Peeling a maximal-weight Verma from a standard filtration

Since $\nu$ is maximal among the weights of $X$, the vector $v$ is ${\mathfrak{n}}^{+}$-fixed: for $x\in {\mathfrak{n}}^{+}$ of weight $\alpha \in Q^{+}\setminus \{0\}$ the vector $xv$, if nonzero, would have weight $\nu +\alpha$, contradicting maximality. By [F2] there is a homomorphism $f:\mathrm{\Delta }(\nu )\to X$ with $f(v_{\nu })=v\mathrm{\ne }0$; let $i$ be the smallest index with $f(\mathrm{\Delta }(\nu ))\subseteq F_i$, which exists because $F_m=X$. By minimality $f(\mathrm{\Delta }(\nu ))$ is not contained in $F_{i-1}$, so the composite $\bar{f}:\mathrm{\Delta }(\nu )\stackrel{f}{\to }{F}_i\twoheadrightarrow F_i\mathrm{/}{F}_{i-1}=\mathrm{\Delta }({\mu }_i)$ is nonzero.

Since $\bar{f}\mathrm{\ne }0$, the weight $\nu$ of $v_{\nu }$ maps to a nonzero vector in $\mathrm{\Delta }({\mu }_i)$, so $\nu$ is a weight of $\mathrm{\Delta }({\mu }_i)$ and therefore $\nu \le {\mu }_i$ by [F2]. On the other hand ${\mu }_i$ is a weight of the subquotient $F_i\mathrm{/}{F}_{i-1}$ of $X$, hence a weight of $X$, and the relation $\nu \le {\mu }_i$ and maximality of $\nu$ force ${\mu }_i=\nu$, so $\bar{f}$ is a nonzero endomorphism of the Verma module $\mathrm{\Delta }(\nu )$. Since $\mathrm{\Delta }(\nu )$ is generated by $v_{\nu }$ and $\bar{f}(v_{\nu })\in \mathrm{\Delta }(\nu {)}_{\nu }=\mathbb{C}{v}_{\nu }$ is nonzero by [F2], the image of $\bar{f}$ contains $v_{\nu }$, so $\bar{f}$ is surjective, and it is injective by [F3]; hence $\bar{f}$ is an isomorphism. Now $f$ itself is injective: if $\ker f\mathrm{\ne }0$, then by [F3] it contains a nonzero ${\mathfrak{n}}^{+}$-fixed vector of some weight $\eta$, which by [F2] provides a nonzero homomorphism $g:\mathrm{\Delta }(\eta )\to \mathrm{\Delta }(\nu )$ with image in $\ker f$; then $\bar{f}g=0$, while $\bar{f}$ is injective and $g\mathrm{\ne }0$, so $\bar{f}g\mathrm{\ne }0$, a contradiction. Hence $f:\mathrm{\Delta }(\nu )\hookrightarrow X$ is injective.

Identify $\mathrm{\Delta }(\nu )$ with its image $f(\mathrm{\Delta }(\nu ))\subseteq F_i$. Since $\bar{f}$ is an isomorphism onto $F_i\mathrm{/}{F}_{i-1}$, one has $F_i=f(\mathrm{\Delta }(\nu ))+F_{i-1}$ and $f(\mathrm{\Delta }(\nu ))\cap F_{i-1}=0$, so $F_i\mathrm{/}f(\mathrm{\Delta }(\nu ))\cong F_{i-1}$. In the quotient $X\mathrm{/}f(\mathrm{\Delta }(\nu ))$ the images of the flag pieces form the chain $0\subseteq F_1\subseteq \cdots \subseteq F_{i-1}=F_i\mathrm{/}f(\mathrm{\Delta }(\nu ))\subseteq F_{i+1}\mathrm{/}f(\mathrm{\Delta }(\nu ))\subseteq \cdots \subseteq X\mathrm{/}f(\mathrm{\Delta }(\nu ))$, whose successive quotients are $\mathrm{\Delta }({\mu }_j)$ for $j\mathrm{\ne }i$ and zero at the repeated step, and $F_j\mathrm{/}f(\mathrm{\Delta }(\nu ))/F_{j-1}\mathrm{/}f(\mathrm{\Delta }(\nu ))\cong F_j\mathrm{/}{F}_{j-1}=\mathrm{\Delta }({\mu }_j)$ for $j>i$; hence $X\mathrm{/}f(\mathrm{\Delta }(\nu ))$ has a Verma flag whose factors are exactly the $\mathrm{\Delta }({\mu }_j)$ with $j\mathrm{\ne }i$, of length $m-1$, and the removed factor is $\mathrm{\Delta }({\mu }_i)=\mathrm{\Delta }(\nu )$.

Steps 2.1 and 3.1 prove that $v$ is a highest-weight vector, that $\mathrm{\Delta }(\nu )\to X$ is injective, and that the cokernel has a Verma flag induced from the given flag by deleting exactly one factor, of length $m-1$.