Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results tagged with smooth-manifolds
Search options not deleted
user 167587
Smooth manifolds and smooth functions between them. For manifolds with additional structure, see more specific tags, such as [riemannian-geometry]. For more topological aspects, see [differential-topology].
14
votes
2
answers
863
views
Must a space that is locally injective image of $\mathbb{R}^n$ be a manifold?
Suppose $X\subseteq\mathbb{R}^m$ s.t. for any $x\in X$ and any open $U\subseteq\mathbb{R}^m$ that contains $x$, there exists a smaller open set $V\subseteq U$ also containing $x$, so that $V\cap X$ is …
- 623
0
votes
Must a space that is locally injective image of $\mathbb{R}^n$ be a manifold?
I feel like this is true for $n=1$. Here is an outline. Suppose $X$ is Hausdorff and
$\{U\subseteq X\mid U\text{ is open and there exists a continuous bijection }f:\mathbb{R}\rightarrow U\}$
form a ba …
- 623