The A2 regular integral-dominant Verma embedding poset

Put $\nu =\lambda +\rho$. Regularity makes the dot orbit free, so its members are indexed by the six elements of $S_3$. For $u\in W$ and a positive root $\alpha$, $⟨u\nu ,{\alpha }^{\vee }⟩=⟨\nu ,(u^{-1}\alpha {)}^{\vee }⟩$ is positive exactly when $u^{-1}\alpha$ is positive; integrality makes every such positive value a positive integer.

Write $s=s_{{\alpha }_1}$, $t=s_{{\alpha }_2}$, and $w_0=sts=tst$. The positive-reflection pairs are $(e,s),(e,t),(e,w_0),(s,st),(s,ts),(t,st),(t,ts),(st,w_0),(ts,w_0)$. The rank inequalities defining Bruhat order give precisely these reflection comparisons (all except $(e,w_0)$ are covers). Thus checking the three positive roots of $A_2$ shows that the reflections satisfying step 1.1 are exactly the pairs $u<s_{\alpha }u$ in Bruhat order. In particular every strong-linkage edge lies in Bruhat order, while every Bruhat cover is one of these positive-integral reflection edges. Taking transitive closures and applying the BGG criterion proves the nonzero-homomorphism equivalence; the cited injectivity lemma turns every such map into an embedding. This proves both directions of the stated embedding equivalence.