Let $\mathbb{Z}_2$ be the group $\{1,-1\}$. I want to construct an action of $\mathbb{Z}_2$ on the torus $T^2$ such that the orbit space is homeomorphic to the sphere $S^2$. Could anyone give some hints?
120677
I understand that a torus is obtained from a sphere by adding a handle. I'm working on a question which is asking if it is possible to obtain a sphere from a quotient of a torus? It seems like this should be possible by perhaps identifying the insides of the torus? But I'm not quite sure how to properly express this.  Help is very much appreciated.
66771
Is there a "nice" way to think about the quotient group $\mathbb{C} / \mathbb{Z}$?  Bonus points for $\mathbb{C}/2\mathbb{Z}$ (or even $\mathbb{C}/n\mathbb{Z}$ for $n$ an integer) and how it relates to $\mathbb{C} / \mathbb{Z}$.  By "nice" I mean something like:   $\mathbb{R}/\mathbb{Z}$ is isomorphic to the circle group via the exponential map $\theta \mapsto e^{i\theta}$, and $\mathbb{C}/\Lambda$ is a complex torus for $\Lambda$ an integer lattice (an integer lattice is a discrete subgroup of the form $\alpha\mathbb{Z} + \beta\mathbb{Z}$ where $\alpha,\beta$ are linearly independent over $\mathbb{R}$.)   Intuitively, it seems like it should be something like a circle or elliptic curve.
78836
Question 5.33 from "Topology and its Applications"  by Baesner is to compute the fundamental group of the torus ($T^2$) with $n$ points removed. I can "see" in my mind that if we remove one point we get a bouquet of two circles. Less clear is what happens when we remove two (or more) points. Any hints?
57119
Let the group $G \subset S_7$, $|G|=8$. Show that there exists $i\in{1,...7}$ such that for all $f\in G,f(i)=i$.  I have attempted this problem, but I am not entirely sure if my thinking is correct. Using orbit stabilizer theorem, I write $|G|=|orb_G(i)||stab_G(i)|$, and since $|G|=8, $ then $|stab_G(i)|$ is equal to either $1,2,4,8$ and $|orb_G(i)|$ equal to $8,4,2,1$ respectively. Now, when the order of the orbit is 8 and stabilizer 1, can I conclude that there exists a function that maps $i$ to $i$?  Any help would be appreciated!
90054
I'm trying to solve the following exercice :  Prove that the first homology group $H_1(E(K); Z)$ of a knot  exterior is an infinite cyclic group which is generated by the class of the meridian.   With $ E(K) $ is the closure of the complement of a tubular neigborhood $ N(K) $ of $ K $ and the merdian is just the boundary of a meridien disk of $ K $ .  Here is the adequate formulation :  Let $K$ denote a knot in $S^3$. $ N(K) $ is the image of an embedding $ S^1 \times D^2 \longrightarrow S^3  $ whose inverse $ f : N(K) \longrightarrow S^1 \times D^2 $ is called a trivialization. A meridien disk of $K$ is $ f^{-1} (D^2 \times \lbrace y \rbrace ) $ for a point $ y \in K $ . $ E(K) = cl(S^3 \setminus N(K)) $ .  I proved that $  E(K) $ is a submanifold with boundary of $ S^3 $ and that $ E(K) \cup N(K) = S^3 $. Then I used the known fact that the Mayer-Vietoris sequence still relevant for a cover of a manifold consisting of two submanifold with boundary to conclude that indeed $ H_1(E(K); \mathbb{Z}) = \mathbb{Z} $.  The issue is that I was not able to get the information on the generator by applying this method.  So my question is how can I exhibit the generator ( knowing that normally there is two generators since the boundary of $ N(K) $ is a torus, a meridien and a longitude ).  Thanks in advance.
41682
Let $P$ denote the quotient space obtained by the action of $\mathbb{Z}\backslash2\mathbb{Z}$ by the antipodal map $z\mapsto\frac{1}{z}$ on the riemann sphere $\hat{\mathbb{C}}$ (identified here with $\mathbb{C}\cup\left\{\infty\right\}$). I identify $P$ with the set:  $\left\{ z\in\mathbb{C}:\left|z\right|&lt;1\right\} \cup\left\{ e^{it}:0\leq t\leq\pi\right\} $   that is to say, I use elements of the above set as the representatives for the equivalence classes in $P$. I am looking for formula for a function $f:P\times P\rightarrow[0,\infty)$ such that $f$ is a metric on $P$. Specifically, I would like a formula for $f$ that I can evaluate by plugging in representative elements in the above set (or something like that, more or less).  Thanks in advance.  Edit: forgive me for sounding desperate, but I cannot make due with an explanation of how to obtain such a formula. I want the formula.   An analogy for you if I have yet to make myself clear: suppose I was asking for the area of a square with side length $s$. The answers I have received so far for my question are akin to saying "multiply $s$ by itself" or "use the area formula for a square". The answer I am looking for is akin to saying "$s^{2}$". I need the formula. And please, no expressions with differentials, nor matrices, or any of that. I need to know how to compute the metric by using the complex numbers in the indicated set that I have identified with $P$.
36882
Let $\pi$ be a group and let $K(\pi, 1)$ be a connected CW complex such that $\pi_1(K(\pi, 1)) = \pi$ and $\pi_q(K(\pi, 1)) = 0$ for $q \neq 1$. Does there exist an isomorphism between $H_*(K(\pi, 1); A)$ and $\text{Tor}_*^{\mathbb{Z}[\pi]}(A, \mathbb{Z})$?
94027
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 $\mathbb R^n \setminus \{\text{straight line}\} \simeq \mathbb R^n \setminus \{\text{axis}\}$?.
86117
Let $\mathbb{Q}$ be the set of rationals, as relative topology to $\mathbb{R}$.   How to show that $\mathbb{Q}$ is homeomorphic to $\mathbb{Q}^2$?  More general, if $T$ is a countable dense space(no isolated poiont) with every singleton closed, how to show that $T^2$ is homeomophic to $T$.
31296
Evidently, $(\mathbb{Z} \oplus \mathbb{Z})/ \langle (2,2) \rangle$ has order $4$, but I think it's infinite.   The four cosets are listed as $(0,0) +  \langle (2,2) \rangle$, $(0,1) +  \langle (2,2) \rangle$, $(1,0)+  \langle (2,2) \rangle$ and $(1,1) +  \langle (2,2) \rangle$. However, $(2,0)$ doesn't appear to be in any of these cosets. Maybe the answer I'm being told is wrong.
7174
I'm trying to solve this problem from Hatcher 3.3.24.  Let $M$ be a closed connected 3-manifold, and write $H_1(M, \mathbb{Z})$ as $\mathbb{Z^r} \oplus T$ where $T$ is torsion. Show that $H_2(M, \mathbb{Z})$ is $\mathbb{Z^r}$ if $M$ is orientable or $\mathbb{Z^{r-1}} \oplus \mathbb{Z_2}$ if $M$ is not orientable. In particular $r \geq 1$ if $M$ is not orientable. Using exercise 6 construct examples showing there are no other restrictions on the homology groups of closed 3-manifolds. (In the nonorientable case consider the manifold $N$ obtained from $S^2 \times I$ by identifying $S^2 \times \{0\}$ with $S^2 \times \{1\}$ via a reflection of $S^2$.)  I think I've solved the first part, here's my proof:   First note the following theorem: If $M$ is a closed connected and orientable n-manifold then $H_n(M, \mathbb{Z}) = \mathbb{Z}$ and $Tors(H_{n-1}(M, \mathbb{Z})) = 0$. If nonorientable, then $H_n(M, \mathbb{Z}) = 0$, $Tors(H_{n-1}(M, \mathbb{Z})) = \mathbb{Z_2}$.  Now note by Poincare duality and UCT we have that $H_2(M) \cong H^1(M) \cong \mathbb{Z^r} \oplus T_0 \cong \mathbb{Z^r}$.   In the nonorientable case consider the exact sequence:  $0 \longrightarrow H_2(M) \longrightarrow H_1(M) \longrightarrow H_0(M) \longrightarrow 0$  But we know $H_1(M) = \mathbb{Z^r} \oplus Z_2$ from the earlier theorem and $H_0(M) = \mathbb{Z}$.   Therefore $H_2(M) \cong ker(\mathbb{Z^r} \oplus Z_2 \rightarrow \mathbb{Z}) \cong \mathbb{Z^{r-1}} \oplus \mathbb{Z_2}$.   For the rest of the question I'm quite lost.
22069
Let $X$ be the quotient space of $S^2$ under the identifications $x\sim -x$ for $x$ in the equator $S^1$. Compute the homology groups $H_i(X)$.   I wrote my solution/attempt below and I would like to know if it it correct and, more importantly, if there is a simpler solution. I tried using cellular homology but I'm not very experienced with it and kept getting nonsense.   Here is my attempt:  $X$ can be written as the following pushout: $$\begin{align*}  \require{AMScd}  \begin{CD}   P^1 @&gt;{i}&gt;&gt; P^2\\   @V{i}VV @VVV \\   P^2 @&gt;&gt;&gt; X  \end{CD} \end{align*}$$ where $i:P^1\hookrightarrow P^2$ is the inclusion. So by a Mayer Vietoris argument, we have $H_i(X,P^2) \approx H_i(P^2, P^1)$. This can be simplified to $\tilde{H}_i(P^2/P^1) \approx \tilde{H}_i(S^2)$. Now we consider the LES of the pair $(X, P^2)$. $$0\to H_2(P^2)\to H_2(X)\to H_2(X, P^2)\to H_1(P^2)\to H_1(X) \to 0$$ We know all homology groups involved except those of $X$, so the LES reads: $$0\to \Bbb{Z}\to H_2(X) \to \Bbb{Z}\to \Bbb{Z}_2\to H_1(X)\to 0$$  The problem lies in identifying the map $\Bbb{Z}\to \Bbb{Z}_2$, call it $f$. The LES tells us that $H_1(X)\approx \Bbb{Z}_2/\text{im }f$. On the other hand, by the Van Kampen theorem for attaching 2-cells, $X$ can be viewed as $P^2$ with a 2-disk attached along the path $aa$, hence $\pi_1(X)\approx \Bbb{Z}_2$. This is abelian, so it coincides with $H_1(X)$. Then $f$ must be the $0$ map. Using that, from the LES we can also compute $H_2(X) \approx \ker f \oplus \Bbb{Z}\approx \Bbb{Z}^2$.
115866
Let be $\mathbb{R}$ the real numbers and $\mathbb{Z}$ the integers. Let $G = \mathbb{R}/\mathbb{Z}$. Determine   $O_\pi(G)$ - the normal maximal $\pi$-group of $G$, $T(G)$ - torsion subgroup.
115349
I am using Do Carmo's Riemannian Geometry, and struggling to solve a problem.  The problem is:  Show that the mapping $F:\mathbb{R}^2\to\mathbb{R}^4$ given by  $$F(x,y)=((r\cos y+a)\cos x,(r\cos y+a)\sin x,r\sin y\cos\frac{x}{2},r\sin y\sin\frac{x}{2}))$$  induces an embedding of the Klein bottle into $\mathbb{R}^4$.  I know that the Klein bottle is defined as a quotient manifold $T^2/G$, where $T^2$ is the 2-torus, $G$ is a group of diffeomorphisms of $T^2$ formed by $\{A,Id\}$, and $A$ is the antipodal map, i.e. $A(p)=-p$.  Besides, I know that $T^2=S^1\times S^1$, where $S^1$ is the unit circle.  Also, I found that $F$ is injective.  However, I don't have a clue about how to attack this problem. I tried googling, but found nothing.  Any help will be appreciated.
81174
I am asking if whether or not a finite group acts freely (as homeomorphisms) on $\mathbb R^n$.  To answer in the negative, it suffices to show: for any homeomorphism $f$ such that $f^d=\text{id}_{\mathbb R^n}$, then $f$ has a fixed point.  I am looking for a complete self-contained answer.
92234
Let us define a properly discontinuous action of a group $G$ on a topological space $X$ as an action such that every $x \in X$ has a neighborhood $U$ such that $gU \cap U \neq \emptyset$ implies $g = e$. I would like to prove that this property is equivalent to, having given $G$ the discrete topology and in the $X$ locally compact Hausdorff case, the map $G \times X \rightarrow X \times X$ given by $(g, x) \mapsto (x, gx)$ being proper (i.e. closed and preimage of compact sets is compact) plus the action being free.  I have managed to prove one direction, that is, if the action is proper and free with $G$ having the discrete topology then it is properly discontinuous. I'm having trouble though with the other direction. Here is an attempt: let's denote by $\rho : G \times X \rightarrow X \times X$ the map $\rho(g, x) = (x, gx)$. Suppose $K \subset X \times X$ is compact. We wish to show $\rho^{-1}(K)$ is compact. Let $(g_i, x_i)$ be a net in $\rho^{-1}(K)$. Then $\rho(g_i, x_i) = (x_i, g_i x_i)$ admits a convergent subnet, so passing to it we may assume $x_i \rightarrow x$ and $g_i x_i \rightarrow y$. Essentially we must now find a way to prove $g_i$ converges, but I can't seem to do this. Any hints?
57897
List all $i$ for which there is a closed orientable $6$-manifold $M$ with $H_i(M) =\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}$.   I am working on an old exam problem and this one stumped me. Progress so far:  I can take $Y=\mathbb{T}^3 \times \mathbb{S}^3$ and get the desired group for $i =1,5,2,4$ I think. But this construction I believe gives $H_3(Y)=\mathbb{Z}^2$. What to do in this case?