Definition. A reduced ordinary homology theory $\tilde{H}_*$ consists of functors $\tilde{H}_q$ from the homotopy category of nondegenerately based spaces to the category of Abelian groups that satisfy the following axioms:
DIMENSION. If $X$ is a point, then $$\tilde{H}_0(X)=\mathbb{Z}\quad \text{and}\quad \tilde{H}_q(X)=0\quad \text{for all}\quad q\neq 0.$$ EXACTNESS. If $i:A\rightarrow X$ is a cofibraion, then the following sequence is exact: $$\tilde{H}_q(A)\rightarrow \tilde{H}_q(X) \rightarrow \tilde{H}_q(X/A).$$ SUSPENSION. There is a natural isomorphism $$\Sigma :\tilde{H}_q(X) \rightarrow \tilde{H}_{q+1}(\Sigma X).$$ ADDITIVITY. If $X$ is the wedge of a set of nondegenerately based spaces $X_i$, then the inclusions $X_i\rightarrow X$ induce an isomorphism $$\bigoplus_i \tilde{H}_*(X_i) \rightarrow \tilde{H}_*(X).$$ WEAK EQUIVALENCE. If $f:X\rightarrow Y$ is a weak equivalence, then $f_*:\tilde{H}_*(X)\rightarrow \tilde{H}_*(Y)$ is an isomorphism.
To prove the Hurewicz theorem, I would like to show that
If $X$ is a wedge of $n$-spheres, then $\tilde{H}_q(X)=0$ for all $q\neq n$.
By the additivity axiom, it suffices to say that $\tilde{H}_q(S^n)=0$, where $S^n$ is an $n$-sphere. I already know the following claim:
For any $q\geq n$, $\tilde{H}_q(S^n)\cong H_{q-n}(*)$.
Here $*$ is the basepoint of $S^n$ and $H_*$ is the ordinary homology theory. If $q>n$, it follows immediately from the above. Therefore we need only check the case of $q<n$. But I could not do it. Is this purely deduced by the axioms?
For your information, I will state the reason why I would like to do. In the proof of the Hurewicz theorem, this book says that $\tilde{H}_n(X)\cong \tilde{H}_n(X^{n+1})$ for any based CW complex $X$. Here $X^q$ is the $q$-skeleton of $X$. To do this, I consider the reduced exact sequence $$\tilde{H}_{n+1}(X^{n+2}/X^{n+1})\xrightarrow{\partial}\tilde{H}_n(X^{n+1})\rightarrow \tilde{H}_n(X^{n+2}) \rightarrow \tilde{H}_n(X^{n+2}/X^{n+1}).$$ If the above was proved, $\tilde{H}_n(X^{n+2}/X^{n+1})$ and $\tilde{H}_{n+1}(X^{n+2}/X^{n+1})$ will be trivial, since $X^{n+2}/X^{n+1}$ is homotopic equivalent to $\bigvee_{j\in J}S^{n+2}$, where $J$ is the set of all $(n+2)$-cells. Then the exact sequence induces an isomorphism $\tilde{H}_n(X^{n+1})\rightarrow \tilde{H}_n(X^{n+2})$ as desired. Is this attempt impracticable? Might there be another way to do this?
Suppose you know that $$ \widetilde{H}_q(S^0) = \begin{cases} \mathbb{Z} & \text{if $q = 0$,}\\ 0 & \text{otherwise.} \end{cases} $$ Note that "otherwise" also includes negative values of $q$. Then this is the base case for an induction proof that $$ \widetilde{H}_q(S^n) = \begin{cases} \mathbb{Z} & \text{if $q = n$,}\\ 0 & \text{otherwise.} \end{cases} $$ The induction step invokes the suspension axiom and the fact that $S^{n+1} = \Sigma S^n$.