Prove that there is no map of degree two from $S^2$ to the torus $T^2$.   I'm struggling with this problem. I've tried lifting the map to the covering space but I'm not sure what to do from there. I keep getting results that I know are wrong. Most examples I can find of problems like this involve maps from $T^2$ to $S^2$, and I think those call for different techniques than what is needed here.
121129
Why every map $f : S^n \to T^n (n&gt;1)$ has topological degree zero?  I don't know anything about covering spaces, and has been told to me that this assertion comes from this theory!  I do appreciate any help.
17047
(This is exercise 2.2.28 from Hatcher) Consider the space obtained from a torus $T^2$, by attaching a Mobius band $M$ via a homeomorphism from the boundary circle of the Mobius band to the circle $S^1\times \{x_0\} \subset T^2$.   We set $A=T^2$, $B=M$, $X:=A\cup B $, $A\cap B\simeq S^1$, so the (reduced) Mayer-Vietoris sequence yields  $$ 0\to \tilde H_2(T^2)\oplus\tilde H_2(M)\to \tilde H_2(X)\to \tilde H_1(S^1)\to \tilde H_1(T^2)\oplus \tilde H_1(M)\to \tilde H_1(X) \to 0$$ and replacing the homology groups we already know we have $$0\to \mathbb Z\overset{\alpha}{\to} \tilde H_2(X) \overset{\beta}{\to} \mathbb Z \overset{\gamma}{\to} \mathbb Z^2\oplus \mathbb Z\overset{\delta}{\to} \tilde H_1(X)\to 0$$  Now, I have to figure out what the maps $\alpha,\beta,\gamma,\delta$ do, and this is where I'm stuck. I'd appreciate a detailed explanation.
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.
72793
I think it wants me to show that there's no bijection between the two sets.  I first tried to show that there's simply no bijection, then I realised that it doesn't work.  If I'm to show there's no bijective morphism that carries multiplication in nonzero real numbers to addition in real numbers, how am I supposed to do it? I'm thinking about doing something with 0, since it's the element of the first set but not the second set.  Thanks!
117295
I am supposed to show that by adding a point to the Moore plane, that the subspace of a $T_4$-space is not necessarily a $T_4$-space. I do know that the Moore plane is NOT a $T_4-$space. Does anybody here know which point could be meant?
28645
I'm trying to prove the next result:     Let $X$ be a connected covering space of the torus   $T:=\mathbb{R}^2/\mathbb{Z}^2$. If $X$ is compact then It is   homeomorphic to a product of two circles.   Probably the most direct way of proving this is to show that there are only $3$ possible connected covering spaces of $T$, to know: $\mathbb{R}\times\mathbb{R}$, $\mathbb{R}\times S^1$ and $S^1\times S^1$ (the last one is the only one which is compact, so the result follows). As far as I know this can be deduced from the classification theorem of covering spaces (see ), however I need to prove first:   $\mathbb{R}^2$ is the universal covering space of $T=\mathbb{R}^2/\mathbb{Z}^2$ The fibers of this covering by the subgroups of $\pi_1(T)$ are precisely those $3$ spaces (up to conjugacy).   I'd appreciate any hints to prove these two facts, or any ideas to prove the main result in a simpler way. Thanks in advance.
53768
Prove: If $f,g:S^{n-1} \to X$ are homotopic maps, then $X\sqcup_fD^n$ and $X\sqcup_gD^n$ are homotopy equivalent.   I think it can be proved by showing they are both deformation retracts of $X\sqcup_H(D^n\times I)$ where $H$ is the homotopy between $f$ and $g$.  However, I have difficult in proving that the deformation retracts are continuous map. In fact, I have difficulty in representing a map in quotient spaces like $X\sqcup_fD^n$. I think a map from $X\sqcup_fD^n$ to $W$ can be represented by two maps: $m_1: X\to W$, $m_2: D^n\to W$, where for $x\in S^{n-1}$, $m_1\circ f(x)=m_2\circ i(x)$.  Then I construct the deformation retract this way: $m_1: X\to X$. For $x\in H(S^{n-1},t)$, $m_1(x)=H(S^{n-1},0)$, otherwise $m_1(x)=x$.  $m_2: D^n\times I\to D^n\times {0}$: $m_2((D^n,t))=(D^n,0)$.  It is easy to verify that $m_1$ and $m_2$ define a map from $X\sqcup_H(D^n\times I)$ to $X\sqcup_fD^n$. As long as this is a continous map, obvioulsy then we find a deformation retract. But it seems such a map is not continous?
65715
Consider $\mathbb{R}^2$ as $\mathbb{R} \times \mathbb{R}$ with the product topology.  I am simply trying to show that the two projections $p_1$ and $p_2$ onto the first and second factor space respectively are not closed mappings.  It seems like this should be easy, but I have not been able to come up with a closed set in $\mathbb{R}^2$ whose projection onto one of the axes is not closed.  I don't really have any work to show...I've really just tried the obvious things like closed rectangles and unions of such, the complement of an open rectangle or union of open rectangles, horizontal and vertical lines, unions of singletons, etc., and haven't come up with anything non-obvious, which I hope is where the answer lies.  It's bothering me that I can't come up with an answer, and I'd appreciate some help.  Thanks.
84916
This is one of the exercises of 'Do Carmo' (Section 3.5, 12)     How do you prove that there are no compact (i.e., bounded and closed in $\mathbb{R}^3$) minimal surfaces?   Thanks!
21953
Can you help me with this?     Let $S^2 := \{x\in \mathbb R^3:||x||_2 = 1\} \subset (\mathbb R^3, ||\cdot||_2)$  and $T:S^2 \to (\mathbb R, |\cdot|)$ be a continuous map.   a) Why does T assume its maximum ($T_{max}$) and its minimum ($T_{min}$)?   b) Is there a point $x_0 \in S^2$ with $T(x_0) = T(-x_0)$?   c) Is there a value $T\in]T_{min},T_{max}[$ that is assumed at only one $x\in S^2$?     a) is easy, it is compact since it's closed and bounded (Heine-Borel).  For b) I know it can't be injective since it maps from a higher to a lower dimension, but I don't know if I can use that to show what I need.  For c) I assume I can use non-injectivity again, but don't know either how. Any hints?
18219
I have a  basic question  Let $f$ be a $\mathcal{C}^1$  map  from  $[0,1]$ into $\mathbb{R}$ vanishing in $0$ and $1$. I want to prove that   $$\int_0^1 f'^2 \geq \pi^2 \int_0^1 f^2 $$  and to see when we have equality.
15614
Could you please verify this (rather simple) proof? I'm a bit new to this kind of reasoning.  The question is:  Let $\mathcal{T}_1$, $\mathcal{T}_2$ be two topologies on some set $X$, when is the identity map $id:X \to X$, $id(x) = x$ continuous?  Here is my reasoning:   If $\mathcal{T}_1 = \mathcal{T}_2$, then for any $S \in \mathcal{T}_2$, $id^{-1} (S) = S \in \mathcal{T}_1$ so then it's continuous.  If $\mathcal{T}_2 \subseteq \mathcal{T}_1$, then it's exactly like above, then it's also continouos.  Now:  If  $\mathcal{T}_1 \subseteq \mathcal{T}_2$ then take some $x \in U \in \mathcal{T}_2 \setminus \mathcal{T}_1$, then $id^{-1}(U) \not\in \mathcal{T}_1$, so it's not continuous.  So the identity map is continuous if $\mathcal{T}_1$ is finer than $\mathcal{T}_2$.   Is this OK?  Is there a better way to show this?  Any extra applicable theory?  Thanks in advance
39678
My professor was lecturing today and he made this statement which I was unable to verify.  (I worded it nicer)  There is no map which is both smooth and onto from $S^1$ to $S^1$$\times$ $S^1$.  When he said this he included the original "one can clearly see" statement which is usually when the most people seem to be confused in his class.  Thus, I was hesitant to ask why.  So, please no one lecture me on the fact that I should have asked anyway.  If it is really that obvious, I would rather look stupid here than in class :)
30057
I was asked to prove that a standard torus(which means we don't consider those pathological cases where it intersects with itself, e.g horn torus) is diffeomorphic to $ \mathbb S^1\times \mathbb S^1$.  I was thinking if we could prove it this way: Since every point on the torus can be uniquely defined with a pair of angles $(\theta_1, \theta_2)$. Then we construct a diffeomorphism $\phi(\theta_1, \theta_2)=(\tilde{\theta}_1 ,\tilde{\theta}_2)$ which maps every point on the torus to every point on $\mathbb S^1 \times \mathbb S^1$. Since the map is definitely bijective and smooth with a smooth inverse. We're basically done...  THERE MUST BE SOMETHING WRONG I THINK.  Thanks a lot for everyone's help!
989
What is an example of an open map $(0,1) \to \mathbb{R}$ which is not continuous? Is it even possible for one to exist? What about in higher dimensions? The simplest example I've been able to think of is the map $e^{1/z}$ from $\mathbb{C}$ to $\mathbb{C}$ (filled in to be $0$ at $0$). There must be a simpler example, using the usual Euclidean topology, right?
70144
Similar to a previous question I asked, this is the other question that I couldn't answer in an exam I gave for my Topology course:  Suppose $g_1,g_2$ are non-negative integers such that $2-2g_2$ divideds $2-2g_1$. I have to construct a covering map $S_{g_1} \rightarrow S_{g_2}$, where $S_g$ is denotes an orientable compact connected surface without boundary with genus $g$.  Now, I attempted to try this problem for surfaces with smaller genera, for example a 2-torus and a 2-sphere. Clearly $2-2g$ is the Euler Characteristic of the said surface, and by classification of surfaces we have to only consider connected sum of 2-tori. So, I have $\chi(S^2)=2$, and $\chi(T^2) = 0$, here we have $2$ dividing $0$, hence I have to construct a covering map from $T^2 \rightarrow S^2$, and I have no idea how to move forward.  Addendum: I have to got great answers to my question here but I had a doubt when it comes to the points being mapped, specifically in Giulio Bresciani's answer. Clearly the handles of the figure is mapped to the handles in the figure above, but I'm unable to understand the mapping of the twist of the cover (above figure) to the extreme handles of the base (the figure below). Any help?
44201
I want to prove the Borsuk-Ulam Theorem:  For every continuous map $f:S^2\rightarrow \mathbb{R}^2$ there exist a pair of antipodal points $x$ and $-x$ in $S^2$ with $f(x)=f(-x)$  I did google search and I got the following hint: My Efforts I know there is quotient map $q:S^2\rightarrow RP^2$ and by property of quotient maps any continuous function from $S^2$ will factor through $q$, so I get a map from $\tilde{f}:\mathbb{R}P^2 \rightarrow \mathbb{R}^{2}$ I have inclusion map, $i:\mathbb{R}^2-\{0\} \rightarrow \mathbb{R}^2$. I have quotient map, $q': \mathbb{R}^2-\{0\}\rightarrow RP^1$. So I get a map $\tilde{i}:RP^1\rightarrow \mathbb{R}^2$ How should I proceed from here ? I am not able to construct a map from $RP^2$ to $RP^1$ For the next step, I think I have to use the fact that $\pi_1(RP^2)=\mathbb{Z}/2\mathbb{Z}$ and $\pi_1(RP^1)=\mathbb{Z}$