Let $I$ be a directed set and let $(A_i)_{i \in I}$ be a collection of abelian groups. Let $A = \varinjlim A_i$ be its direct limit. Suppose its maps are $\rho_{ij} : A_i \to A_j$ for $i \leq j$. I don't quite understand how the group operation on $A$ is defined. In exercise 8 of Dummit and Foote, page 268, this whole thing is carried out. I've got (a) and (b), but in (c) they assume the maps are group homomorphisms and define the group operation $A \times A \to A$ in the following way: say $a \in A_i, b \in A_j$. Define $\overline{a} + \overline{b} = \overline{\rho_{ik}(a) + \rho_{jk}(b)}$ for any $k \in I$ such that $k \geq i,j$. So my question is: why can we chose any $k$ such that $k \geq i,j$?
I'm sure this is going to be trivial. I tried working it out and I failed.
Never mind, I realized that it really is easy.
Let $a \in A_i$ and let $b \in A_j$. Let $k,k' \geq i,j$. I want to show that $\rho_{ik}(a) + \rho_{jk}(b) \sim \rho_{ik'}(a) + \rho_{jk'}(b)$. Well, let $\ell \in I$ be such that $\ell \geq k$ and $\ell \geq k'$. Then we get that \begin{equation} \begin{split} \rho_{k \ell}(\rho_{ik}(a) + \rho_{jk}(b)) \\ &= \rho_{k \ell}(\rho_{ik}(a)) + \rho_{k \ell}(\rho_{jk}(b)) \\ &= \rho_{i \ell}(a) + \rho_{j \ell}(b) \\ &= \rho_{k' \ell}(\rho_{i k'}(a)) + \rho_{j k'}(b)) \end{split} \end{equation}