TITLE: Reduced Homology on unreduced suspension
QUESTION [6 upvotes]: I want to establish the result for any homology theory, there is a natural isomorphism which gets us $\tilde{H}_{p+1}(\Sigma X) \cong \tilde{H}_p(X)$. Most places use the Mayer-Vietoris sequence to get this, but since I haven't covered that yet, I would like to do it using the Eilenberg-Steenrod axioms for homology. 
Edit: Note that I'm defining $\Sigma X$ as the space where I shrink $X \times \{0\}$ and $X \times\{1\}$ to points. Wikipedia calls this $SX$.

REPLY [3 votes]: Disclaimer: The following is a proof that does not use the Mayer-Vietoris sequence, but neither does it strictly lie within the framework of the Eilenberg-Steenrod axioms for homology (as there is mention of chain complexes), which is being requested for by the OP.

Let $ X $ be a (non-empty) topological space. Contracting the base $ X \times \{ 0 \} $ of the cone $ C X $, we obtain the unreduced suspension $ \Sigma X $. In other words,
  $$
\Sigma X \stackrel{\text{homeo}}{\cong} C X / (X \times \{ 0 \}).
$$
  Consider the short exact sequence of chain complexes
  $$
    0
\to {\Delta_{\bullet}}(X \times \{ 0 \})
\to {\Delta_{\bullet}}(C X)
\to {\Delta_{\bullet}}(C X,X \times \{ 0 \})
\to 0,
$$
  which in dimension $ -1 $ is
  $$
0 \to \mathbb{Z} \stackrel{\text{id}}{\to} \mathbb{Z} \to 0 \to 0.
$$
  There is a corresponding long exact sequence of reduced homology groups:
  $$
    \cdots
\to {\tilde{H}_{n + 1}}(C X)
\to {\tilde{H}_{n + 1}}(C X,X \times \{ 0 \})
\to {\tilde{H}_{n}}(X \times \{ 0 \})
\to {\tilde{H}_{n}}(C X)
\to \cdots.
$$
  As $ C X $ is contractible, we have $ {\tilde{H}_{n}}(C X) = 0 $ for all $ n \in \mathbb{Z} $. Hence,
  $$
\forall n \in \mathbb{Z}: \quad
{\tilde{H}_{n + 1}}(C X,X \times \{ 0 \}) \cong {\tilde{H}_{n}}(X \times \{ 0 \}).
$$
  This, however, yields
  \begin{align*}
\forall n \in \mathbb{Z}: \quad
        {\tilde{H}_{n + 1}}(\Sigma X)
& \cong {\tilde{H}_{n + 1}}(C X / (X \times \{ 0 \})) \quad
        (\text{As $ \Sigma X \stackrel{\text{homeo}}{\cong} C X / (X \times \{ 0 \}) $.}) \\
& \cong {H_{n + 1}}(C X,X \times \{ 0 \}) \quad
        (\text{As $ (C X,X \times \{ 0 \}) $ is a good pair.}) \\
& \cong {\tilde{H}_{n + 1}}(C X,X \times \{ 0 \}) \quad
        (\text{As $ X \neq \varnothing $.}) \\
& \cong {\tilde{H}_{n}}(X \times \{ 0 \}) \quad (\text{As explained above.}) \\
& \cong {\tilde{H}_{n}}(X).
\end{align*}
  The claim is therefore established.

The definition of a good pair can be found on Page 114 of Allen Hatcher’s Algebraic Topology. On Page 118 of the same book, there is an explanation of the fact that
$$
\forall n \in \mathbb{Z}: \quad
{\tilde{H}_{n}}(X,A) \cong {H_{n}}(X,A)
$$
for all pairs $ (X,A) $ such that $ A \neq \varnothing $.