I have a few questions about the proof here of the $n$th reduced homology group of $X$ being isomorphic to the $(n+1)$st reduced homology group of the suspension $SX$ of $X$.

I will repeat the argument here:

Viewing $SX$ as the union of two cones $CX_N$ and $CX_S$ with their bases identified, consider the pair $(SX, CX_N)$. By the long exact sequence of reduced homology groups, we have the long exact sequence

$\cdots \rightarrow \tilde{H}_n(CX_N) \rightarrow \tilde{H}_n(SX) \rightarrow \tilde{H}_n(SX,CX_N) \rightarrow \tilde{H}_{n-1}(CX_N) \rightarrow \cdots \rightarrow \tilde{H}_0(CX_N) \rightarrow \tilde{H}_0(SX) \rightarrow \tilde{H}_0(SX,CX_N) \rightarrow 0$

But, $CX_N$ is contractible, so all of its reduced homology groups are trivial, giving $\tilde{H}_n(SX, CX_N) \cong \tilde{H}_n(SX)$ for all $n$. Furthermore, by the Excision Theorem, we have $\tilde{H}_n(SX – N, CX_N – N) \cong \tilde{H}_n(SX, CX_N)$ for all $n$. Since $X \simeq CX_N – N$, we get $\tilde{H}_n(X) \cong \tilde{H}_n(CX_N – N)$ for all $n$. Lastly, by the long exact sequence of reduced homology groups for the pair $(SX-N,CX_N – N)$ and the fact that $SX-N$ is contractible, we have $\tilde{H}_n(SX-N, CX_N-N) \cong \tilde{H}_{n-1}(CX_N-N)$ for all $n \geq 1$. Putting together all of our isomorphisms, the desired result follows.

Presumably, $N$ denotes the north tip point of the cone $CX_N$ in $SX$.

My questions:

- Why is $SX – N$ contractible? Similarly, why is $X \simeq CX_N – N$?
- To apply the Excision Theorem, we would need to know that the closure of $N$ is contained in the interior of $CX_N$. Why is this?

Thanks!

## Best Answer

We have

$SX = X \times I/\sim$, where $\sim$ identifies $X \times \{0\}$ to a point $S$ and $X \times \{1\}$ to a point $N$

$CX_N = X \times [\frac 1 2 ,1]/ X \times \{1\} \subset SX$

$CX_S= X \times [0,\frac 1 2]/ X \times \{0\} \subset SX$

Thus

$SX \setminus N \approx X \times [0,1)/X \times \{0\}$ which is contractible to $S$.

$CX_N \setminus N = X \times [\frac 1 2,1) \simeq X$.

$\{N\}$ is closed and $\operatorname{int} CX_N = CX_N \setminus X' = X \times (\frac 1 2 ,1]/ X \times \{1\}$, where $X' = CX_N \cap CX_S = X \times \{\frac 1 2\}$ is the common base of both cones.

To see this note that $ X \times (\frac 1 2 ,1]/ X \times \{1\}$ is open in $SX$ because its preimage under the quotient map $p : X \times I \to SX$ is $X \times (\frac 1 2 ,1]$ which is sopen in $X \times I$. Thus $X \times (\frac 1 2 ,1]/ X \times \{1\} \subset \operatorname{int} CX_N$. The points in $X'$ are no interior points of $CX_N$, thus $X \times (\frac 1 2 ,1]/ X \times \{1\} = \operatorname{int} CX_N$.