Let $X$ be a complex manifold, $D$ a divisor (that is globally the zero locus of a function) and $U$ its complement. Recall Beilinson's "how to glue perverse sheaves":
Given a perverse sheaf $\mathcal{M}$ we write $\mathcal{M}_U$ and $\mathcal{M}_Z$ for its restriction to $U,Z$, and form $$\hspace{10mm}\Psi^{\text{un}}(\mathcal{M}_U)\ \to\ \ \Xi^{\text{un}}(\mathcal{M}_U)\oplus \Phi^{\text{un}}(\mathcal{M}_Z)\ \to\ \Psi^{\text{un}}(\mathcal{M}_U).\hspace{10mm}(\star)$$ This sequence is exact and $\mathcal{M}$ is recovered its middle cohomology sheaf. Likewise, given any pair of perverse sheaves $\mathcal{M}_U,\mathcal{M}_{Z,\Phi}$ pushed forward from $U,Z$ with maps $$\Psi^{\text{un}}(\mathcal{M}_U)\ \to\ \mathcal{M}_{Z,\Phi} \ \to\ \Psi^{\text{un}}(\mathcal{M}_U)$$ whose composition is $1-t$, where $t$ is the monodromy operator, we may apply this procedure to get a sequence $(\star)$ whose middle cohomology we take to give a perverse sheaf $\mathcal{M}$.
My question is about understanding this geometrically. There are lots of questions about how to think geometrically about the (unipotent) nearby $\Psi^{\text{un}}$ and vanishing $\Phi^{\text{un}}$ cycles functors. In the spirit of those questions,
- How do you think about the (unipotent) maximal extension functor $\Xi^{\text{un}}$?
- How do you think about all four maps in $(\star)$, as geometrically as possible?
The explanation should ideally be as concrete as in the linked answers, and should make it plausible that the middle cohomology sheaf is $\mathcal{M}$.
