# 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?

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).