^{n-1})=\left\{ \begin{array}{ll} \mathbb{Z},\ \ \ k=n-1\\ 0,\ \ \ 1\leq k\leq n-2\\ 0,\ \ \ k=0 \end{array} \right. =\left\{ \begin{array}{ll} \mathbb{Z},\ \ \ k+1=n\\ 0,\ \ \ 2\leq k+1\leq n-1\\ 0,\ \ \ k+1=1 \end{array} \right. $$
Therefore, $H_{k}(\mathbb{S}^{n})$ is known, since for all $k>0$, the reduced homology is the same as the regular topology, and we know what $k=0$ is, as mentioned as beginning. $$H_{k}(\mathbb{S}^{n})=\left\{ \begin{array}{ll} \mathbb{Z}, k=n\\ 0, 1\leq k\leq n-1\\ \mathbb{Z}, k=0. \end{array} \right. $$
Doing this inductively, this trick is actually doing this below graph:
Thus, the only thing you need to do is to compute $H_{k}(\mathbb{S}^{1})$ to use the inductive graph and $H_{k}(\mathbb{S}^{0})$ as a special case.
The computation of $H_{k}(\mathbb{S}^{1})$ is regular. To compute $H_{k}(\mathbb{S}^{0})$, note that $\mathbb{S}^{0}$ is just two points. I claim that $$H_{k}(2\ pts)=H_{k}(pt)\oplus H_{k}(pt)=\left\{ \begin{array}{ll} \mathbb{Z}\oplus\mathbb{Z},\ \ \ k=0\\ 0,\ \ \ k>0 \end{array} \right. $$
This can be proved easily using excision isomorphism: $$H_{k}(2pts, pt)\cong_{excision}H_{k}(pt,\varnothing)=H_{k}(pt),$$ and then just do the long exact sequence thing.