Reputation
Next privilege 25,000 Rep.
Access to site analytics
Badges
5 63 155
Newest
 Necromancer
Impact
~613k people reached

Jan
29
awarded  Necromancer
Jan
28
comment If the fibers of a submersion are connected, does it mean that any 2 sections are homotopic (locally on the base)?
then add "since a given point can be mapped to any other point of connected manifold by a diffeomorphism".
Jan
28
accepted Wild half-line in a Euclidean space
Jan
28
comment If the fibers of a submersion are connected, does it mean that any 2 sections are homotopic (locally on the base)?
I would not look for a ref. Instead I would say that the statement can be reduced to the case of projection map $\mathbb R^{n+m}\to \mathbb R^{n}$ where it is obvious.
Jan
27
revised Actions on Sⁿ with quotient Sⁿ
deleted 78 characters in body
Jan
25
comment Wild half-line in a Euclidean space
@YCor I do not know --- I guess there are no examples for $m\le 3$. Anyway I am happy with this one --- it solves all my problems.
Jan
25
answered Wild half-line in a Euclidean space
Jan
25
comment Wild half-line in a Euclidean space
@YCor, but the answer is the same.
Jan
25
comment Wild half-line in a Euclidean space
@YCor it is a 1-dimensional subcomplex in the triangulation coming from the suspension, but (by obvious reason) it is not subcomplex in the standard triangulation of sphere.
Jan
25
comment Wild half-line in a Euclidean space
@YCor double suspension over $X$ is the joint $\mathbb{S}^1*X$ and $\mathbb{S}^1$ is its equator.
Jan
25
comment Wild half-line in a Euclidean space
@YCor The double suspension over Poincaré sphere is homeomorphic to $\mathbb{S}^5$. The complement of the double-suspension-equator in it is not simply connected.
Jan
25
revised Wild half-line in a Euclidean space
added 406 characters in body
Jan
24
revised Wild half-line in a Euclidean space
added 1 character in body; edited title
Jan
24
comment Wild half-line in a Euclidean space
@YCor the cone is infinite.
Jan
24
asked Wild half-line in a Euclidean space
Jan
21
comment Hausdorff convergence of submanifolds in Riemannian manifolds
@sva No, I use the other theorem of Nash --- if $N$ is large, you can do the same for $C^\infty$ embeddings.
Jan
21
answered Hausdorff convergence of submanifolds in Riemannian manifolds
Jan
20
comment Example of compact $CD(K,\infty)$ space, but doubling condition fails
Take the product measure, $CD(0,\infty)$ can be checked directly, or you may think of it as a limit space of n-dimensional rectangles for $n\to\infty$.
Jan
20
comment Example of compact $CD(K,\infty)$ space, but doubling condition fails
Hilbert cube. (too short for an answer)
Jan
19
revised On closed simple curve with curvature at most 1
deleted 46 characters in body