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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!

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    $\begingroup$ Yes $N$ is the north pole of $SX$. You can call it whatever you want. It is the class $ \{0\} \times X$ in $CX = [0 , 1] \times X / \{0\} \times X$. $\endgroup$ Commented Nov 2, 2021 at 10:32
  • $\begingroup$ @infinitelooper Thanks! I edited the question. $\endgroup$ Commented Nov 2, 2021 at 10:36
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    $\begingroup$ What did you try ? Can you show that $CX_N$ is contractible ? $\endgroup$ Commented Nov 2, 2021 at 10:45
  • $\begingroup$ @infinitelooper I think I can show why $SX - N$ is contractible. I think the explicit homotopy would be $h:SX - N \times [0,1] \rightarrow SX - N : ([x,s],t) \mapsto [x,st]$. This would give that the identity map on $SX - N$ is homotopic to a constant map. $\endgroup$ Commented Nov 2, 2021 at 10:51
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    $\begingroup$ Nice. Now $SX - N$ is homeomorphic to $CX_N$. Can you find why ? $\endgroup$ Commented Nov 2, 2021 at 11:08

1 Answer 1

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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 $X \times (\frac 1 2 ,1]/ X \times \{1\} = \operatorname{int} CX_N$.

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  • $\begingroup$ I'm struggling to see why the interior of $CX_N$ would only exclude the "base" of $CX_N$ and not also exclude the "border" of the cone, including the tip point. Wouldn't any open ball centered at a point on the border of the cone or at the tip point not entirely be contained in the cone? $\endgroup$ Commented Nov 4, 2021 at 19:50
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    $\begingroup$ @michiganbiker898 See my update. $\endgroup$
    – Paul Frost
    Commented Nov 4, 2021 at 23:06

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