I'm taking a basic topology course, and studying for my exam, I'm stuck with an exercise that I'm not sure how to do it. Says the following:  Show that the $ \mathbb{R}^ 2 - {(0,0)} $ plane is a homeomorphism at $ S ^ 1 \times (0, \infty) $  I took the function  $f:\mathbb{R}^ 2 - {(0,0)} \rightarrow S ^ 1 \times (0, \infty) $, $f(rcost,rsent)= ((cost, sent), r)$  But I'm not sure if I defined it correctly, even I do not know how to prove it is a homeomorphism. Could you give me a suggestion? Greetings.
45980
Are there any specific 'tricks' or 'techniques' in finding homeomorphisms between topological or metric spaces?  I'm trying to construct a homeomorphism between $\mathbb{R} \times S^1 \to \mathbb{R}^2 \setminus \{(0,0)\}$ but I'm having trouble even visualizing how this is going to work. I'm thinking some sort of stereographic projection from the circle and then including the $\mathbb{R}$ somehow but I'm stuck. Any ideas?
74631
Show that $\mathbb{CP}^2$ is homeomorphic to the orbit space of $(S^2\times S^2)/\mathbb{Z}_2$. The action is given by $(x,y) \mapsto (y,x)$. My attempt: I can see that $S^2\times S^2$ is a subset of $S^5$. I am unsuccessfully trying to restrict the action of $S^1$ on $S^5$ to $S^2\times S^2$ to get the required map.
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I have the following question:  Is $(\mathbb{R}^2, lexicographic) \cong (\mathbb{R}, discrete)\times (\mathbb{R}, usual)$?  Note: the symbol $\cong$ denotes a homeomorphism between topological spaces  What I think:  I think that in fact there is a homeomorphism between those spaces, because is possible to express the lexicographic order as union of vertical lines with open extremes, I mean this is a basis for the lexicographic order. If I consider the Identical function I have the homeomorphism. Am I right?
18246
Consider $f(x)=mx$ and $g(x) =bx$ in which $m&gt;0$ and $b&gt;0$. Does there exist any homeomorphism $h$ such that $h \circ f$ = $g \circ h$? If exists find the domain of $h$.   Obviously I tried $h(mx)=bh(x)$. But I don't know how to go for solving it! In addition, I have no idea about how to prove homeomorphism existence in such problems.
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Let $f:X \to Y$ be a homeomorphism between topological spaces $X$ and $Y$. Let $A \subseteq X$ and $B = f(A) \subseteq Y$ (given the subspace topology) and let $g: A \to B$ be the restriction of $f$ given by $g(a) = f(a)$ for $a \in A$.  Showing that $g$ is a homeomorphism.   $g$ injective   Let $g(a_1) = g(a_2)$, $a_1, a_2 \in A$.  $\implies f(a_1) = f(a_2)$  $\implies a_1 = a_2$ as $f$ homeomorphic.   g surjective   Let $y \in B$  $\implies y = f(a)$ for some $a \in A$.  $\implies y = g(a)$ for some $a \in A$.   $g$ continuous   Let $U \subseteq B$ be an open set in $B$.  $f(a) = g(a)$ for all $a \in A =&gt; f^{-1}(b) = g^{-1}(b)$ for all $b \in B$.  So $g^{-1}(U) = f^{-1}(U)$ which is open in $A$ as $f$ homeomorphic.   $g^{-1}$ continuous   Let $U \subseteq A$ be an open set in $A$.  $g(U) = f(U)$ which is open in $B$ as $f$ homeomorphic.  So $g$ is homeomorphic.  Is my understanding correct, this seems almost too straightforward so I was wondering have I overlooked something?
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In Jeffery M. Lee's Manifolds and Differential Geometry Exercise 1.77:    For smooth manifolds $X$ and $Y$, show that if $f_0: X \rightarrow Y$ and $f_1: X \rightarrow Y$ are $C^r$ homotopic then there is a $C^r$ map $H: X \times \mathbb{R} \rightarrow Y$ such that $H(x,s) = f_0(x)$ for $s \le 0$ and $H(x,s) = f_1(x)$ for $s \ge 1$    The definition of homotopy had been given only for $H: X\times [0,1] \rightarrow \mathbb{R}$.    But I came up with an example which explains the difficulty with this exercise. Let $X = \mathbb{R}$ and $Y = \mathbb{R}$, $f_0(x) = 0$, $f_1(x) = 1$. Both $f_0$ and $f_1$ are smooth maps, and homotopic via $H(x,s) = s$ where the domain of $H$ is $\mathbb{R} \times [0,1]$. I don't see how you could extend the domain of $H$ to be $\mathbb{R} \times \mathbb{R}$ as the exercise requires since the partial derivative in the $x$ direction would be zero for $s \notin [0,1]$ and suddenly nonzero for $s \in [0,1]$    Somehow I need a general way to make the partial derivatives in the $x$ direction zero at the boundaries.
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I'm trying to prove that $\mathbb R^n \setminus \{\text{straight line}\}$ is homeomorphic to $\mathbb R^n\setminus\text{\{an axis\}}$. Their complements, a straight line and an axis, are isomorphic, since the bijective map we find is just a translation from one line to the other. So I thought it could be a good idea to prove that $G\cong H \Leftrightarrow G^c\cong H^c$, but I'm not sure that this is generally true. If it is true, which map could we take to prove it? And if it is false, how should I say that $\mathbb R^n \setminus \{\text{straight line}\} \cong \mathbb R^n \setminus \{\text{axis}\}?$ Or at the very least, that $\mathbb R^n \setminus \{\text{straight line}\} \simeq \mathbb R^n \setminus \{\text{axis}\}$?.
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Suppose C is a Cantor set in the Euclidean plane, or even in R^3. Suppose h is a homeomorphism of C onto itself. Can h be extended to a homeomorphism of the whole space? What about if h preserves the ´end points´ of the Cantor set? Seems that the latter is a necessary condition but I don´t know how to prove it.
43323
I am self-studying topology from Munkres.  One exercise asks, in part, to show that the spaces $(0,1)$ and $(0,1]$ are not homeomorphic.  An apparent solution is as follows: If you remove a point, $x$, from $(0,1)$, you get a disconnected space; however, you can remove the point $\{1\}$ from the space $(0,1]$ and the space will still be connected.  This apparently means the spaces aren't homeomorphic. I don't quite see the connection to the definition of homeomorphic spaces (that two spaces are homeomorphic if there is a bicontinuous function between them).   So, what is the connection between the "bicontinuous" definition of homeomorphisms and the "removing a point" procedure?  Sorry for missing something obvious.
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We have two spaces $X=\{(x,1/n):n\neq 0, n\in\mathbb{Z}, x\in\mathbb{R}\}$ and $Y=\{(x,n):n\neq 0, n\in\mathbb{Z}, x\in\mathbb{R}\}$. On both spaces we introduce the equivalent relation $(x,y)\sim (x&#39;,y&#39;)$ if $x=x&#39;$ and $y=y&#39;$ or $x=x&#39;=0$. That is, all points on the $y$ axis are collapsed to the same point.  We are asked whether $X/\sim$ and $Y/\sim$ are homeomorphic in quotient topologies.  It is easy to show that the original spaces are homeomorphic. However, I don't know how to answer the question about the quotient spaces.   My guess is that they might not be homeomorphic and some problem might occur at the $origin$ but I am not sure.  Any hint would be helpful! Thanks!
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Let $f\colon X\to X$ be a homeomorphism between a CW-complex $X$ and iteself.  Let $M_f=X\times [0,1]/(x,0)\sim (f(x),1)$, mapping torus of $X$ from $f$.  I want to calculate the fundamental group $\pi_1(M_f)$ of $M_f$ in terms of $\pi_1(X)$ and $f_*\colon \pi_1(X)\to \pi_1(X)$.  Are there any hint to do this?  Note: This is not a homework problem.
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I'm having trouble solving the following exercise in Hatcher's Algebraic Topology(1.3 #5):  "Let $X$ be the subspace of $R^2$ consisting of the four sides of the square $[0,1] \times [0,1]$ together with the segments of the vertical lines $x=\frac{1}{2},\frac{1}{3},\frac{1}{4},\ldots$ inside the square. Show that for every covering space $\tilde{X}\to X$ there is some neighbourhood of the left edge of $X$ that lifts homeomorphically to $\tilde{X}$. Deduce that $X$ has no simply-connected cover."  I tried piecing together open sets in $\tilde{X}$ which were homeomorphic to open square shaped $\epsilon$-neighbourhoods of points in the left edge of $X$, but I couldn't glue them together coherently.  My second idea was to use the path lifting property to lift the left edge, but I couldn't extend it to a lift of an open neighbourhood of the left edge.  Edit: Some more detail on my first attempt:  Suppose I have two open squares $U_1$ and $U_2$ in $X$ (containing the left edge) with nonempty intersection, and that have preimages $\coprod_i U_{1,i}$ and $\coprod_j U_{2,j}$ with each piece of that mapping homeomorphically onto $U_1$ and $U_2$ respectively.  Suppose I've chosen a specific $U_{1,i}$, and I want to choose a specific $U_{2,j}$ so that their union maps homeomorphically to $U_1$ union $U_2$. Such a choice can be specified by a point in the chosen $U_{1,i}$ that maps into $U_2$, but for each such point, there could be a different choice of $j$.  My intution tells me that to make this choice, one needs to appeal to the bad local nature of the left edge of $X$, (so that you're certain to get some of the vertical lines inside of the desired open set) but I'm not sure how to proceed with that.
123213
Let $S^1\subseteq\mathbb{C}$ be the unit circle and let $U\subseteq S^1$ be open. How to show that there exist a continuous function $$\theta:U\to\mathbb{R}$$ such that $$e^{i\theta(z)}=z$$ for all $z\in U$ if and only if $U\neq S^1$?  I have been able to show the $\Rightarrow$ direction but not the other one.  $(\Rightarrow)$ If such a continuous $\theta:S^1\to\mathbb{R}$ exist, then it is injective and so $\theta(S^1)$ is a subset of $\mathbb{R}$ that is homeomorphic to $S^1$. Now, $S^1$ is compact and connected so $\theta(S^1)$ is a closed interval $[a,b]$. But removing a point from $[a,b]$ gives two connected components, while removing a point from $S^1$ still gives a connected space. Hence, $[a,b]\not\cong S^1$, so $\theta$ doesn't exist.  How would you show the $\Leftarrow$ direction?  Edit:  This is an exercise in topology with no assumed prior knowledge of complex analysis. Note that we only need to show continuity of $\theta$ (not analyticity). The question is phrased in terms of complex numbers for simplicity of notation, but it is to be interpreted in terms of subspace of $\mathbb{R}^2$.
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I am trying to construct a diffeomorphism between $S^1 = \{x^2 + y^2 = 1; x,y \in \mathbb{R}\}$ with subspace topology and $\mathbb{R P}^1 = \{[x,y]: x,y \in \mathbb{R}; x \vee y \not = 0  \}$ with quotient topology and I am a little stuck.  I have shown that both are smooth manifolds, and I used stereographic projection for $S^1$, but now I am runing into trouble when I give the homeomorphism between $S^1$ and $\mathbb{RP}^1$ as the map that takes a line in $\mathbb{RP}^1$ to the point in $S^1$ that you get when letting the parallel line go through the respective pole used in the stereographic projection.   If I use the south and north poles I get a potential homeomorphism, but I cannot capture the horizontal line in my image, but when I pick say north and east then my map is not well defined as I get different lines for the same point in $S^1$. Can somebody give me a hint how to make this construction work, or is it better to move to a different representation of $S^1$ ?
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I'm working through Massey's "Basic Course in AT."  One of the problems is prove that a homeomorphism of the closed disk maps the boundary to the boundary and the interior to the interior.  How would one prove this?  I can't seem to get this one problem.
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Suppose we have a bijective continuous map $\mathbb{R}^n\to\mathbb{R}^n$ (relative to the standard topology). Must this map be a homeomorphism?  I have little doubt about this. I think that if it happens, I guess it's true, I've heard it is true, but I can not prove it.
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I need to show that any open interval is homeomorphic to the real line.  I know that $f(x)=a+e^x$ will work for the mapping $f:R \to (a,\infty)$ and $f(x)=b-e^{-x}$ will work for the mapping $f:R \to (-\infty,b).$  Without using two functions, how can I prove the result in general?