From Dummit & Foote, pg. 268:
Let $I$ be an index set with a partial order. Suppose for every pair of indices $i, j \in I$ with $i \leq j$ there is a map $\rho_{ij} : A_i \to A_j$ such that the following hold:
i. $\rho_{jk} \circ \rho_{ij} = \rho_{ik}$
2. $\rho_{ii} = 1$ for all $i \in I$.
Let $B$ be the disjoint union of all the $A_i$. Define a relation $\sim$ on $B$ by $a \sim b$ if and only if there exists $k$ with $i,j \leq k$ and $\rho_{ik}(a) = \rho_{jk}(b)$.
It's easy to show that the relation is reflexive and symmetric but I can't see how we can derive a relationship between $a \sim b $ and $b\sim c$ since the order on the index set $I$ is partial and not neccessarily total. So how do I prove that this relation is indeed transitive as well?
The result is false as stated.
Let $I=\{2,3,5,6,15\}$, partially ordered by $i\preceq j$ iff $i\mid j$. For $i\in I$ let $A_i=\{i\}$. For $i,j\in I$ with $i\preceq j$ let $\rho_{ij}:A_i\to A_j$ be the only possible map: $\rho_{ij}(i)=j$.
Then $2\sim 3$, since $\rho_{2,6}(2)=\rho_{3,6}(3)=6$, and $3\sim 5$, since $\rho_{3,15}(3)=\rho_{5,15}(5)=15$, but $2\not\sim 5$.
It’s not sufficient that $I$ be partially ordered: $I$ must be directed by $\le$, so that for any $i,j\in I$ there is a $k\in I$ such that $i,j\le k$. Once you have that hypothesis, the result is pretty straightforward; leave a question if you have any trouble with it.