I am trying to show that the tangent bundle of $S^2$ not diffeomorphic to $S^2\times \mathbb{R}^2$. This is from an exam, where there is a hint stating that this is more than showing that $TS^2$ is non-trivial.  I know how to show the hairy ball theorem, according to which $TS^n$ is non-trivial iff n is even.  I also know that a vector bundle $\pi:E\rightarrow M$ of rank $m$ on a smooth manifold $M$ is trivial (by definition) iff there exists a diffeomorphism $f:E\rightarrow M\times \mathbb{R}^m$ such that for every $p\in M$, $f$ induces a vector space isomorphism $f:\pi^{-1}(p)\rightarrow \{p\}\times \mathbb{R}^m$.  So I see that showing that $TS^2$ is non-trivial only guarantees that $TS^2$ is not diffeomorphic to $S^2\times \mathbb{R}^2$ via a diffeomorphism satisfying the property above, but it is not enough to conclude that there isn't any diffeomorphism.  How can I show this then?
120531
I have seen the claim that $T\mathbb{S}^2$ and $\mathbb{S}^2 \times \mathbb{R}^2$ are not diffeomorphic, but I have only ever seen the proof that they are not isomorphic as vector bundles (which is a cute application of the hairy ball theorem). How would I prove that they are indeed not diffeomorphic (or better, not homeomorphic)?
117050
I would like to prove that the map $f: S^n \times S^m \to 2S^{m+n+1}: ((x_1,..,x_{n+1}), (y_1,...,y_{m+1})) \to (x_1,...,x_{n+1},y_1,...,y_{m+1})$  is an imersion. Here $2S^{m+n+1}$ is the $m+n+1$ dimensional sphere with radius $\sqrt2$.   I know that I have to prove that the map $(f_\star)_p : T_p(S^n \times S^m) \to T_{f(p)}(2S^{m+n+1}) : [\gamma] \to [f \circ \gamma]$ is injective but I fail to do this. My initial idea was the following:   Assume that $(f_\star)_p([\gamma_1]) = (f_\star)_p([\gamma_2])$. It holds that $[f \circ \gamma_1] = [f \circ \gamma_2]$ and thus $f \circ \gamma_1  \sim f \circ \gamma_2$. So there is a chart $(U,\phi: U \to \mathbb{R}^p)$ with $ U \subset \mathbb{R}^{n+m+2}$ open such that $(\phi \circ (f \circ \gamma_1))'(0) = (\phi \circ (f \circ \gamma_2))'(0)$. It is easy to see that $f:S^n \times S^m \to f(S^n \times S^m)$ is a homeomorphism and thus $\phi \circ f$ is a chart for $S^n \times S^m$. Thereby $\gamma_1 \sim \gamma_2$ and thus $[\gamma_1] = [\gamma_2]$.   I don't think that this is true because we do not know if $\phi \circ f$ is globally good defined. Can someone help me?   Thanks!
70761
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.
33887
This is problem 4.19 in Lee's Topological Manifolds.  I have been working on this problem for a couple days now, and just a need a hint in the right direction.  If $M$ and $N$ are two $n$-manifolds, and $B_1\subset M$ and $B_2\subset N$ are two open, regular coordinate balls (definition below), the connected sum $M\#N$ is the quotient of the disjoint union $(M-B_1)\displaystyle\sqcup (N-B_2)$ by the relation that identifies points on the spherical boundaries of each component via some homeomorphism $h$.  Now I want to show that there are two open sets $U,V\subset M\#N$, such that:   $U\cong M-\{p\}$ and $V\cong N-\{q\}$, for some points $p\in M$ and $q\in N$ $U\cap V\cong S^{n-1}\times\mathbb{R}$ $U\cup V=M\#N$   Here's my intuition: take a larger coordinate ball $D$ in $N$, containing $B_2$, which works because $B_2$ is regular. $(M-B_1)\sqcup (D-B_2)$ is a saturated open set, so its image in $M\#N$ is open. This is what I want to be $U$.   Now I've managed to define a homeomorphism $g$ from $D-B_2$ to $\overline{\mathbb{B}}_1(0)-\{0\}$ (the punctured closed unit ball). If I can then map that punctured unit ball to $\overline{B}_1-\{p\}$, I can try and show the map $f:\ (M-B_1)\sqcup(D-B_2)\rightarrow M-\{p\}$  $$ f(x)=\begin{cases} x&amp; x\in M-B_1\\ g(x)&amp; x\in D-B_2\\ \end{cases} $$  is a quotient map, and then use the uniqueness of quotients to show $U$ is homeomorphic to $M-\{p\}$.  However, this is too hard for me to do, because I need to somehow incorporate $h$ in the mix, so that the identifications of $f$ match those of $h$.  Can someone provide a hint here? I feel like this is a lot of work for one problem, and I understand intuitively what to do here, but getting all the details right is proving to be too much.  I'm also open to hints about part 2. as well (3. is easy).  Definition: A coordinate ball $B\subset M$ is called regular if it has a neighborhood $B'\supset B$, such that there is a homeomorphism $k: B'\rightarrow \mathbb{B}_s(0)$ (this is the open ball of radius $s$ around $0$). Under this homeomorphism, $B$ goes to $\mathbb{B}_r(0)$ for some $0&lt;r&lt;s$, and $\overline{B}$ goes to $\overline{\mathbb{B}}_r(0)$.
43533
Let $n \geq 1$, and $M \subset \mathbb{R}^{n+2}$ be a smooth $n$-dimensional manifold which is closed in $\mathbb{R}^{n+2}$.  Show that for each $m \in M$, there exists a line $L \subset \mathbb{R}^{n+2}$ such that $L\cap M = \{m\}$.   I tried to consider $m$ as the origin of $RP^{n+1}$ space and use Sard's Theorem, but it seems not to work.
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.
6822
Let $f\colon S^1\to S^1$ be any smooth map. Prove that there exists a smooth map $g\colon\mathbb{R}\to\mathbb{R}$ such that $f(\cos(t),\sin(t))=(\cos(g(t)),\sin(g(t)))$ and satisfying $g(2π)=g(0)+2\pi q$.  The book told me to show that $g(2\pi)=g(0)+2\pi q$, then extend $g(t+2\pi) =g(t)+2\pi q$.  But I don't understand what they mean. How can I get function $g$ from $f(\cos(t),\sin(t))=(\cos(g(t)),\sin(g(t)))$, and what is $g$?
112444
At the moment I'm taking an algebraic topology course. In our current exercise we have to prove the following: Consider any topological space $X$ and any two maps $f, g : X \to S^n$, such that $f(x) \neq -g(x)$. It follows that $f$ is homotopic to $g$.  My idea is to use the mapping cylinders $M_f, M_g$ to construct a homotopy via $$M := M_f \amalg M_g/ (x, 0)_f \sim (x,0)_g$$ But this doesn't seem to work. In addition I don't know how $f(x) \neq -g(x)$ is going to help me (I can see how this could help if $n=1$, but I have no clue how this would work for $n&gt;1$).  I would appreciate any hints towards a solution.  Thanks in advance.
5062
Let $M$ be a smooth ($C^\infty$) manifold. Let $\mathfrak{X}(M)$ be a set of all vector fields on $M$ and let $\mathfrak{F}(M)$ be a set of all real smooth functions on $M$. $\mathfrak{X}(M)$ is a real vector space and it is also a module over $\mathfrak{F}(M)$. We know that partial derivatives constitute a basis for tangent space at any point $p\in M$. Is there some sort of basis for $\mathfrak{X}(M)$ (as a vector space or as a module)? Do partial derivatives constitute a basis here as well? I think this question  is similar to mine, but because of the way it's written, I'm not really sure I understand the question and I'm not sure we're asking about the same thing.
122553
Let $X$ be the subspace of $\mathbb{R}^2$ that is the union of the circles $C_n$ of radius $n$ and center $(n,0)$ for $n \in \mathbb{N}$.  Show that $X$ and $\bigvee_\infty S^1$ are homotopy equivalent, but not homeomorphic.   It is unclear to me why the "obvious" maps between the two spaces do not give a homeomorphism.   It seems like the difference would have to be a result of the behavior of the map near the origin.  Could I get a hint?
21325
What's the easiest way to see this? The only thing I could think of was to try to patch together trivializations. I couldn't find a way to make that work. Thank you! edit: For the record, here's why I asked about this special case of the more general result about fiber bundles over contractible spaces. In the much beloved book by Bott and Tu, it's claimed that the Leray Hirsch theorem can be proven in the same way the Kunneth theorem is proven:  induct on the size of a finite good cover for the base space, then apply the Mayer Vietoris sequence and the Poincare lemma for the induction step. It's assumed that there exists a finite good cover for the base space, but it's not assumed that this cover is a refinement of the cover of the base space, which gives the local trivializations of the fiber bundle. Therefore, to apply the Poincare lemma in the induction step, it seems that you need to know that the result I asked about is true. Since fiber bundles had just been introduced in the text, I thought there may be a short, elementary proof that the authors had taken for granted.
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 :)
123187
I am trying to work out the details of following example from page 101 of Tu's An Introduction to Manifolds:     Example 9.3. Let $\Gamma$ be the graph of the function $f(x) =  \sin(1/x)$ on the interval $]0, 1[$, and let $S$ be the union of   $\Gamma$ and the open interval $I=\{(0,y)\in \mathbb{R^2} |−1&lt;y&lt;1\}$.      The subset $S$ of $\mathbb{R^2}$ is not a regular submanifold for the   following reason: if $p$ is in the interval $I$, then there is no   adapted chart containing $p$, since any sufficiently small   neighborhood $U$ of $p$ in $\mathbb{R^2}$ intersects $S$ in infinitely   many components.   Using Tu's definitions, I need to show that given a neighborhood $U$ of $p$, there exists no homeomorphism $\phi$ from $U$ onto an open set of $V \subset\mathbb{R^2}$ with the property that $U \cap S$ is the pre-image with respect to $\phi$ of the $x$ or $y$ axes intersected with $V$.   I am not sure where the fact that $U \cap S$ has infinitely many components comes into play. I tried to derive a contradiction using this fact; but even if restrict my attention to connected neighborhoods $U$ of $p$, the intersection of the connected set $\phi(U)$ with the $x$ or $y$ axes might have infinitely many components (I think), so there's no contradiction with the fact that homeomorphism preserves components.  I would appreciate any help!
93169
Let $B^n$ be a n-dimensional disc (ball) with boundary $S^{n-1}$. Prove that  $B^n/S^{n-1}$ is homeomorphic to $S^n$.  Could someone check my proving, please?  Let  put the ball $B^n=\{x\in\mathbb R^n\mid |x|\leq 1\}$ into $\mathbb R^{n+1}$ using $x\mapsto (x,0)$. Then we can use homeomorphism $(x,0)\mapsto (x,\sqrt{1-|x|})$. We fall into the half of the sphere $S^n$. The boundary of the ball after this will turn just into points of the form $(*,0)$. If you pull it to the point then we will get suspension $S^{n-1}$ or $S^n$.
65740
Let $M$ be a smooth manifold and let $N$ be a manifold with boundary, both with the same dimension $n$. If $dF_{p}$ is an isomorphism, then $F\left(p\right)\in\mbox{int}N  $.  I am trying to prove this theorem to prove a result about smooth embeddings.  Here is how I am thinking the problem could be solved. Assume thet $F(p)$ is a boundary point of $N$. Then there exists a chart $(V,\psi)$ at $F(p)$ such that $\psi(V) $ is an open subset of the upper half space $\mathbb{H}^{n}$. I guess we have to use some fact about $M$ having no boundary and the the fact that $dF_p$ is an isomorphism to show that there is a contradiction, but I cannot figure that out.
45537
Simple question but I am confused.  What is the boundary of $S^2\times\mathbb{R}$? Is it just $S^2$?  What would be the general way to evaluate the boundary of a product manifold?  Thanks for the replies!
112655
I am trying to show that a smooth function $f:S^1\times S^1\to \mathbb  R$ must have more than two critical points. Since $f$ attains maximum and minimum, it must have at least two critical points. How would one show that they can't be two?   If one considers the gradient vector field $\nabla f$ then by the Poincare Hopf index theorem its index is equal to the Euler characteristic of the torus, i.e. it is $0$. Therefore $\nabla f$ has an even number of zeros.   This question has been posted  but it hasn't been answered.