First we prove the equivalence of (1) and (2).  Suppose (1) holds and let $i \in I$. Corresponding to the map $f_i: M_i \to M$, we can choose $g: M_i \to Q$ as in (1).  Since $M_i$ and $Q$ are of finite presentation, so is $\Coker(g)$.  Then by Lemma \ref{lemma-domination}, $f_i : M_i \to M$ factors through $g: M_i \to Q$, say $f_i = h \circ g$ for some $h: Q \to M$.  Then since $Q$ is finitely presented, $h$ factors through $M_j \to M$ for some $j \geq i$, say $h = f_j \circ h'$ for some $h': Q \to M_j$.  In total we have a commutative diagram $$ \xymatrix{   &  M  & \\ M_i \ar[dr]_g \ar[ur]^{f_i} \ar[rr]^{f_{ij}} & & M_j \ar[ul]_{f_j} \\   & Q  \ar[ur]_{h'} & } $$ Thus $f_{ij}$ dominates $g$.  But $g$ dominates $f_i$, so $f_{ij}$ dominates $f_i$. \medskip\noindent Conversely, suppose (2) holds.  Let $P$ be of finite presentation and $f: P \to M$ a module map.  Then $f$ factors through $f_i: M_i \to M$ for some $i \in I$, say $f = f_i \circ g'$ for some $g': P \to M_i$. Choose by (2) a $j \geq i$ such that $f_{ij}$ dominates $f_i$.  We have a commutative diagram $$ \xymatrix{ P \ar[d]_{g'} \ar[r]^{f}           & M  \\ M_i \ar[ur]^{f_i} \ar[r]_{f_{ij}} & M_j \ar[u]_{f_j} } $$ From the diagram and the fact that $f_{ij}$ dominates $f_i$, we find that $f$ and $f_{ij} \circ g'$ dominate each other.  Hence taking $g = f_{ij} \circ g' : P \to M_j$ works. \medskip\noindent Next we prove (2) is equivalent to (3).  Let $i \in I$.  It is always true that $f_i$ dominates $f_{ik}$ for $k \geq i$, since $f_i$ factors through $f_{ik}$.  If (2) holds, choose $j \geq i$ such that $f_{ij}$ dominates $f_i$.  Then since domination is a transitive relation, $f_{ij}$ dominates $f_{ik}$ for $k \geq i$. All $M_i$ are of finite presentation, so $\Coker(f_{ik})$ is of finite presentation for $k \geq i$.  By Lemma \ref{lemma-domination}, $f_{ij}$ factors through $f_{ik}$ for all $k \geq i$. Thus (2) implies (3).  On the other hand, if (3) holds then for any $R$-module $N$, $f_{ij} \otimes_R \text{id}_N$ factors through $f_{ik} \otimes_R \text{id}_N$ for $k \geq i$.  So $\Ker(f_{ik} \otimes_R \text{id}_N) \subset \Ker(f_{ij} \otimes_R \text{id}_N)$ for $k \geq i$.  But $\Ker(f_i \otimes_R \text{id}_N: M_i \otimes_R N \to M \otimes_R N)$ is the union of $\Ker(f_{ik} \otimes_R \text{id}_N)$ for $k \geq i$.  Thus $\Ker(f_i \otimes_R \text{id}_N) \subset \Ker(f_{ij} \otimes_R \text{id}_N)$ for any $R$-module $N$, which by definition means $f_{ij}$ dominates $f_i$. \medskip\noindent It is trivial that (3) implies (4) implies (5).  We show (5) implies (3).  Let $N = \prod_{s \in I} M_s$. If (5) holds, then given $i \in I$ choose $j \geq i$ such that $$ \Im( \Hom(M_j, N) \to  \Hom(M_i, N)) = \Im( \Hom(M_k, N) \to  \Hom(M_i, N)) $$ for all $k \geq j$.  Passing the product over $s \in I$ outside of the $\Hom$'s and looking at the maps on each component of the product, this says $$ \Im( \Hom(M_j, M_s) \to  \Hom(M_i, M_s)) = \Im( \Hom(M_k, M_s) \to   \Hom(M_i, M_s)) $$ for all $k \geq j$ and $s \in I$.  Taking $s = j$ we have $$ \Im( \Hom(M_j, M_j) \to  \Hom(M_i, M_j)) = \Im( \Hom(M_k, M_j) \to   \Hom(M_i, M_j)) $$ for all $k \geq j$.  Since $f_{ij}$ is the image of $\text{id} \in  \Hom(M_j, M_j)$ under $\Hom(M_j, M_j) \to  \Hom(M_i, M_j)$, this shows  that for any $k \geq j$ there is $h \in \Hom(M_k, M_j)$ such that $f_{ij} = h \circ f_{ik}$.  If $j \geq k$ then we can take $h = f_{kj}$. Hence (3) holds.