Like me, Daniel asked to restrict to acyclic diagrams. However, I realized that this restriction is ineffectual for my entire question. (Thus, his algorithm can't need it.) An exact sequenceYou can convert any diagram into an acyclic one using the fact that$$0 \longrightarrow A \stackrel{f}{\longrightarrow} B stackrel{\phi}{\longrightarrow} A' \longrightarrow 0$$makes $f$ \phi$into an isomorphism. If$\mathcal{D}$is a diagram, so you can thereafter treat an early should first triangulate all commutative polygons in to make commutative triangles. Then make three copies$A, A', A''$of each object$A$as the same as A \in \mathcal{D}$ together with isomorphisms$$A \stackrel{\phi}{\longrightarrow} A' \stackrel{\phi'}{\longrightarrow} A'.$$Then a late object homomorphism $f:A \to B$ in an acyclic diagram that has these four terms. So you $\mathcal{D}$ can cheat; you be expressed acyclically as this commutative square:$$\begin{matrix} A & \stackrel{f}{\longrightarrow} & B' \\\downarrow && \downarrow \\A' & \stackrel{f'}{\longrightarrow} & B'' \end{matrix}$$Finally, a commutative triangle $h = f \circ g$ can encode any diagram be expressed as a commutative square $h' \circ \phi = f' \circ g$. Or if $f$ and $g$ are an acyclic oneexact pair, you can require that $f'$ and $g$ make an exact pair.
This answer is a response to Daniel Litt's answer above. First, let me distill his point. Given a diagram of finite-dimensional vector spaces, every term $A$ has a dimension which is a non-negative integer. In addition, every morphism $f:A \to B$ has a non-negative rank which is at most $\min(\dim A,\dim B)$. If a pair of morphisms $$A \stackrel{f}{\longrightarrow} B \stackrel{g}{\longrightarrow} C,$$ then you get a linear relation $$\mathrm{rank}\ f + \mathrm{rank}\ g = \dim B.$$ Thus, there is an integer programming problem to express whether the dimensions and ranks are feasible. Since such an integer programming problem is homogeneous, you can instead treat it as a rational linear programming problem and later clear denominators. There is an algorithm to determine feasibility, even a polynomial time algorithm. As Daniel points out, this is enough to establish the nine lemma in the special case of finite-dimensional vector spaces. The argument/algorithm even works in many categories other than finite-dimensional vector spaces. For instance it works for finite abelian groups.
Like me, Daniel asked to restrict to acyclic diagrams. However, I realized that this restriction is ineffectual for my entire question. (So in particular Thus, his algorithm can't need it.) An exact sequence $$0 \longrightarrow A \stackrel{f}{\longrightarrow} B \longrightarrow 0$$ makes $f$ into an isomorphism, so you can thereafter treat an early object $A$ as the same as a late object $B$ in an acyclic diagram that has these four terms. So you can cheat; you can encode any diagram as an acyclic one.
Daniel also says without explanation that if there is a solution to the dimension and rank equations, then there are ways to fill in all of the maps. But without a lot more work, I don't think that this inference is reasonable. The hard part is satisfying commutative triangles. It is certainly not true that there is simply a canonical choice for each map using a skeleton $\{\mathbb{F}^n\}$\{k^n\}$of the category of finite-dimensional vector spaces. Because, if$f$and$g$each have some rank, then their composition$f \circ g$might also have some desired rank, and that rank depends on the choices of$f$and$g$. One interesting remark here is that $$\mathrm{rank}\ f \circ g \le \min(\mathrm{rank}\ f,\mathrm{rank}\ g),$$ and there is a similar inequality on the other side. However, I think that there is more going on than that. Finally, it is worth giving a simple example to show that feasibility in finite-dimensional vector spaces is not the same as feasibility in vector spaces. Given the exact sequence $$0 \to longrightarrow A \to longrightarrow A \to longrightarrow A \to longrightarrow 0,$$ you can conclude (using the dimension equations that Daniel suggests) that$A = 0$if it is finite-dimensional. But if it is infinite-dimensional, then there are non-trivial solutions. 1 This answer is a response to Daniel Litt's answer above. First, let me distill his point. Given a diagram of finite-dimensional vector spaces, every term$A$has a dimension which is a non-negative integer. In addition, every morphism$f:A \to B$has a non-negative rank which is at most$\min(\dim A,\dim B)$. If a pair of morphisms $$A \stackrel{f}{\longrightarrow} B \stackrel{g}{\longrightarrow} C,$$ then you get a linear relation $$\mathrm{rank}\ f + \mathrm{rank}\ g = \dim B.$$ Thus, there is an integer programming problem to express whether the dimensions and ranks are feasible. Since such an integer programming problem is homogeneous, you can instead treat it as a rational linear programming problem and later clear denominators. There is an algorithm to determine feasibility, even a polynomial time algorithm. As Daniel points out, this is enough to establish the nine lemma in the special case of finite-dimensional vector spaces. The argument/algorithm even works in many categories other than finite-dimensional vector spaces. For instance it works for finite abelian groups. Like me, Daniel asked to restrict to acyclic diagrams. However, I realized that this restriction is ineffectual for my entire question. (So in particular his algorithm can't need it.) An exact sequence $$0 \longrightarrow A \stackrel{f}{\longrightarrow} B \longrightarrow 0$$ makes$f$into an isomorphism, so you can thereafter treat an early object$A$as the same as a late object$B$in an acyclic diagram that has these four terms. So you can cheat; you can encode any diagram as an acyclic one. Daniel also says without explanation that if there is a solution to the dimension and rank equations, then there are ways to fill in all of the maps. But without a lot more work, I don't think that this inference is reasonable. The hard part is satisfying commutative triangles. It is certainly not true that there is simply a canonical choice for each map using a skeleton$\{\mathbb{F}^n\}$of the category of finite-dimensional vector spaces. Because, if$f$and$g$each have some rank, then their composition$f \circ g$might also have some desired rank, and that rank depends on the choices of$f$and$g$. One interesting remark here is that $$\mathrm{rank}\ f \circ g \le \min(\mathrm{rank}\ f,\mathrm{rank}\ g),$$ and there is a similar inequality on the other side. However, I think that there is more going on than that. Finally, it is worth giving a simple example to show that feasibility in finite-dimensional vector spaces is not the same as feasibility in vector spaces. Given the exact sequence $$0 \to A \to A \to A \to 0,$$ you can conclude (using the dimension equations that Daniel suggests) that$A = 0\$ if it is finite-dimensional. But if it is infinite-dimensional, then there are non-trivial solutions.