Properties of a general element of the intersection of two Schubert cycles

Question

We have the following lemma:

Lemma

Let $\Sigma_a(\mathcal{V}),\Sigma_b(\mathcal{W})$ be two Schubert cycles defined relative to transverse flags $\mathcal{V}$ and $\mathcal{W}$. If $\Lambda \in \Sigma_a(\mathcal{V}) \cap \Sigma_b(\mathcal{W})$ is a general point of their intersection, then:

1. $\Lambda$ doesn't lie in any strictly smaller Schubert cycle $\Sigma_{a'}(\mathcal{V}) \subsetneq \Sigma_a(\mathcal{V})$.
2. The flags induced by $\mathcal{V}$ and $\mathcal{W}$ on $\Lambda$ are transverse.

By the first part of the Lemma it follows that the induced flags $\Lambda^{\mathcal{V}}$ and $\Lambda^{\mathcal{W}}$ are explicitly $\Lambda_i^{\mathcal{V}}=\Lambda \cap V_{n-k+i-a_i}$ and $\Lambda_i^{\mathcal{W}}=\Lambda \cap W_{n-k+i-b_i}$, for $i=1,\dots,k$.

The proof of the point $1$ is clear to me. As for the point $2$, there is a step is which not clear to me. Explicitly we have to show that for general $\Lambda$ in the intersection, we have $\Lambda \cap V_{n-k+i-a_i} \cap W_{n-i-b_{k-i}}=0$. For this purpose we consider the incidence correspondence $$ \phi=\{(\Lambda,[v]) \in (\Sigma_a \cap \Sigma_b)\times \mathbb{P}(V_{n-k+i-a_i}\cap V_{n-i-b_{k-i}})\; | v \in \Lambda\} $$ This is a projective variety and if we show that $\mathrm{dim}(\phi) < \mathrm{dim}(\Sigma_a \cap \Sigma_b)$, then we are done, since the projection $\phi \to \Sigma_a \cap \Sigma_b$ cannot be dominant. Now it is said that

By the first part of the lemma we can replace $\phi$ by the preimage of the complement $U$ of $\mathbb{P}(V_{n-k+i-a_i-1} \cap W_{n-i-b_{k-i}}) \cup \mathbb{P}(V_{n-k+i-a_i} \cap W_{n-i-b_{k-i}-1})$ in $\mathbb{P}(V_{n-k+i-a_i}\cap W_{n-i-b_{k-i}})$.

I thought that maybe this is because the preimage of $U$ is an open dense subset of $\phi$, and hence this preimage must have the same dimension as $\phi$. But, if this is true I cannot see why this follows from the first point of the lemma. Moreover, if my assumption is not true, I cannot see why we can substitute $\phi$ by the preimage of $U$.

The lemma is taken from "3264 and all that" by Harris and Eisenbud (page 140).

Answer 1

Define $\tilde{a}$ by $\widetilde{a_i} = a_i+1$. Define also $\widetilde{b}$ by $\widetilde{b_{k-i}} = b_{k-i}+1$. Then it is clear that $\Sigma_{\widetilde{a}}$ is a strict Schubert sub-cycle of $\Sigma_a$ and $\Sigma_{\widetilde{b}}$ is a strict Schubert sub-cycle of $\Sigma_b$.

By the first part of the lemma, we know that $\Sigma_{\widetilde{a}} \cap \Sigma_b$ and $\Sigma_{a} \cap \Sigma_{\widetilde{b}}$ are strict subvarieties of $\Sigma_{a} \cap \Sigma_b$.

But the preimage of $$\mathbb{P}(V_{n-k+i-a_i-1}) \cap \mathbb{P}(W_{n-i-b_{k-i}}) \cup \mathbb{P}(V_{n-k+i-a_i}) \cap \mathbb{P}(W_{n-i-b_{k-i}-1})$$ in $\phi$ maps to $\Sigma_{\widetilde{a}} \cap \Sigma_b \cup \Sigma_{a} \cap \Sigma_{\widetilde{b}}$. Hence, it doesn't matter that you withdraw this subset in $\phi$ to check that the map $\phi \rightarrow \Sigma_a \cap \Sigma_b$ is dominant or not.