Theorem: Let $SX$ be the suspension of $X$. Then, $\wt{H}_n(SX) \cong \wt{H}_{n-1}(X)$ for all $n$.

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Proof: Let $SX = X \times I$ with $X \times \set{0}$ and $X \times \set{1}$ both identified to (distinct) points. Let $A = X \times [0,\frac{3}{4}]$ and $B = X \times [\frac{1}{4},1]$ (any overlap will do). Then, $A \cup B$ is certainly $SX$.

$$A$$ and $$B$$ are both contractible, having $$\wt{H}_n(A) = \wt{H}_n(B) = 0$$ for all $$n$$, and $$A \cap B$$ deformation retracts onto $$X$$, so $$\wt{H}_n(A \cap B) = \wt{H}_n(X)$$. Meyer-Vietoris gives:$$

\dots \rightarrow 0 \rightarrow \wt{H}n(SX) \rightarrow \wt{H}{n-1}(X) \rightarrow 0 \rightarrow \dots

$$

for all $n$, and the statement is immediate.