In a manifold, $\angle xpy>\frac{\pi}{2}$, for $q$ on $px$ or $py$, $B_q(r)$ homeomorphic to $B_p(r)$?

Question

Let M be an n-dimensional Riemannian manifold without boundary, with sectional curvature $\geqslant -1$. For a point $p\in M$, suppose there exist $l, \delta>0$, $x,y \in M$ with $d(p,x),d(p,y)>l$ and a geodesic $px$ and $py$ with angle $\angle xpy>\frac{\pi}{2}+\delta$. Let $q$ be a point on geodesic $px$ or $py$, Question: is there $r>0$, which depends only on $n,l,\delta$ such that $B_q(r)$ is homeomorphic to $B_p(r)$?

Equivalently, we can state the question in the following way:

Let $M_i$ be a sequence of Riemannian manifolds with $sec \geqslant -1$ and diameter $\leqslant D$. Suppose $(M_i,p_i)$ Gromov-Hausdorff converge (possibly collapse) to $(X,p)$ (we know it's an Alexandrov space). Suppose there exist $l>0, \delta>0$, $x,y\in X$ with $\angle xpy> \frac{pi}{2}+\delta$. lift $x,y$ to $M_i$, we get $x_i,y_i\in M_i$. with$\angle x_i p_i y_i >\frac{\pi}{2}+\delta$. Let $q_i$ be a point on geodesic $p_ix_i$ or $p_iy_i$. Question: Is there $r>0$, such that such that $B_{q_i}(r)$ is homeomorphic to $B_{p_i}(r)$?

Answer 1

There is a simple counterexample. Consider warped product $M=\mathbb{R}{\times}_{\exp}\mathbb{S}^1$; that is, $\mathbb{R}{\times}\mathbb{S}^1$ with the metric $(\begin{smallmatrix}1&0\\ 0&e^{2\cdot x}\end{smallmatrix})$ at $(x,y)\in\mathbb{R}{\times}\mathbb{S}^1$. Note that for any $r>0$ there are arbitrary close points $p,q$ on a line $\mathbb{R}{\times}\{y\}$ such that $B_q(r)$ is homeomorphic to a disc and $B_p(r)$ is homeomorphic to a cylinder.

However by Morse lemma, there is a neighborhood of $p$ with size of order $\delta\cdot l$ is a homeomorphic to a product $(a,b)\times L$, where the first coordinates is the distance function $\mathrm{dist}_x$. (So morally your are right.)