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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].
3
votes
Accepted
functions which covers(good covers) manifolds
For (1), the answer is almost always no; the combinatorial properties of open covers of $\mathbb{R}$ are far too restricted. Suppose that $M$ is compact and connected and such an $f$ and $\{U_\alpha\ …
10
votes
Are there two non-diffeomorphic smooth manifolds with the same homology groups?
More surprisingly, you can find smooth manifolds which are homeomorphic (and in particular, have the same homology) but are not diffeomorphic! The best-known examples are exotic spheres.
43
votes
5
answers
5k
views
How can you tell if a space is homotopy equivalent to a manifold?
Is there some criterion for whether a space has the homotopy type of a closed manifold (smooth or topological)? Poincare duality is an obvious necessary condition, but it's almost certainly not suffi …
18
votes
Accepted
Pullbacks as manifolds versus ones as topological spaces
Here's a counterexample. Let $X=Y=\mathbb{R}$, and let $Y'$ be a point. Let $f:\mathbb{R}\to\mathbb{R}$ be a smooth map such that $f^{-1}(\{0\})=\{1,1/2,1/3,\dots\}\cup\{0\}$, and let $f'$ map $Y'$ …
35
votes
What are the open subsets of $\mathbb{R}^n$ that are diffeomorphic to $\mathbb{R}^n$
You can certainly have a set diffeomorphic to $\Bbb R^n$ but not star-shaped. For example, for $n=2$, the Riemann mapping theorem implies that any simply connected open set is diffeomorphic to the pl …
17
votes
Spectrum of Ring of Smooth Functions on $\mathbb{R}^n$
A trivial way you can tell this is false is that Spec of any ring is (quasi)compact, but $\mathbb{R}^n$ is not (if $n>0$).
A bit less trivially, this is still false if you replace $\mathbb{R}^n$ with …
3
votes
Accepted
A metric associated with a continuous surjective map $f:X\to Y$
Here's an example where the $d_f$-topology is not locally compact. Let $Y=[0,1]$ and let $$X=[0,1]\times \{0\}\cup \{(q,1/n):q\in[0,1]\cap\mathbb{Q}\text{ and $n$ is the minimal denominator of $q$}\} …
8
votes
Accepted
Compactly supported cohomology of homotopy equivalent manifolds
Let $M$ be a punctured torus and $N$ be a twice-punctured plane. Then $M$ and $N$ are homotopy equivalent, but their one-point compactifications are not (the first being a torus and the second having …