Colimits and $\text{Hom}(-,G)$

Question

$\DeclareMathOperator{\Hom}{Hom}\DeclareMathOperator{\colim}{colim}$ Let $(A_i)_i$ be an inverse system of groups. If $G$ is a group, then we have a projective system of set $(\Hom (A_i,G))_i$. I would like to understand $$\colim_i\Hom(A_i,G).$$ Assume that for each $f_i:A_i\rightarrow G$, there exists a $j\geq i$ such that $$A_j\rightarrow A_i\rightarrow G$$ is trivial. Can we conclude that $\colim_i\Hom(A_i,G)$ is a singleton?

Answer 1

Yes, this version is correct: the elements of the colimit are the equivalence classes of homomorphisms $\varphi: A_i \rightarrow G$ (for any $i$) under the equivalence $[\varphi]=[\varphi \circ p_{i,j}]$ where $p_{i, j}: A_j \rightarrow A_i$ are the maps from the inverse system.

By your assumption, every homomorphism is thus equivalent to the trivial morphism $A_j \rightarrow G$ for some $j$.

So suppose that $\varphi: A_i \rightarrow G,$ $\psi: A_{i'}\rightarrow G$ are two homomorphisms. By the above, $\varphi$ is equivalent to the trivial map $A_j \rightarrow G$ for some $j$, and $\psi$ is equivalent to the trivial map $A_{j'} \rightarrow G$ for some $j'$. Taking $j'' \geq j, j',$ one can then see that the trivial maps $A_j \rightarrow G, A_{j'}\rightarrow G$ are both equivalent to the trivial map $A_{j''}\rightarrow G$. Thus, $\varphi$ and $\psi$ are ultimately equivalent, and the colimit is thus a singleton.