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?

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?