# Tag Info

1 vote

### Local diffeomorphism everywhere vs. global diffeomorphism

Smooth covering maps provide examples of local diffeomorphisms that are not bijections. One simple class of examples are self-covers of the unit circle in $\mathbb{C}$ given by $z\mapsto z^n$ ($n$ an ...
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### Let $A=\{(x,y) \in \Bbb S^n \times \Bbb S^n \mid x \ne y \}$. Let $f: \Bbb S^n \to A, \ x \mapsto(x,-x).$ Show that $f$ is a homotopy equivalence.

We have to find a homotopy inverse for $f$. It seems obvious that a nice candidate is $$g : A \to S^n, g(x,y) = x .$$ In fact $g \circ f = id$. The map $r = f \circ g$ is given by $$r(x,y) = (x,-x) .$$...
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### Set of probability measures with support in open $D \subset \mathbb{R}$ is open in the weak topology?

Try this. Let $D = (-1,1)$, an open set. Let $\mu_n$ be the probatility measure with mass $1/n$ at point $5$ and mass $(n-1)/n$ at point $0$. Then $\mu_n$ converges to the unit-point-mass at $0$. ...
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### Can every manifold with torus boundary be cut?

I'm not sure quite what you mean by "cut". If your question is whether there is always a properly embedded surface $S$ in $M$ such that $M\backslash \backslash S$ has boundary consisting of ...
1 vote

### Let $X$ be a locally path connected compact space. Show that $X$ has only finitely many path components.

In a locally path connected space the path connected components are open. Thus the collection of the path connected components is an open cover of $X$ and from compactness of $X$ has a finite subcover....
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### How can I see that a parametric surface is "pointy"?

This surface can also be written as $f^{-1}(0)$ where $f:\mathbb{R}^3\to \mathbb{R}$ is the map $f(x,y,z)=x^3-y^2$. The surface is 'pointy' because the partial derivatives of $f$ vanish along the line ...
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### Prove that a maximal atlas is the union of all compatible atlases if and only if it contains all admissible charts to atlases on a Manifold M

An atlas is a collection of compatible charts. It is maximal if for every chart $(U,x)$ in it, any other chart compatible with it belongs itself to the chart. So by definition is the union of all the ...
1 vote
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### A question related to the homotopy of two maps

A well-known result from algebraic topology is that any such $h$ can be deformed to an injective map. See cellular approximation theorem. Then $h$ (up to homotopy) can't be surjective, otherwise we'...
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### A simple curve of positive area

Look up Osgood_curve at Wikipedia, https://en.wikipedia.org/wiki/Osgood_curve In particular I copy the following picture and text from there to make this answer self-contained. (Note that the ...
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### Question about product of compact sets

Ok (too long for a comment, but I read your definition of a neighborhood) then do a picture for the plane first as it is easier to visualize (so $d=1$). A neighborhood of a point $p=(a,b)$ is a circle ...
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### Question about product of compact sets

$X$ is a box. A box is a product of open sets in $A$ and $B$ respectively. The definition of being open in the product implies the existence of just such a box. Note the product topology is the box ...
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### Why are subsets of $\mathbb{R}^n$ with soft inequalities closed?

$\def\R{\Bbb R}$ If you’re happy with the fact that preimages of closed sets under continuous functions are closed then I like the following approach. The functions $\pi(x,y)=x$ and $f(x,y)=x^2-y$ ...
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### Interpretation of homeomorphism

Your approach does not work properly. Of course you can restrict to subspaces $A, B \subset \mathbb R^n$. You try to decribe a deformation of $A$ into $B$ inside $\mathbb R^n$. Here are some problems. ...
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### In metric spaces, are two elements equal if and only if they're indistinguishable?

No mystery: In a metric space, if $x\ne y$ then $r:=d(x,y)>0$ and $x,y$ are distinguishable (e.g. $B(x,r)$ is not a neighborhood of $y$). In a pseudometric space, as you said twice, $x,y$ are ...
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### Let $f:\Bbb R^2 \to [0,\infty)$ be defined as $f(x,y)=x^2+y^2$. Show that $f$ is a quotient map.

$f$ is continous and surjective so it's enough to show that f is an open map. Remember that $\{B(x,r)|x\in \mathbb R^2 ,r\in [0,\infty)\}$ is a base to the topology on $\mathbb R^2$ so it's enough to ...
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### Question on Proposition 2.22 of Hatcher's Algebraic Topology: How to induce a deformation retraction on the quotient space

All you have to know is that if $p : Y \to Z$ is a quotient map, then also $p \times id_I : Y \times I \to Z \times I$ is a quotient map. This is a well-known theorem from general topology. In ...
• 61.3k
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### Closure of real interval in a topology with intervals with rational ends as a bsis.

The closure is $[-\sqrt{2},\sqrt{2}]$, $\pm \sqrt{2}$ are indeed limit points since we can find points in $(-\sqrt{2},\sqrt{2})$ in any neighbourhood of $\pm \sqrt{2}$, and obviously there can not be ...
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### Closed and bounded subsets of $\mathbb{C}^n$

The simple answer: there is a clear $\Bbb C\cong\Bbb R^2$ isometry, a $\Bbb C^n\cong\Bbb R^{2n}$ isometry. So closed and bounded subsets of one turn into closed and bounded subsets of the other, and ...
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1 vote
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### State whether the following set is open or closed.

You are correct to say that $A$ is not open since any nonempty open set must contain an open rectangle/disk, and all such sets are uncountable. One strategy to show that it's closed is to prove that ...
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### When is there a one to one correspondence between closed sets in the Zariski topology of the prime spectrum of a ring and radical ideals of that ring?

You're correct, there is a bijective correspondence between closed subsets of $\operatorname{Spec} A$ and radical ideals $I\subsetneq A$ and this always holds. This correspondence gets discussed a lot ...
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### Prove that $f_0$ is not an inner point of $A$ relative to $\lVert\cdot\rVert_1$

Hint Can you construct a sequence $f_n$ of continuous functions such that $f_n \notin A$ for all $n$ $\|f_n-f_0\|_1 \to 0$ as $n \to \infty$ If such a sequence exists, why does it prove the ...
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