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

algebraic-topologygeneral-topologyhomology-cohomology

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!

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