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

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 $$B_{p_i}(r)$$?

Contractable and Simply Connected Doubling Spaces Homeomorphic to Euclidean Space

Is there a characterization of all simply connected, contractable doubling metric spaces which are homeomorphic to a simply connected subset of Euclidean space?

Homeomorphic extension to totally disconnected sets

Dear Mathoverflow Community,

I am looking for a reference for the following topological fact:

Fact

Let $$E$$ and $$F$$ be two totally disconnected compact subsets of the plane (can assume perfect if you want). Then every homeomorphism $$f: \mathbb{R}^2 \setminus E \to \mathbb{R}^2 \setminus F$$ extends to a homeomorphism of $$\mathbb{R}^2$$ onto itself.

Thank you, Malik

Are triangulations of compact manifolds PL homeomorphic?

I have frequently come across the statement “Any two triangulations of a compact n-manifold are related by bistellar moves” attributed to Pachner via Lickorish’s paper ‘Simplicial moves on complexes and manifolds’. The theorem this refer to is the following: Closed combinatorial n-manifolds are PL homeomorphic if and only if they are bistellar equivalent.

My question is: Considering that Hauptvermutung is not true for manifolds of dimension more than 3, how can we justify this statement?

What is the one dimensional image that the torus with one circular hole on its surface homeomorphic to?

I am really poor of geometry imagination and struggled to figure this out.

Prove that spheres in 2 and 3 dimensions are not homeomorphic.

The spaces $$S^1$$ and $$S^2$$, with their topologies inherited from $$\mathbb R^2$$ and $$\mathbb R^3$$, respectively, are not homeomorphic. Here $$S^1$$ and $$S^2$$ are spheres in $$2$$ and $$3$$ dimension.

Topological groups with homeomorphic underlying spaces, isomorphic abstract groups and homotopy equivalent classifying spaces

Define the classifying space $$BG$$ of a well-pointed topological group $$G$$ as the fat realization of the nerve of $$G$$.

Let $$G$$, $$H$$ be well-pointed topological groups. Assume that there is a continuous group homomorphism $$f:G\rightarrow H$$ that induces a homotopy equivalence on the underlying topological spaces. It is claimed in an answer to this question that the induced map $$BG\rightarrow BH$$ is a homotopy equivalence. It is important here that we have a map between groups that simultaneously respects algebra and topology.

The question is basically what happens if we do not have such a map. Let $$G$$, $$H$$ be well-pointed topological groups. Assume that their underlying abstract groups are isomorphic and that their underlying topological spaces are homeomorphic. This by itself does not mean that $$G$$ and $$H$$ are isomorphic as topological groups (a pro-choice example is given by $$p$$-adic rationals for different $$p$$).

But what happens if we assume in addition that the classifying spaces of $$G$$ and $$H$$ are homotopy equivalent?

Note that a theorem of Notbohm says that two compact Lie groups are isomorphic as Lie groups iff their classifying spaces are homotopy equivalent so at least in some contexts, this question has a positive answer.

Are 0,8 and 9 homeomorphic topological space?

Consider the topological spaces “0”,”8″ and “9” in $$\mathbb{R}^{2}$$. Are they homeomorphic?

I have an approach that doesnt look very rigorous to me. I wanted to know how to formalize this if its correct.

• 0 and 8 are not homeomorphic since excluding one point of 0 the space is still connected, but excluding the “tangent point” of 8, we have a disconnected space.

• Same idea for 8 and 9.

• The space 9 is union of one circle and one arc. The arc is homeomorphic to the circle, so we can view 9 as a union of two circles, then 8 and 9 are homeomorphic

PS: the topology of the spaces is induced by topology of $$\mathbb{R}^{2}$$.

Homeomorphic characterization of the real line?

Let $$A$$ be a path-connected subset of $$\mathbb R^2$$ such that the removal of any singleton from $$A$$ splits $$A$$ into two open connected components, each of which is path-connected.

Is $$A$$ necessarily homeomorphic to $$\mathbb{R}$$?

Connected and homogeneous $T_2$-space not homeomorphic to a subset of $\mathbb{R}^n$

What is an example of an connected and homogeneous $$T_2$$-space $$(X,\tau)$$ with $$|X|=2^{\aleph_0}$$ such that for no $$n\in\mathbb{R}$$ the space $$(X,\tau)$$ is homeomorphic to a subspace of $$\mathbb{R}^n$$?