Verma filtrations are not closed under quotients

The inclusion $M(-2)\hookrightarrow M(0)$ of [F1] is the map sending the highest-weight generator of $M(-2)$ to the singular vector $f{v}_0\in M(0)$, and $M(-2)=L(-2)$ is simple, so the quotient $M(0)\mathrm{/}M(-2)$ has weights only in weight $0$: the $h$-coordinates of the weights of $M(0)$ are $0,-2,-4,\dots$ and those of $M(-2)$ are $-2,-4,\dots$, and $M(0{)}_0=\mathbb{C}{v}_0$. The quotient is therefore the one-dimensional simple module $L(0)=\mathbb{C}{v}_0$ with $v_0$ the image of the highest-weight generator, and the sequence $0\to \mathrm{\Delta }(-2)\to \mathrm{\Delta }(0)\to L(0)\to 0$ is nonsplit: a splitting would exhibit $L(0)$ as a one-dimensional submodule of $M(0)$, necessarily spanned by the weight-zero vector $v_0$, but the submodule generated by $v_0$ is all of the infinite-dimensional module $M(0)$.

Both $\mathrm{\Delta }(-2)=M(-2)$ and $\mathrm{\Delta }(0)=M(0)$ are Verma-filtered: each has the one-step flag $0\subseteq \mathrm{\Delta }(\lambda )$.

The simple module $L(0)$ is not Verma-filtered. If it had a finite Verma flag, [F2] would give $[L(0)]={\sum }_{\mu }{m}_{\mu }[\mathrm{\Delta }(\mu )]$ with all $m_{\mu }\ge 0$. On the other hand the exact sequence of step 1.1 gives $[L(0)]=[\mathrm{\Delta }(0)]-[\mathrm{\Delta }(-2)]$ in $K_0(\mathcal{O})$. Since the standard classes form a $\mathbb{Z}$-basis, comparing the two expressions forces $m_0=1$ and $m_{-2}=-1$, contradicting $m_{-2}\ge 0$.

Thus $\mathrm{\Delta }(-2)$ and $\mathrm{\Delta }(0)$ are Verma-filtered while their quotient $L(0)=M(0)\mathrm{/}\mathrm{\Delta }(-2)$ is not, refuting the statement; the quotient in the standard-filtration theorem for projectives is a direct summand rather than an arbitrary quotient, which is what makes that argument work.