Bruhat covers give canonical Verma embeddings, and composites are inclusions

For every $w\in W$, fix an embedding $j_w\text{\hspace{0.17em}⁣}:M(w\circ \lambda )\hookrightarrow M(\lambda )$ with image $S_w$, taking $j_e$ to be the identity. There are only finitely many choices. For every comparable pair $u\ge v$, define ${\iota }_{u\to v}=j_v^{-1}\circ j_u$, where $j_v^{-1}$ is the inverse from $S_v$ to $M(v\circ \lambda )$; [F1] gives $S_u\subseteq S_v$, so this is well-defined and $j_v{\iota }_{u\to v}=j_u$. These are precisely the literal submodule inclusions transported to the abstract Verma copies. For a cover $x⊳y$, the map is nonzero and injective and spans the one-dimensional Hom space by [F2]. Its image is proper: otherwise the two Verma modules would have the same highest weight, contradicting $x\circ \lambda \mathrm{\ne }y\circ \lambda$ by [F3].

For $x⊳m⊳y$, the defining equations give $j_y{\iota }_{m\to y}{\iota }_{x\to m}=j_m{\iota }_{x\to m}=j_x=j_y{\iota }_{x\to y}$. Since $j_y$ is injective, ${\iota }_{m\to y}{\iota }_{x\to m}={\iota }_{x\to y}$. The same argument for $m^{\mathrm{\prime }}$ shows that the two diamond composites are equal as maps, rather than merely proportional.

More generally, for $x\ge y\ge z$ one has $j_z{\iota }_{y\to z}{\iota }_{x\to y}=j_y{\iota }_{x\to y}=j_x=j_z{\iota }_{x\to z}$, so injectivity of $j_z$ proves ${\iota }_{y\to z}{\iota }_{x\to y}={\iota }_{x\to z}$. Iterating this equality gives path independence, including the claimed length-gap-two case. The normalization depends on the chosen $j_w$ on abstract copies; the submodules $S_w$ and their literal inclusions are canonical. Arbitrarily rescaled cover maps need not have equal diamond composites.