# Reduced homology of the suspension

This is an exercise from Algebraic Topology book of Hatcher:

Exercise 2.1.20, pg. 132: Show that $$\tilde H_n(X) ≈ \tilde H_{n+1}(SX)$$ for all $$n$$, where $$SX$$ is the suspension of $$X$$.

I offer a somehow alternative proof, which can be geometrically checked by the picture of the suspension (of the circle):

Note that the $$SX$$ can be realized as the union of two cones $$CX_N$$ and $$CX_S$$ with their bases identified.

For the pair $$(SX, CX_N)$$, we have a long exact sequence: $$\ldots \to H_i(CX_N) \to H_i(SX) \xrightarrow{f} H_i(SX,CX_N) \to H_{i-1}(CX_N) \to \ldots$$ Then $$f$$ is clearly an isomorphism since $$CX_N$$ is contractible.

By Excision Theorem, the inclusions $$(SX-N, CX_N- N) \hookrightarrow (SX, CX_N)$$ induce isomorphisms $$H_i(SX-N,CX_N- N) \cong H_i(SX, CX_N)$$

Also for the pair $$(SX-N,CX_N- N)$$, consider the long exact sequence $$\ldots \to H_i(SX-N) \to H_i(SX-N, CX_N -N) \xrightarrow{g} H_i(CX_N -N) \to H_{i-1}(SX-N) \to \ldots$$

Since $$SX -N$$ is contractible, $$g$$ is also an isomorphism.

Now, we are done by gathering isomorphisms $$f$$ and $$g$$ with the fact $$CX_N -N \simeq X$$.

Is there any gap in the proof?

(2) $$\tilde{H}_n = H_n$$ for $$n > 0$$.
(3) $$\tilde{H}_0(\ast) = 0$$.