Let $A,B$ be two homeomorph topological space under homeomorphism $\varphi$  and let $\sim$ be an equivalence relation on $A$. Then     Does $A\simeq_\varphi B$ imply $A/{\sim}\,\, \simeq B/\varphi(\sim)$?   Where $\varphi(\sim)$ is defined as follow $$x,y\in A,\quad x\sim y \iff \varphi(x)\,\varphi(\sim)\,\varphi(y).$$
122139
Let $f:X\to Y$ be a homeomorphism between topological spaces. Suppose we have an equivalence relation $\sim$ defined on $X$. Define an equivalence $\sim'$ on $Y$ by $y_1\sim'y_2$ iff $x_1\sim x_2$ where $y_i=f(x_i)$     Is $X/{\sim}$ homeomorphic to $Y/{\sim'}$?   I have a well defined continuous map $F:X/{\sim}\to Y/{\sim'}$ given by $[x]\mapsto[f(x)]$. This can easily be seen to be bijective. I am having trouble proving that this is open. I know there is some diagram chasing involved but I am confused.  We have the following commutative diagram - $\require{AMScd}$ \begin{CD}     X @&gt;f&gt;&gt; Y\\     @Vq_XVV\circlearrowright @VVq_YV\\     X/{\sim}@&gt;&gt;F&gt; Y/{\sim'} \end{CD}  If $U$ is open in $X/\sim$ then $q_X^{-1}(U)$ is open in $X$ and $f(q_X^{-1}(U))$ is open in $Y$. To prove that $F(U)$ is open in $Y/\sim'$ we need to show that $q_Y^{-1}(F(U))$ is open in $Y$. This will be true if $f(q_X^{-1}(U))=q_Y^{-1}(F(U))$. But I am having trouble proving this is equality. Any help will be appreciated.  Thank you.
40602
For all $x,y\in\mathbb{R}$   define that $x\equiv y$   if $x^{2}=y^{2}$  . Then $\equiv$  is an equivalence relation on $\mathbb{R}$  , there are infinitely many equivalence classes, one of them consists of one element and the rest consist of two elements.  Solution  True. To show that \equiv  is reflexive we need to show that \forall x\in\mathbb{R}  :x=x.  Let x\in\mathbb{R}  , x\equiv\mbox{\ensuremath{x}}  if x^{2}=x^{2},   which is obvious.  [x]  ={y\in\mathbb{R}  |x\equiv y}  =[0]={y\in\mathbb{R}|0\equiv y}={0}  . Hence y^{2}  =0^{2}   =0   which implies that y=0  .
5456
$X$ is a Hausdorff space and $\sim$ is an equivalence relation.  If the quotient map is open, then $X/{\sim}$ is a Hausdorff space if and only if $\sim$ is a closed subset of the product space $X \times X$.  Necessity is obvious, but I don't know how to prove the other side. That is, $\sim$ is a closed subset of the product space $X \times X$ $\Rightarrow$ $X/{\sim}$ is a Hausdorff space. Any advices and comments will be appreciated.
118422
Here is the problem and my attempted solution. Problem: Let $m^*(E) &gt; 0$. Define the Rational Equivalence Relation on $E$ by $x \sim y$ if, any only if, $x - y \in \mathbb{Q}$. Show that the choice set $(c_E)$ for the Rational Equivalence Relation must be uncountably infinite. (my) Solution: Let $m^*(E) &gt; 0$. Define the Rational Equivalence Relation on $E$ by $x \sim y$ if, any only if, $x - y \in \mathbb{Q}$. By way of contradiction, suppose that $c_E$ is countably infinite. Notice, one property of $c_E$ is the following: (i) For any $e \in E, \exists c \in c_E: e - c = q$, for $q \in \mathbb{Q}$. Therefore, by (i) if $c_E$ is countably infinite so is $E$. Since the Lebesgue Outer Measure of a countably infinite set is zero we obtain: $$m^*(E) = 0,$$ a contradiction. Hence, our assumption that $c_E$ is countably infinite is false. Therefore, $c_E$ must be uncountably infinite, as desired.
35909
In the book of first order logic I'm reading they say that if $(\varphi_1,...,\varphi_n)$ is a proof from a set of formulae $\Sigma$, then so is $(\varphi_1,...,\varphi_n,\forall x \varphi_j)$ for all $1\leq j \leq n$. They assume that as an axiom but it doesn't make sense to me. Why if $\varphi$ is true then so is $\forall x \varphi$? For example if $\varphi$ is $x=1$ that wouldn't mean that $\forall x : x=1$ , or would it?  Maybe what I don't understand is: why $(\Sigma \vdash \varphi \implies \Sigma \vdash \forall x \varphi)$ is true but $(\varphi \implies \forall x \varphi)$ is not true?  The book is this:
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I'm trying to find two compact, nonhomeomorphic subsets of the plane, say $X$ and $Y$, such that $X \times [0,1]$ is homeomorphic to $Y \times [0,1]$. I can not think of how a homeomorphism arises when you product with the interval.
69825
Suppose we have   $$Hom(x_0, y) \simeq Hom(x_1, y)\ \forall y \in \mathcal{C}$$  Does it follow that $x_0 \simeq x_1$ in $\mathcal{C}$?   Now suppose that $F, G: \mathcal{C} \rightarrow \mathcal{D}$ are two functors such that $Fx \simeq Gx\ \forall x\in \mathcal{C}$. Does it follow that any natural transformation $\epsilon: F \Rightarrow G$ is a natural isomorphism?
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I'm having trouble in proving the following:  If $U$ is a neighborhood of $e$, there exists $V$ neighborhood of $e$ such that $V^{-1}V \subseteq U$  Can someone please verify if what I wrote makes sense? And if possible show the way to go further, because I'm stucked.  My attempt:   Let $G$ a topological group.  (Definition: A topological group $G$ is a topological space with a group structure such that the applications $m: G \times G \to G$ and $r:G \to G$, defined by $m(x,y) = x.y$ and $r(x) = x^{-1}$ are continuous.)  $V^{-1}V = \{ x^{-1}y ; x,y \in V\}$   $U$ is a neighborhood of $e$, then is $V^{-1}V$ a neighborhood of $e$?  Since $r$ is homeomorphism and the right-translation $\delta_y:G \to G, \delta_y(x)=x.y$ is homeomorphism, the application $\delta_y \circ r :G \to G$ is homeomorphism. Let $x \in U$. Then $\delta_y \circ r(x) = x^{-1}y$ is in an open set, because of continuity. Why is it a neighborhood of $e$?
103874
I was given this question on my midterm. Currently I am studying for finals and am still unsure how to properly solve this question.  Let X and Y be two sets and f be a map from X to Y be a bijection. Prove that, when we consider X and Y with their respective T1-topologies (cofinite topology), this map is a homeomorphism.   I know that I must show that f is continuous in order to show X and Y are homeomorphic. On the midterm I tried showing that open sets in Y have preimage that are open in X and closed sets in Y have preimage that are closed in X. However I'm unsure how to approach the sets in X,Y that are neither open nor closed.  So my question is, would it be easier to show continuity by first showing local continuity at every point in X and using that to prove the continuity of f? Or is this the wrong way to approach this problem?
13382
A topological space $X$ is simply connected if it is pathwise connected and each closed path $u : I = [0,1] \to X$ is path homotopic to the constant path at $x_0 = u(0) = u(1)$. Recall that  A closed path $u$ is one with $u(0) = u(1)$. A closed path will also be called a loop and we write $\mathcal L'(X)$ for the set of all loops in $X$. If we want to be more precise, we say $u$ is a loop based at $x_0 \in X$ provided $x_0 = u(0) = u(1)$. By  $\mathcal L'(X,x_0)$ we denote the set of all loops in $X$ which are based at $x_0$. Also note that each constant path is a loop.  Two paths  $u, v : I \to X$ with the same initial point $x_0 = u(0) = v(0)$ and the same terminal point $x_1 = u(1) = v(1)$ are path homotopic, $u \simeq_p v$, if there exists a  homotopy $H : I \times I \to X$ such that $H(t,0) = u(t)$ and $H(t,1) = v(t)$ for all $t$ and $H(0,s) = x_0, H(1,s) = x_1$ for all $s$ (such a homotopy $H$ is stationary on the boundary $\partial I = \{0,1\}$; it is called a path homotopy). It is well-known that $\simeq_p$  is an equivalence relation on the set $\mathcal P(X)$ of all paths in $X$. Clearly $\simeq_p$ restricts to an equivalence relation on $\mathcal L'(X)$ and to an equivalence relation on each $\mathcal L'(X,x_0)$.  The set of equivalence classes $\mathcal L'(X,x_0)/\simeq_p$ is nothing else than the fundamental group $\pi_1(X,x_0)$ with base point $x_0$.   Question: Find a list, as complete as possible, of properties characterizing topological spaces which are simply connected. Each of these properties can then be taken as an alternative definition of simply connected. Note that many questions in this forum deal with aspects of simple connectedness. The answer to the present question will hopefully cover most of them.
113622
Let $X,Y$ be topological spaces.  Let $p:X\rightarrow Y$ be a quotient map.  Then, does there exist an equivalence relation $\sim$ on $X$ such that $X/\sim$ is homeomorphic to $Y$?
69743
I was searching for an easy example of a quotient space $X/{\sim}$ which is not second countable even though $X$ is second countable. I have found an example in the following answer, but I do not quite understand it:      Let $X:=\mathbb{R}$ and consider the equivalence relation $\sim$ defined by $x\sim y \Leftrightarrow x=y \text{ or }x,y\in\mathbb{Z}$.  Now it is claimed that the space is not first countable at the point $[0]\in X/{\sim}$. Why is this the case? I would be very glad, if someone can explain it to me in more detail.
87213
Let $S$ be the equivalence relation defined on $\wp(\{1, 2, 3, 4\})$ defined by: $$XSY\text{ if and only if } |X|\equiv|Y|\;\mod 2$$ Write down the equivalence classes of S.  I understand that equivalence classes have relations where it is reflexive, symmetric and transitive but how are you supposed to write equivalence classes?
41208
This problem is bugging me for some time.     $f:(X,\mathcal{T}) \to (Y,\mathcal{T}')$ is a continuous bijection where $(X,\mathcal{T})$ is compact and $(Y,\mathcal{T}')$ is $T_2$ (i.e. Hausdorff and $T_1$). Then show that $f$ must be a homeomorphism.   So far I have failed to understand any relation between the compactness of domain and $T_2$-ness of the range. I tried in vain to prove that $f(U)$ is open in $(Y,\mathcal{T}')$ whenever $U$ is so in $(X,\mathcal{T})$. I know $f$ makes $X$ and $Y$ respectively $T_2$ and compact and hence makes both of them $T_4$ spaces. But that is not coming to any use. I would be glad if someone helps me with a solution. Thank you.
109455
I am reading  of the Book of Proof, and I'm stuck at the Exercise 10 of section 11.2. It is as follows.  Suppose $R$ and $S$ are two equivalence relations on a set $A$. Prove that $R \cap S$ is also an equivalence relation.  Thanks for helps!
109272
I want to find all group homomorphisms $\varphi: \mathbb{Z}/ m \mathbb{Z} \to \mathbb{Z}/n\mathbb{Z}$, with $m$ and $n$ natural numbers.  Clearly $\varphi(0)=0$ since the identity in $\mathbb{Z}/m\mathbb{Z}$ must map to the identity in $\mathbb{Z}/n\mathbb{Z}$, and it follows that $m \varphi(1)=0$, so is $mx \equiv 0 \bmod n$?  This as far as I get. How can I find all the homomorphisms?
54242
$\phi:\mathbb{R}\to\mathbb{R}$ be a diffeomorphism with the following property  $a\in\mathbb{R},|a|&lt;\frac{1}{10}$  (i)$\phi(a)=0$  (ii) $\phi(x)=x$ when $|x|&gt;1$  and how to generalize that in $\mathbb{R}^n$   please help.