The Problem: Let $I$ be a nonempty index set with a partial order $\leq$. For each $i\in I$, let $A_i$ be an additive abelian group. Suppose for every $i, j\in I$ there is some $k\in I$ such that $k\geq i, j$.
Suppose for every pair of indices $i, j$ with $i\leq j$ there is a map $\rho_{ij}: A_i\to A_j$ such that: i) $\rho_{jk}\circ\rho_{ij}=\rho_{ik}$ whenever $i\leq j\leq k$, and ii) $\rho_{ii}=1$ for all $i\in I$. Suppose $a\sim b$ iff there exists $k$ with $i, j\leq k$ and $\rho_{ik}(a)=\rho_{jk}(b)$, for $a\in A_i$ and $b\in A_j$. This is an equivalence relation on $\cup A_i$. Let $A$ denote the set of all the equivalence classes and let $\overline{x}$ denote the class of $x$.
Assume all $\rho_{ij}$ are group homomorphisms. For $a\in A_i$, $b\in A_j$, define the operation $\overline{a}+\overline{b}=\overline{\rho_{ik}(a)+\rho_{jk}(b)}$. This makes $A$ into an abelian group. If $C$ is any abelian group such that for each $i\in I$ there is a homomorphism $\varphi_i: A_i\to C$ with $\varphi_i=\varphi_j\circ\rho_{ij}$ whenever $i\leq j$, prove there is a unique homomorphism $\mathbf{\varphi: A\to C}$ such that $\mathbf{\varphi\circ\rho_i=\varphi_i}$ for all $\mathbf{i\in I}$.
Source: Abstract Algebra, $\mathit{3^{rd}}$ edition by Dummit and Foote.
My Attempt: Define $\varphi_A\to C$ by $\overline{a}\mapsto \varphi_i(a)$, where $a\in A_i$. We must first show that $\varphi$ is well defined.
Suppose $a\sim a'$, i.e., there is some $k\geq i, j$ such that $\rho_{ik}(a)=\rho_{jk}(a')$. Note $\varphi(\overline{a})=\varphi_i(a)$, $\varphi(\overline{a'})=\varphi_j(a')$. We must show $\varphi_i(a)=\varphi_j(a')$.
Case I: Suppose $i\geq j$; then $\varphi_j=\varphi_i\circ\rho_{ji}$, thus $\varphi_j(a')=\varphi_i\circ\rho_{ji}(a')$. Recall $\rho_{ik}\circ\rho_{ji}=\rho_{jk}$, thus $\rho_{jk}(a')=\rho_{ik}\circ\rho_{ji}(a')$, i.e., $\rho_{ik}(a)=\rho_{ik}\circ\rho_{ji}(a')$. Screeching halt.
Obviously I want $\rho_{ik}$ to be injective, because then $\rho_{ik}(a)=\rho_{ik}\circ\rho_{ji}(a')$ would imply $a=\rho_{ji}(a')$ and the rest would follow easily. But is that the case? Any help would be greatly appreciated.