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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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1 vote
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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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Colimits for gluing schemes and the functor of points 1

You can write the first gluing as a pushout. The second can be seen as a coequalizer diagram $$\coprod_{i,j} U_{i,j}\rightrightarrows \coprod _iU_i\to X$$ The upper horizontal map includes each $U_{i,...
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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.

This is not a complete answer. Just idea. Let $X=S^n\times S^n$ and $A=\{(x,x)|x\in S^n\}$. We want to prove that $X-A\simeq S^n$. My idea of for a proof: We can show that $A\cong \{x_0\}\times S^n$ ...
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Proving a corollary to the Jordan Curve Theorem

Note that this is essentially the Lovers & Haters problem in disguise. Ref e.g. Lovers and haters via the Jordan curve theorem . While not strictly necessary, the problem dramatically simplifies ...
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Let $x_1,\dots,x_n$ be distinct elements of a Hausdorff space, prove there are pairwise disjoint open sets $U_1,\dots,U_n$ s.t. $x_i \in U_i$

Your approach has the right idea, but it is inappropriate to do it as you write: for each $x_i$ we can choose open neighborhoods $U_i^j$ of $x_i$ where $j\in \{1,2,\dots,n\}\setminus \{i\}$ such that ...
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Possible corollary of Jordan's curve theorem

Here's a solution using winding numbers combined with a theorem of Janiszewski (at very end). A useful reference is sections 3 and 4 of chapter 11 of Beardon's Complex Analysis. The more advanced ...
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1 vote
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Are there two computable numbers in the unit interval that map to the same point under a space-filling curve?

The Hilbert curve definitely has the property that it maps certain pairs of computable numbers to the same point in the plane. Moreover, to see this we only need to consider rather few properties of ...
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3 votes

Being homeomorphic is a relation between two topological spaces. How to understand the result that any open interval is homeomorphic to R?

This is an example of an abuse of notation: it's extremely common in mathematics to name a topological space $X$ by naming a set and leaving the topology to be implicitly deduced from context, and in ...
2 votes

Origin of the $\bar{\mathbb{R}}$ notation

You could regard it as a simple abuse of notation, but it is a motivated abuse of notation. In the topological space $\mathbb R \cup \{-\infty,+\infty\}$ with the order topology, the closure of the ...
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1 vote
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Topology of the direct sum $\bigoplus_{n \in \mathbb N} \mathbb R$

No, the topology in (2) is finer. For instance, the set $\{(f_n)_{n\in\Bbb N}\,:\,\forall n, \lvert f_n\rvert<2^{-n}\}$ isn't open in (1), but it is in (2): notice that its intersection with each $\...
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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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1 vote

Let $q :X \to Y$ be a quotient map, $f: X \to Z$ a continuous map and $h:Y \to Z$ a map such that $f = h \circ q$. Is $h$ is continuous also?

Since $q$ is a quotient map, for $U\subseteq Z$ open we have $f^{-1}(U)\subseteq X$ open and hence $q(f^{-1}(U))\subseteq Y$ open. On the other hand we have $h^{-1}(U)=q(f^{-1}(U))$ since $f^{-1}(U)=q^...
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1 vote
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Closure contained in open set of $\mathbb R^n$

I don't think this is true without further conditions on $X$. Suppose: $$X = \big\{(x,y) \in \mathbb{R}^2 : y > 0\big\} \cup \{(0,0)\},$$ i.e. that $X$ consists of everything strictly above the $x$-...
3 votes
Accepted

The monic quintic polynomials with one real root form a manifold

Implicit here is that you are supposed to topologize the set of monic quintics as $\mathbb{R}^5$ via the coefficient map as you describe, so the entire set of monic quintics is trivially a manifold, ...
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Property of closure in topology

Equivalence of the two following definitions: $x\in\overline A\iff \forall U\in\mathcal T\left(x\in U\implies U\cap A\neq \emptyset\right)$ $x\in\overline A\iff x\in\bigcap_{U\in\mathcal T,\;A\subset ...
2 votes
Accepted

Factorization of a nullhomotopic map

Yes, at least for path connected $Y$. Consider the unbased path space $Y^I$, the set of all continuous maps $u : I = [0,1] \to Y$ endowed with the compact-open topology. The map $r : Y^I \to Y,p(u) = ...
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Why are subsets of $\mathbb{R}^n$ with soft inequalities closed?

You can use continous functions. If you have a continous function $f: \mathbb R^n \to \mathbb R$ and a close set $C \subseteq \mathbb R$ then $f^{-1}(C)$ is closed. For example you looked at the set $\...
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2 votes
Accepted

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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Why are subsets of $\mathbb{R}^n$ with soft inequalities closed?

I think the best way to see this is working with the idea of a boundary. A boundary point for a set $S$ in $\mathbb{R}^n$ is a point $x$ such that any ball with center $x$ intersects both $S$ and $\...
1 vote
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The connected component containing $a$ is the subset of the intersection of clopen sets containing $a$

Your proof is correct, but it can be simplified. You do not need the collection $\Lambda_a$. Simply take $C$ (which is by the way, also a member of $\Lambda_a$) instead of an arbitrary $B \in \...
2 votes
Accepted

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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1 vote
Accepted

Property of closure in topology

Since you defined $\overline A$ as the intersection of closed sets containing $A$, for any $x\in X$ we have: $x\notin\overline A\Longleftrightarrow$ there exists a closed set $F\supset A$ such that $x\...
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1 vote
Accepted

Proving that something is not a manifold; assuming a topology, or for every atlas?

You don't have to talk about atlases at all to define a topological manifold: a manifold is a topological space satisfying some topological conditions, namely 1) Hausdorff, 2) locally Euclidean, 3) ...
3 votes
Accepted

neccesary conditions for quotient maps

You prove that if $U=p^{-1}(V)$ with $V\subseteq Y$ open, then $p(U)$ is open. However, not all open of $X$ is the form $p^{-1}(V)$, with $V$ open of $Y$.
1 vote

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.

Suppose $S$ is open and saturated with respect to $f$ in $\mathbb R^2$ and let $0\neq p\in S.$ Rotate $S$ so that without loss of generality, $p=(0,b).$ Then, $f^{-1}(f(\{p\}))=\{x^2+y^2=b^2\}\...
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2 votes

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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1 vote

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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2 votes
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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 ...
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2 votes
Accepted

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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0 votes

Compactness of function space

On compactness and completeness of $I^I$ and $C(I,I)$ Ok, I know this is already answered, but, working on it, I clarified myself some points I would like to share. Let $X=Y=I$, where $I=[0,1]$. Let $\...
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1 vote

Let $f:\mathbb{R}^n \rightarrow \mathbb{R}$ be continuous. Show that $\lbrace x : f(x) \neq 0 \rbrace$ is open.

Another way to visualize what @AnotherUser said: \begin{align*} \{x\in\mathbb{R}^{n} \mid f(x) \neq 0\} & = \{x\in\mathbb{R}^{n} \mid f(x) > 0\}\cup\{x\in\mathbb{R}^{n} \mid f(x) < 0\}\\\\ &...
4 votes
Accepted

Let $f:\mathbb{R}^n \rightarrow \mathbb{R}$ be continuous. Show that $\lbrace x : f(x) \neq 0 \rbrace$ is open.

Other than the fact that $f(x)-0=f(x)$ rather than $0$, it is correct. Alternatively, you can say that that set is open since it is equal to $f^{-1}(\mathbb{R}\setminus\{0\})$, $\mathbb{R}\setminus\{0\...
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1 vote

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
Accepted

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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3 votes
Accepted

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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2 votes

Compactness of the boundary of a subset

It is not generaly true, For example define $ \mathbb H=\mathbb R×\left]-1,+\infty\right[$ let $X=\mathbb H× \{0,1\}$ be the product space of the open half plane with the standard topology with $\{0,1\...
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Sets with nonempty infinitieth derived set

Embedding the countable ordinal $\omega^{\omega}$ as a bounded subset of the real line $\mathbb R$ (and then adding the unique limit point) gives an example of a compact set whose "infinitieth ...
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1 vote

$X=A\cup B$ be an open cover of $X$. If $X,A,B$ are simply connected , then $A\cap B$ path-connected?

Here is an interesting example to think about that doesn't prove or disprove the question. Let $X$ be the quasi-circle shown in the figure, a closed subspace of $\mathbb R^2$ consisting of a portion ...
1 vote

Prove that $f_0$ is not an inner point of $A$ relative to $\lVert\cdot\rVert_1$

Let $\,\varepsilon\!\in\left]0,1\right[\,$ and consider the following function : $f_\varepsilon(x)=\begin{cases}-\dfrac 1\varepsilon x+1\qquad\text{if }x\in\big[0,\varepsilon\big]\\0\qquad\qquad\quad\...
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1 vote
Accepted

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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