Here is the problem 1-8 in Lee's introduction to smooth manifolds:          An angle function on a subset $U \subset S^1$ is a continuous     function $ \theta: U\to \mathbb R$ such that $e^{i\theta(z)}=z$ for all $z \in U$. Show that there exists an angle function on an open subset $U\subset S^1$ if and only if $U \neq S^1.$ For any such  angle function, show that $(U,\theta)$ is a smooth coordinate chart for $S^1$ with its standard smooth structure.      I wonder if $U\ne S^1$ is sufficient for the existence of such a continuous angle function(I feel that the statement should be "there exists an angle function on an open subset $U\subset S^1$ if and only if $U$ is not dense in $S^1.$").    Let's take a special case when $U=S^1-\{-1\}$. Suppose there is such a continuous angle function $ \theta: U\to \mathbb R$ such that $e^{i\theta(z)}=z$ for all $z \in U$. Then we must have $\theta(z)=arg(z)+2k\pi$. However, It seems that there will always be a "jump" somewhere on the circle making $\theta$ discontinuous. For example, if we define $\theta(z)=\arg(z)$, then when $z$ approaches $-1$ from the first quadrant and the four quadrant separately, $\theta(z)$ approaches to $\pi$ or $-\pi$.(This argument turns out to be wrong, see the answers and comments given below by repliers)    Since professor Lee doesn't make any corrections to this problem, it should be correct as given. But how to define such a continuous angle function for arbitrary proper open subset of $S^1$? Thanks in advance!
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 $\mathbb{R}^2$.
73855
For , $(x,y)\in \mathbb R^2$ with $(x,y)\not =(0,0)$ , let $\theta=\theta(x,y)$ be the unique real number such that $-\pi&lt;\theta \le\pi$ and $(x,y)=(r\cos \theta , r\sin \theta)$ , where $r=\sqrt{x^2+y^2}$. Then the resulting function $\theta :\mathbb R^2 \setminus\{(0,0)\}\to \mathbb R$ is (A) differentiable. (B) continuous but not differentiable. (C) bounded but not continuous. (D) neither bounded nor continuous.  Here , $\theta=\arctan\left(\frac{y}{x}\right)$. So it is differentiable in $\mathbb R^2 \setminus \{(0,0)\}$. So option (A) is correct. Am I correct ?
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!
114828
Let $f: \mathbb{R} \to D \subseteq \mathbb{R}$ be a continuous function. Is there a function $f$ that satisfies the following property? $\forall y \in D$, there are exactly 3 $ x_1,x_2,x_3 \in \mathbb{R}$ such that $f(x_1) = f(x_2) = f(x_3) = y$.
90575
I have the following problem.      Let $f:D\to \mathbb C$ be a holomorphic function, where $D=\{z:|z|\leq 1\}.$ Let $$f(z)=\sum_{n=0}^\infty c_nz^n.$$ Let $l_2(A)$ denote the Lebesgue measure of a set $A\subseteq \mathbb C$ and $G=f(D).$ Prove that $$l_2(G)=\pi\sum_{n=1}^\infty n|c_n|^2.$$   After a long struggle I managed to come up with the following formula $$l_2(G)=\iint_D |f\,&#39;(z)|^2dxdy.$$ It looks like the right thing to use because $$f\,&#39;(z)=\sum_{n=1}^\infty nc_nz^{n-1},$$ which is similar to what I have to prove. I understand that the $\pi$ will appear when I integrate something over the angle $\phi$ in polar coordinates. But I don't know how to find the square of the absolute value of the right-hand side to even start integrating...  EDIT: The function is supposed to be  (one-to-one). (I'm not perfectly sure I'm translating the term correctly. The (Polish) word in the statement of the problem was "jednolistna". I have not met it before.)
70476
A subset $M \in \mathbb{R}^n$ is called a smooth k-dimensional submanifold of $\mathbb{R}^n$, $k \leq n$, if any point $x \in M$ has a neighborhood $O_x$ in $\mathbb{R}^n$ in which $M$ is described in one of the following ways:  (1) there exists a smooth vector-function $$F:O_x \to \mathbb{R}^{n-k}, \text{  } rank \frac{dF}{dx}|_q=n-k $$ such that $O_x \cap M = F^{-1}(0)$  (2) there exists a smooth vector-function $$f:V_0 \to \mathbb{R}^{n}$$ from a neighborhood of the origin $0 \in V_0 \subset \mathbb{R}^k$  with $$f(0)= x \text{   and   } rank \frac{df}{dx}|_0=k $$ such that $f: V_0 \to O_x \cap M$ is homeomorphism.  (3) there exists a smooth vector-function $$\Phi:O_x \to O_0 \subset \mathbb{R}^{n},$$ onto a neighborhood of the origin  $0 \in O_0 \subset \mathbb{R}^n$  with $$\text{  } rank \frac{d \Phi}{dx}|_x=n $$ such that $ \Phi(O_x \cap M)= \mathbb{R}^k \cap O_0$   I am trying to show the equivalence of these definitions and my first question is why is each definition imposing itself on the rank?
110358
In Milnor's book Topology from the Differentiable Viewpoint there's the following problem:  Problem $6$ (Brouwer). Show that any map $S^n\to S^n$ with degree different from $(-1)^{n+1}$ must have a fixed point.  My solution: Assume that the map $f:S^n\to S^n$ has no fixed points. Let $a:S^n\to S^n$ denote the antipodal map $a(x)=-x$. Then the map $a\circ f$ is homotopic to the identity as follows:  Since $x$ is never mapped to $-x$ (by assumption), there exist a unique shortest great circle arc from $a(f(x))$ to $x$, simply take the straight line homotopy flowing along such arches to get a homotopy $a\circ f\simeq\operatorname{id}$.  Now we have: $$1=\deg(\operatorname{id})=\deg(a\circ f)=\deg(a)\deg(f)=(-1)^{n+1}\deg(f)$$ and thus $\deg(f)=(-1)^{n+1}$.  My questions:   Is this solution correct? How to show that the homotopy is continuous? It seems intuitively true to me, but I'm not sure about how to proceed to show it. Maybe saying that it is the flow of some good tangent field? Are there other (elegant/short/interesting) proofs for this fact?
105431
Let $f:(a,b)\to\mathbb{R}$ be a continuous function with continuous derivative. Suppose $f$ has a unique critical point $x_0\in (a,b)$. If $f$ has a local maximum at $x_0$, then $f$ must have a global maximum at $x_0$. Intuitively, if $f$ had a global maximum somewhere else, the graph would have to turn around, leading to a second critical point.  I would like to know if a similar results holds in two dimensions.     Does there exist a connected open subset $U\subset\mathbb{R}^2$ and a function $f:U\to\mathbb{R}$ with continuous first partial derivatives, such that $f$ has a unique critical point $u_0\in U$, which is a local maximum but not a global maximum?
57757
Let $M$ be a smooth manifold. Let $A$ and $B$ be disjoint closed sets of $M$. Show there exists a smooth function $f$ such that $f^{-1}(0)=A$ and $f^{-1}(1)=B$.   This is my idea so far,   Since $A$ and $B$ are disjoint closed subsets $\{M-A,M-B\}$ is an open cover for $M$. Therefore there exists a partition of unity $\{\psi_{1},\psi_{2}\}$ with $\psi_{1}$ supported in $M-A$ and $\psi_{2}$ supported in $M-B$. Furthermore $\psi_{1}+\psi_{2}=1$. Then $\psi_{1}(1-\psi_{2})$ is zero on $A$ and $1$ on $B$. I get the inclusions $A\subset f^{-1}(0)$ and $B\subset f^{-1}(1)$. I've tried adding a few more open sets to the cover like $M-(A\cup B)$ and $(M-A)-B$ and $(M-B)-A$ but I still can't ensure that the function only vanishes at $A$.   Then I moved on to attempt to find any bump function that is 1 only on $A$ and ran into the same problem.   Am I missing something here or do I just need to be clever with how I define a function. Any hints would be much appreciated. Thanks.
89826
Let $\Omega \subseteq \mathbb{C}$ be open and $\gamma:[\alpha,\beta]\rightarrow \Omega$ be a piecewise continuously differentiable and closed path. Why does $z^{-1}$ have no antiderivative on $\mathbb{C}\setminus0$? Why is $\int\limits_\gamma z^{-1} dz = 2\pi i \cdot \text{ind}_\gamma (0)$, where $\text{ind}_\gamma$ denotes the winding number.
45352
I'm working in Lee's book on topological manifolds and have gotten stumped on the first question in chapter 5, the chapter on cell complexes.  The problem is:   Let $D$ and $D'$ be two closed cells not necessarily of the same dimension.     Show that any continuous map $f:\partial D \to \partial D'$ can be extended to a continuous map $F:D \to D'$ such that $F(Int \; D) \subseteq Int \; D'$. Given points $p \in Int \; D$ and $p' \in Int \; D'$, show that $F$ can be chosen so $F(p) = p'$. Show that if $f$ is a homeomorphism then $F$ can be chosen to be a homeomorphism.   I proved the first part roughly as follows: First suppose $D$ and $D'$ are convex and each contain $0$ in their interior (in their respective ambient spaces).  Then every element other than $0$ in $D$ can be expressed uniquely in the form $\lambda q$ where $q \in \partial D$ and $\lambda\in (0,1]$ (an equivalence relation on the chords connecting $0$ to boundary points partitions $D- \{0\}$).  I then define the map $F(\lambda q) = \lambda f(q)$ which is continuous since if $\lambda_n q_n \to \lambda q$ in $D$ then $F(\lambda_n q_n) \to F(\lambda q)$.  Lastly $\lambda q \in Int \; D$ implies $\lambda &lt;1$ and so $\lambda f(q)$ is an interior point of $D'$ since $f(q)$ is a boundary point and $D'$ is convex.    Now if we suppose $D$ and $D'$ are arbitrary closed cells there are homeomorphisms $g_1: \overline{\mathbb{B}^n} \to D$ and $g_2: \overline{\mathbb{B}^m} \to D'$ (where possibly $m=n$) then we have that $g_2^{-1}\circ f \circ g_1$ is a continuous map between the boundaries of two closed balls and so by the first part can be extended to a continuous map $F:\overline{\mathbb{B}^n} \to \overline{\mathbb{B}^m}$.  The mapping $g_2 \circ F \circ g_1^{-1}$ is a continuous map that satisfies the desired claim.    My issue is that parts 2 and 3 don't seem (to me at least) to follow that easily from the proof I constructed.  Maybe my proof is wrong and I'm not seeing it, but more likely I think that I'm missing the spirit in which Lee is intending for us to approach this problem.
112640
Let $X = \{(x,y) \in \mathbb{R^2}: x &gt; 0, y = sin(1/x)\}$. Is this set open, closed, or neither, and what is the boundary, interior, and closure of $X$?  I just want to make sure I am thinking about these problems in a reasonable manner. I am also trying to apply this to other concepts I have learned.  Is $X$ open?  X is open if every $x \in X$ is an interior point. Take an arbitrary $p \in X$ so that $p = (p_x, p_y)$ then $p_x &gt; 0$ and $p_y = sin(1/p_x)$. Define an open ball $B_r(p)$, and take $q \in B_r(p)$ be an arbitrary point. Is $q \in X$? Is it possible for $q$ to not be in X? Since $y = sin(1/x)$ is defined and is continuous for all $x \neq 0$,then there will always exist a $q_y = sin(1/q_x)$, provided $q_x \neq 0$. How do I know $q_x \neq 0$? Otherwise, I propose that X is open.   Alternatively, X is open if whenever $f: \mathbb{R^2 \to R}$ is continuous and $U \subseteq R$ so that if $U$ is open, then $f^{-1}(U)$ is also open. Could I construct such a $U$ to prove that it's open? My only concern is that if $U = (-1,1)$, then $f^{-1}(U) = (-\infty, \infty) = \mathbb{R}$ which is not open nor closed.   Is $X$ closed?   I am not sure about this one. I want to say that is not closed, since $X^c$ is not open. It is not open because $(0,0)$ defines a boundary point which lies in the set $X^c$.  Thanks!
119136
Let $U \subset \mathbb{C}$ be an open set such that {$ z\in\mathbb{C};|z|\leq2$}$ \subset U$ and $f:U\rightarrow\mathbb{C}$ a holomorphic function.   Show that there exists infinitely many natural numbers $n \in \mathbb{N}$ such that $f(\frac{1}{n})\neq(\frac{1}{n+1})$  So i have been told to start by comparing with the function $g(z)=(\frac{z}{z+1})$  however i am still not getting very far with this. Would i have to prove by a contradiction?   Completely stuck and unsure, some help would be fantastic, many thanks!
97399
Specifically, the question is as follows:     Let $\mathbb{D}$ be the open unit disk in $\mathbb{C}$ and suppose that $f:\bar{\mathbb{D}}\rightarrow\mathbb{C}$ is a continuous function such that both $\Re f$ and $\Im f$ are harmonic.       (a) Show that   $$f(re^{i\theta})=\frac{1}{2\pi}\int_{-\pi}^\pi f(e^{it})P_r(\theta-t)dt$$   for all $re^{i\theta}$ in $\mathbb{D}$.       (b) Show that $f$ is holomorphic in $\mathbb{D}$ if and only if   $$\int_{-\pi}^\pi f(e^{it})e^{int}dt=0$$   for all $n\geq1$.   I believe (a) is a direct consequence of Poisson's representation theorem for harmonic functions, stated as follows:  Let $\mathbb{D}$ be the open unit disk and $f:\partial\mathbb{D}\rightarrow\mathbb{R}$ be continuous. Then there exists a continuous function $u:\bar{\mathbb{D}}\rightarrow\mathbb{R}$ such that $u(z)=f(z)$ if $z\in\partial\mathbb{D}$, $u$ is harmonic on $\mathbb{D}$, and $u(re^{i\theta})=\frac{1}{2\pi}\int_{-\pi}^\pi P_r(\theta-t)f(e^{it})dt$ and $u$ is unique.  It seems to me that this can just be applied to the real and imaginary parts of the function $f$ from the question to derive the desired result, but maybe I'm making a careless mistake?  For (b), it's not obvious to me how part (a) assists in the solution. This is a homework question, so I am looking for assistance with understanding it, but not a full solution. Thanks.
112038
Let $f:\mathbb{C}\to\mathbb{C}$ be a complex valued function of the form $f(x,y)=u(x,y)+iv(x,y)$. Suppose that $u(x,y)=3x^2y$. Then  $f$ cannot be holomorphic on $\mathbb{C}$ for any choice of $v$.  $f$ is holomorphic on $\mathbb{C}$ for a suitable choice of $v$.  $f$ is holomorphic on $\mathbb{C}$ for all choices of $v$.  $u$ is not differentiable.    () well, I have calculated by applying CR equation, getting the option $1$ is correct? could anyone tell me  just am I right?thank you.
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!
10588
Let $\Omega\subset\mathbb{R}^2$ be an open subset such that $\partial\Omega$ is a closed, simple curve.   I'm trying to find an example of an $u:\overline{\Omega}\to\mathbb{R}$ such that $\Sigma:=\text{graph}(u)$ is a minimal surface and, yet, there exists another minimal surface $\Sigma'$ with $\partial\Sigma'=\partial\Sigma$ and $\text{Area}(\Sigma')&lt;\text{Area}(\Sigma)$.  Does such an example exist?