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 $X$.
This question is extensively discussed on Natural isomorphism $\tilde H_i(X) \xrightarrow{\cong} \tilde H_{i+1}(\Sigma X)$ where $\Sigma X$ is the suspension of $X$. and Reduced Homology on unreduced suspension.
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?
Best Answer
Your proof is essentially correct. However, you have to use three facts:
(1) There exists a long exact sequence for the reduced homology groups, and this has to be used in your proof.
(2) $\tilde{H}_n = H_n$ for $n > 0$.
(3) $\tilde{H}_0(\ast) = 0$.