# Showing the relationship between the reduced homology of $X$ and the reduced homology of the suspension of $X$

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!

• Yes $N$ is the north pole of $SX$. You can call it whatever you want. It is the class $\{0\} \times X$ in $CX = [0 , 1] \times X / \{0\} \times X$. Commented Nov 2, 2021 at 10:32
• @infinitelooper Thanks! I edited the question. Commented Nov 2, 2021 at 10:36
• What did you try ? Can you show that $CX_N$ is contractible ? Commented Nov 2, 2021 at 10:45
• @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. Commented Nov 2, 2021 at 10:51
• Nice. Now $SX - N$ is homeomorphic to $CX_N$. Can you find why ? Commented Nov 2, 2021 at 11:08

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

• 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? Commented Nov 4, 2021 at 19:50
• @michiganbiker898 See my update. Commented Nov 4, 2021 at 23:06