# Tag Info

## New answers tagged geometric-topology

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Alternately, stereographic projection maps circles to circles, and so for topological reasons $f$ maps open disks in the planes into open discs on $S^n - \{p\}$ (regions with circular boundary). But the set of open disks in the plane is a basis for the standard topology on the plane, and hence (once you know it exists) $f^{-1}$ is continuous. Similarly, so ...

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Hint: To show that $f$ is bijective, one can often (as can be managed here) compute $f^{-1}$ explicitly. Then, to show that $f$ is a homeomorphism, by definition it remains to show that $f$ and $f^{-1}$ are continuous, but they are both visibly compositions of continuous functions.

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In finite (fixed) dimension all the norms are equivalent. The function $$x\longmapsto\frac{x}{\|x\|_2}$$ is continuous and bijective form $S_1$ (compact by Heine-Borel) to $S_2$, so the inverse is continuous.

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Hint: In both cases we can use the law of cosines. This says that if you have a triangle with side lengths $a$, $b$, and $c$, with the angle opposite the side of length $c$ as $\theta$, then you have the relation $$c^2=a^2+b^2-2ab\cos\theta$$ Notice that in each situation, you have $3$ out of the $4$ variables in this formula, so you can solve for the last ...

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A Primer on Mapping Class Groups by Benson Farb and Dan Margalit is a good bet. You can also take a look at the answers to a similar question on MathOverflow.

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Since $X=S^n\times\mathbb{R}\cong S^n\times (0,1)$, you can think of $X$ as an n-dimensional annulus or shell, without boundary. Usually, $X$ retracts onto $S^n$. But after removing a point, $X\setminus\{*\}$ retracts onto $S^n$, with a bump. This "bump" is caused by the removed point and looks like a copy of $S^n$ (you can easily visualize this for ...

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If $k$ is a positive integer, then the random variable $X$ takes on the value $k$ with probability $p(1-p)^{k-1}$. It follows by the Law of the Unconscious Statistician that $$E(e^{aY})=\sum_{k=1}^\infty e^{ak} p(1-p)^{k-1}.$$ This is an infinite geometric series with first term equal to $pe^a$ and common ratio $r=(1-p)e^a$. If $r\lt 1$, the series ...

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Another way to look at the question is: when an $\mathbb R^m$ embedded if $M$ can be represented as $f^{-1}(\Delta)$ for such a transversal $f$? As commented before, the embedding must be closed (hence $M$ is not compact). Let us first look at the case $N=\mathbb R^n$. Prop.1 A submanifold $X\subset M$ diffeomorphic to $\mathbb R^m$ that is of the form $\ ... 0 I finally found an answer to my question:$\lambda$is up to sign always$2$if$n$is odd and$0$if$n$is even. A sketch of a proof can be found in Levine's "Lectures on groups of homotopy spheres" on page 64. 0 I think the answer to your question will vary widely depending on$n$. For example, for$n=3,5$and$7$(mod 8),$\pi_{n-1}SO_{n}$is known to be$\mathbb{Z}/2$, which is torsion (hence$\lambda=0$). There is a complete list of the homotopy groups$\pi_{n-1}SO_{n}$in the book "Topology of Lie groups" by Mimura Toda. 8 The intuition is basically that$4$dimensions is large enough that$4$-manifolds exhibit great variety (one example: there are$4$manifolds with arbitrary finitely-generated fundamental groups), but small enough that the high-dimensional-manifold surgery theory apparatus doesn't apply to smooth manifolds, only topological manifolds. The key idea for the ... 1 A$M_{\kappa}$-simplicial complex$K$can be viewed as a collection of geodesic simplices$\{ S_{\lambda} \mid \lambda \in \Lambda \}$glued together by isometries; let$\mathcal{S}$denote the set of all faces of the$S_{\lambda}$'s. The relation of isometry$\simeq$on$\mathcal{S}$, that is$S_{\lambda_1} \simeq S_{\lambda_2}$if they are isometric, is an ... 1 An map of manifold pairs $$f: (N, \partial N)\to (M, \partial M)$$ is called proper if$f^{-1}(\partial M)=\partial N$. Here I am assuming that$N$is compact. Now, apply this to maps which are immersions and$N\$ which is a closed interval.

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