How to construct the Moore spectrum? I am trying to understand how the Moore spectrum is constructed. And in reading Foundations of Stable Homotopy Theory by David Barnes and Constanze Roitzheim, I see that in example 8.4.7 (pg 340) they show how to construct the Moore spectrum. The construction goes something like this:

Let $G$ be an abelian group. Assume that $G$ is obtained by the following short exact sequence: $$0 \to F_2 \xrightarrow{\rho} F_1 \to G \to 0$$ where $F_2$ is the free group on the set $I_2$ and $F_1$ is the free group on the set $I_1$. The set $I_1$ is a generating set of $G$ and $\rho(I_2)$ is the corresponding set of relations. No issues here since we can always find a group presentation.
Next we construct a cofiber sequence:
$$\bigvee_{I_2} \mathbb{S} \xrightarrow{r}\bigvee_{I_1}\mathbb{S} \to M(G)$$
where $\pi_0 (r)=\rho$.

Now I can see that $F_2=\bigoplus_{I_2} \mathbb{Z}$, $F_1=\bigoplus_{I_1} \mathbb{Z}$ and since $\pi_0(\mathbb{S})=\mathbb{Z}$ and $\pi_0$ preserves coproducts, I can see that $\rho$ sits between the following:
$$\pi_0\left(\bigvee_{I_2} \mathbb{S}\right) \xrightarrow{\rho} \pi_0\left(\bigvee_{I_1}\mathbb{S}\right)$$.
I don't however see that such an $r$ ought to exist such that $\pi_0(r)=\rho$. Why can we do this?
 A: What you're missing is that $[\mathbb{S},\mathbb{S}]=\mathbb{Z}$. Let now $R$ be the infinite matrix of integers representing $\rho$. Note that since $\rho$ takes value in $\bigoplus_{I_1}\mathbb{Z}\subseteq \prod_{I_1}\mathbb{Z}$, all of its columns have only finitely many non-zero values. So, for every column of $R$ we can construct a map $\mathbb{S}\to \bigvee_{I_1}\mathbb{S}$ sending $1\in\pi_0(\mathbb{S})$ to the column in $\bigoplus_{I_1}\mathbb{Z}=\pi_0\bigvee_{I_1}\mathbb{S}$ (for example by composing $\mathbb{S}\to \bigvee_{I_1'}\mathbb{S}\to \bigvee_{I_1}\mathbb{S}$, where $I'_1$ is the finite subset of $I_1$ where the column is non-zero and using that finite wedges are categorical products and so $[\mathbb{S},\bigvee_{I_1'}\mathbb{S}]=\prod_{I_1'}[\mathbb{S},\mathbb{S}]=\prod_{I_1'}\mathbb{Z}$).
Finally, using that the wedge is the categorical coproduct we can put it all together in a map
$$\bigvee_{I_2}\mathbb{S}\to \bigvee_{I_1}\mathbb{S}$$
as required.
In fact what I'm doing is essentially proving that the map
$$\pi_0:\left[\bigvee_{I_2}\mathbb{S},\bigvee_{I_1}\mathbb{S}\right]\to \mathrm{Hom}\left(\bigoplus_{I_2}\mathbb{Z},\bigoplus_{I_1}\mathbb{Z}\right)$$
is an isomorphism (injectivity is quite easy to show).
