I am trying to understand how the Moore spectrum is constructed. And in reading [Foundations of Stable Homotopy Theory](https://www.kent.ac.uk/smsas/personal/csrr/stablemodelcatsCUP.pdf) 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?



How to construct the Moore spectrum?
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).


