Finding an example of a functor $\alpha:\mathbb{N}\to Mod(\mathbb{Z})$ such that the colimit of $\alpha$ is $0$.

Question

$\DeclareMathOperator{\colim}{colim}$ $\DeclareMathOperator{\Hom}{Hom}$ $\DeclareMathOperator{\Mod}{Mod}$ As described in the title, I'm trying to construct an example of a functor $\alpha: \mathbb{N}\to Mod(\mathbb{Z})$ such that $\colim \alpha \simeq 0$, i.e. a sequence of $\mathbb{Z}-$module and morphisms $\alpha(0)\to\alpha(1)\to\alpha(2)\to\cdots$ with trivial colimit. But there's a catch, no $\alpha(n)$ can be $0$, and similarly none of the morphisms $\alpha(i)\to\alpha(j)$ can be $0$.

I don't really have any how to start, I'm not really interested in just having an example, I'm more interested in understanding how one goes about making one up. All I can say is that since $\Hom_{\Mod(\mathbb{Z})}(\colim_{i}\alpha(i),M)\simeq \lim_i\Hom_{\Mod(\mathbb{Z})}(\alpha(i),M)$, if $\colim_i\alpha(i)\simeq 0$, then we must have $\lim_i\Hom_{\Mod(\mathbb{Z})}(\alpha(i),M)\simeq 0$ for any $M\in\Mod(\mathbb{Z})$.

Answer 1

Another example, which matches the intuition of Aphelli's comment: Define $\mathbb{Z}[\mathbb{N}]$ to be functions from $\mathbb{N} \to \mathbb{Z}$ which are zero except at finitely many points. We can make it into an abelian group by defining addition pointwise.

Then define $\alpha(n) = \mathbb{Z}[\mathbb{N}]$ for each $n$, and let the $n$'th arrow be the projection mapping the first $n$ coordinates to zero (thinking of elements in $\mathbb{Z}[\mathbb{N}]$ as sequences $(a_0, a_1, \ldots)$).

Then commutativity of the triangles in a cocone on $\alpha$ tells you that all maps to the vertex of the cocone must be zero.

Answer 2

Consider $\alpha(i)\equiv\mathbb Z\oplus\mathbb Z$ and the morphisms from $\alpha(i)$ to $\alpha(i+1)$ to be$$\begin{cases} P_1(x,y)=x & \text{ if } i \text{ is even } \\ P_2(x,y)=y & \text{ if } i \text{ is odd} \end{cases}$$

In particular we have $P_1\circ P_2=0$ and $P_2\circ P_1=0$, and hence we must have if there is a compatible system of morphisms from $\alpha(i)$ to $M$,the morphism from $\alpha(i)$ to $M$ factors through the morphism from $\alpha(i+2)$ to $M$ hence must be trivial.