# Tag Info

## New answers tagged geometric-topology

1

Yes there are. If $\tau_1$ is the standard topology and $\tau_2$ is the discrete topology, i.e., $\tau_2=\mathscr P(\mathbb R)$ (that means that every subset of $\mathbb R$ is $\tau_2-$open), then the identity map $$I : (\mathbb R,\tau_1) \to (\mathbb R,\tau_2),$$ is not continuous since, if $U\subset\mathbb R$ is not $\tau_1-$open, while however it is ...

0

Since your question lacks some details on what you know and what you have tried, I will make some assumptions about what you know in order to give you the quickest route to an answer, which is to apply the theorem on the classification of surfaces. To apply that theorem, you need to compute the orientability and the Euler characteristic of the surface. ...

0

Edit: This is totally not correct, although I will leave it up for posterity's sake. It is clear that this can't be right; as suggested in the comments, if we look at $B_3$ which has two generators $\sigma_i$ satisfying $$\sigma_1\sigma_2\sigma_1 = \sigma_2\sigma_1\sigma_2$$ then if we "forgot" the last strand as I suggested by setting $\sigma_2 = 1$, ...

7

Yes. Your space is homeomorphic to the standard unit sphere $S^2\subseteq\mathbb R^3$ with three open disks removed. These disks can be chosen so that the rotation of $\mathbb R^3$ by $\frac{2\pi}3$ around some axis cyclically permutes them. This rotation, however, has two fixed points, where the axis of rotation intersects $S^2$. Compose it with a ...

1

No: look at an open cube . Edit (October 2, 2014) To clarify matters (since the question mentioned just the word "manifold"): $\bullet$ The closure of the open cube is not a "topological manifold" but a "topological manifold with boundary" $\bullet \bullet$ The closure of the open cube is not a "differential manifold" nor even a " differential ...

2

No, that's not true. Imagine three open triangles in $\mathbb{R}^2$ "meeting" at a vertex (they don't actually contain the point, but it's in their common closure). The triangles are also linked by an open annulus. See this picture: Then the point at the center prevents this space from being a topological manifold (with boundary). It can be given the ...

0

No. Consider $X=\{|y|>|x|\}\cup\{|y|>1\}\cup\{|x|>1\}$. This is an open and connected set. Its closure contains $0$ which has no neighborhood in $X$ homeomorphic to any $\mathbb R^n$.

0

Take your favourite properly embedded compressible surface, and compress it. You have changed the Euler characteristic, but not the homology class.

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