Let $\mathsf{A}$ be an Abelian category (perhaps vector spaces or modules over your favorite ring), and let $\mathsf{A}(x,y)$ denote the set of morphisms in $\mathsf{A}$ from an object $x$ to another object $y$.



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> Does there exist a non-trivial partial order on the morphism-sets of $\mathsf{A}$ which behaves well with respect to composition and happens to be monotone with rank, whenever rank is defined?
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More specifically, do there exist partial orders $<$ so that for $f,f':x \to y$ and $g, g':y \to z$, we have:


1. $f < f'$ and $g < g'$ implies $gf < g'f'$ in $\mathsf{A}(x,z)$, and
2. if $f < f'$ then there is a monomorphism from the image of $f$ to image of $f'$?


Probably the answer is no, but I'm not sure how to prove that no such thing could possibly exist --- even in the category of vector spaces over $\mathbb{R}$ or $\mathbb{C}$.



Poset-enrichment of abelian categories
Suppose that $V, W$ are vector spaces and $f, g : V \to W$ are two parallel morphisms such that $f \le g$, for some preorder $\le$ satisfying your conditions. 



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> **Claim:** $f$ is a scalar multiple of $g$.
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*Proof.* The first condition implies that if $v : 1 \to V$ is any vector in $V$ (here $1$ denotes the $1$-dimensional vector space) then $f \circ v \le g \circ v$. The second condition implies that there is a monomorphism from the image of $f \circ v$ to the image of $g \circ v$ (**Edit:** here I interpret the condition to mean that the monomorphism must be compatible with the inclusion into $W$, but as Eric Wofsey's answer shows, this extra assumption is unnecessary). This is equivalent to the condition that for all $v \in V$ there is a scalar $\lambda(v)$, possibly zero, such that


$$f(v) = \lambda(v) g(v).$$


Observe that


$$f(v + v') = \lambda(v + v') (g(v) + g(v')) = f(v) + f(v') = \lambda(v) g(v) + \lambda(v') g(v').$$


Hence if $g(v)$ and $g(v')$ are linearly independent it follows that $\lambda(v) = \lambda(v') = \lambda(v + v')$. Now we split into cases.


**Case:** $\text{im}(g)$ has dimension at least $2$. Then the equivalence relation on vectors in $\text{im}(g)$ generated by "linearly independent" identifies every vector, and it follows that $\lambda(v)$ is a constant $\lambda$ and $f = \lambda g$. 


**Case:** $\text{im}(g)$ has dimension $1$. Then $\text{im}(f)$ is contained in $\text{im}(g)$, and hence has dimension at most $1$. Picking $v\_0 \in V$ such that $g(v\_0) \neq 0$, we then have $f(v\_0) = \lambda(v\_0) g(v\_0)$, and hence it again follows that $f = \lambda g$ (where $\lambda = \lambda(v\_0)$).


**Case:** $\text{im}(g)$ has dimension $0$. Then $f = g = 0$.


This concludes the proof. $\Box$


So possibilities are limited. Over any field, we can take $f \le g$ iff $f = \lambda g$ for some $\lambda$. This is a preorder, but it's pretty boring: linear maps which are nonzero scalar multiples of each other are isomorphic, and the only order relation which isn't an isomorphism is that $0 \le f$ for any nonzero $f$. Over $\mathbb{R}$ we can take $f \le g$ iff $f = \lambda g$ for some $0 \le \lambda \le 1$. This is even a partial order, but it's still not very interesting. 


