Let $G$ be a complex connected reductive Lie group, $\theta$ an involution, and $K = G^\theta$ the fixed-point subgroup. Then $K$ has finitely many orbits on $G/B$, one of which is open and (quite often) several of which are closed.

We can pick a $\theta$-stable Borel $B$ and maximal torus $T$, and using Springer's map $$ \varphi : K\backslash G \to G, \qquad Kg \mapsto \theta(g^{-1}) g $$ we get a map of the same name $$ \varphi: K\backslash G/B \to B\backslash G/B \qquad \cong W := N(T)/T. $$ This map isn't usually injective. For example, if $G = SL(2)$ and $K = SO(2)$, then the LHS has three elements (North and South poles and the rest) while the RHS has only two.

If we fix a closed orbit $v \in K \backslash G/B$, and restrict $\varphi$ to the order ideal {$u \in K \backslash G/B : u \geq v $}, does it become injective?

Motivation: the poset $K \backslash G/B$ is somewhat mysterious, whereas the poset of "twisted involutions" {$w \in W : (\theta\cdot w)^2 = 1$} is at least less so. I need to sum over chains in this order ideal and would rather do it inside $W$.