Injectivity of images of modules to direct limits

Question

Suppose $A_i$ is a $R$-module, with $i\in I$, $I$ being an indexed set. We assume there is a $R$-module homomorphism $\phi_{ij}:A_i\rightarrow A_j$ for every $i,j\in I$, $i\leq j$. This homomorphism has the following properties:

1. $\phi_{ik}=\phi_{jk}\circ\phi_{ij}$ for all $i\leq j\leq k$.
2. $\phi_{ii}$ is the identity of $A_i$

Now we take the equivalence relation $\sim$, with $a_i\sim a_j$ iff there exists a $k\in I,i,j\leq k$ such that $\phi_{ik}(a_i)=\phi_{jk}(a_j)$ with $a_i\in A_i$ and $a_j\in A_j$. Now take the direct limit $$\varinjlim A_i = \bigsqcup_{i}A_i {\bigg /}\sim$$ Now define $\phi_i: A_i \rightarrow \varinjlim A_i, a\mapsto [a]$. In words, $a$ gets sent to its equivalence class.
This is basic stuff, surrounding direct limits.

My question is: if we assume $\phi_{ij}$ is injective, how do I prove $\phi_i$ is injective as well?

EDIT: I got a hint: "In other words, we can view every $A_i$ as a subset of $\varinjlim A_i$."

Answer 1

Write $[a]_i$ for the equivalence class of $a\in A_i$ in the direct limit. We need to prove that $[a]_i=[0]_j$ for some $j$ entails $a=0$. This means that there is $k\ge i,j$ with $\phi_{ik}(a)=\phi_{jk}(0)=0$. Therefore $a=0$ by injectivity of $\phi_{ik}$.