Relationship between homology of suspension of $X$ and $X$ The exercise is the following:

Show that, for any homology theory (satisfying the usual axioms), there is a natural isomorphism $ \tilde{H_i}(X) \rightarrow \tilde{H}_{i+1}(\Sigma X)$.

Well, I tried using the long exact sequence:
$$...\rightarrow \tilde{H}_{i+1}(X_{1/2}) \rightarrow \tilde{H}_{i+1}(\Sigma X) \rightarrow H_{i+1}(\Sigma X, X_{1/2}) \rightarrow \tilde{H}_{i}(X_{1/2}) \rightarrow \tilde{H}_{i}(\Sigma X) \rightarrow ...$$
where $X_{1/2}$ is $\{1/2\} \times X$ in the suspension. Then, I tried to compute $H_{i+1}(\Sigma X, X_{1/2})$. For that, I enlarged a bit $X_{1/2}$ to $\overline{X}_{1/2}:=[\frac{1}{4}, \frac{3}{4}] \times X$ and by excision (cutting off a small neighbourhood of $X_{1/2}$) and the long exact sequence for the pair $(C, X_{3/4})$, where $C$ is the upper part of the "cone" that is left, I managed to prove that:
$$H_{i+1}(\Sigma X, X_{1/2}) \cong H_{i}(X) \oplus H_i(X)$$
but I got stuck after this.
 A: I'm going to use a different notation mostly because I'm largely copying this out of an old homework of mine.
Let $C_+^n$ and $C_-^n$ be the cones in $\Sigma X$, let $U$ be a neighborhood around the cone point with $\overline{U}\subset Int(C_-^n)$. Then $U$ can be excised, so 
$$\widetilde{H}_q(\Sigma X,C_-^n)\simeq \widetilde{H}_q(\Sigma X\setminus U,C_-^n\setminus U).$$ 
Since $(\Sigma X\setminus U,C_-^n\setminus U)$ deformation retracts to $(C_+^n,X)$ we get 
$$\widetilde{H}_q(\Sigma X,C_-^n)\simeq\widetilde{H}_q(C_+^n,X)$$
Since $CX$ is contractible, $C_+^n\simeq C_-^n\simeq \{pt.\}$, and thus $\widetilde{H}_\ast(C_+^n)\simeq\widetilde{H}_\ast(C_-^n)\simeq 0$. Then the long exact sequence
$$ \dots\to \widetilde{H}_q(C_-^n)\to\widetilde{H}_q(\Sigma X)\to \widetilde{H}_q(\Sigma X,C_-^n)\to\widetilde{H}_{q-1}(C_-^n)\to\dots $$
gives
$$ \dots\to 0 \to\widetilde{H}_q(\Sigma X)\to \widetilde{H}_q(\Sigma X,C_-^n)\to 0 \to\dots $$
so that $\widetilde{H}_q(\Sigma X)\simeq\widetilde{H}_q(\Sigma X, C_-^n)$. Similarly, the long exact sequence
$$ \dots\to \widetilde{H}_q(C_+^n)\to\widetilde{H}_q(C_+^n,X)\to \widetilde{H}_{q-1}(X)\to\widetilde{H}_{q-1}(C_+^n)\to\dots $$
gives
$$ \dots\to 0 \to\widetilde{H}_q(C_+^n,X)\to \widetilde{H}_{q-1}(X)\to 0 \to\dots $$
so that $ \widetilde{H}_q(C_+^n,X)\simeq \widetilde{H}_{q-1}(X) $. 
Thus, $\widetilde{H}_q(\Sigma X)\simeq\widetilde{H}_q(\Sigma X, C_-^n)\simeq\widetilde{H}_q(C_+^n,X)\simeq \widetilde{H}_{q-1}(X) $.
A: Let $CX$ be the cone on $X$. We can write $\Sigma X$ as the following pushout: 
$$\require{AMScd}
\begin{CD}
(x_0\times I)\cup (X\times\{1\}) @>>> * \\
@VVV @VVV \\
CX @>>> \Sigma X
\end{CD}$$
where the left map is the inclusion. The reduced  Mayer Vietoris sequence corresponding to that pushout contains the following segment:
$$\tilde{H}_{i+1}(*)\oplus \tilde{H}_{i+1}(CX) \to \tilde{H}_{i+1}(\Sigma X) \to \tilde{H}_{i}((x_0\times I)\cup (X\times\{1\})) \to \tilde{H}_{i}(*)\oplus \tilde{H}_{i}(CX)$$
Note that $(x_0\times I)\cup (X\times\{1\})$ deformation retracts onto $X$, and $CX$ is contractible. Then the above sequence reads
$$0\to \tilde{H}_{i+1}(\Sigma X) \to \tilde{H}_{i}(X)\to 0.$$
This gives the desired isomorphism.
A: Let $\Sigma X = X \times [0,1]$, with $X \times \{0\}$ identified to a point and $X \times \{1\}$ identified to a point.
Let $A = X \times [0,\frac{3}{4}], B = X \times [\frac{1}{4},1]$ with the same identifications.
Then $A$ and $B$ are contractible, and $A \cap B \simeq X$.
Reduced Mayer-Vietoris gives the long exact sequence
$$\cdots \rightarrow \tilde H_{i+1}(A) \oplus \tilde H_{i+1}(B) \rightarrow \tilde H_{i+1}(\Sigma X) \rightarrow \tilde H_i(A \cap B) \rightarrow \tilde H_i(A) \oplus \tilde H_i(B) \rightarrow \cdots$$
and for all $i$, this is
$$\cdots \rightarrow 0 \rightarrow \tilde H_{i+1}(\Sigma X) \rightarrow \tilde H_i(X) \rightarrow 0 \rightarrow \cdots$$
hence giving the desired isomorphism.
