of basis on $V$ and $Q$. Thus it suffices to prove this in the case where $F = \hat{F}(\lambda)$, in which case $X_\lambda = \hat{X_\lambda}$ and $F_V = {F^{\mathrm{std}}}_V$.

We must verify that for every $\phi \in X_\lambda{\big /}S$, that $\phi ((F_{V/S})_j) \subset (F_Q)_{\lambda'_j}$. However, this is straightforward, as $(F_{V/S})_j$ is the image under the quotient map of $(F_V)_{k_j}$ which maps to $(F_Q)_{\lambda_{k_j}} =(F_Q)_{\lambda'_j}$. Thus $X_\lambda{\big /}S \subset
X_{\lambda'}$.

Moreover, every element of $X_{\lambda'}$ can be seen to have a unique lifting to $\{\phi \in X_\lambda\ |\ S \supset \ker \phi\}$. To see this, note that by a change of coordinates which preserves $F_V$ we can make $S$ is a coordinate subspace, in which case, this is obvious. Thus $X_\lambda{\big /}S = X_{\lambda'}$.

### Genericity of $S$

The one remaining fact we will need is that when $S$ is the special subspace $S = \ker \phi$, $\phi \in \bigcap_{i=1}^s X_{\lambda^i}$, then the flags $F_S$ and $F_{V/S}$ are actually [*generic*]{} flags. It is a priori conceivable that by choosing this particular $S$ (its definition involves the flags $F^i$), we could have undone all the genericity we had before, thereby ending up in a position where all the flags $F^i_{V/S}$ are not 