I am taking a first course in differential geometry, and we have looked at vector bundles and principle G bundles (I suppose that the following will hold for fibre bundles in general where the base is a manifold). My query concerns a point in several of the proof we have looked at, namely that we can find an open cover of the base manifold $B$, say $\{U_\alpha \}$, where the $U_\alpha \subset B$ are both coordinate charts and trivialising neighborhoods.  By trivialising neighborhood, I mean that there is a diffeomorphism $\Phi_\alpha$ such that $\Phi_\alpha : \pi^{-1}(U_\alpha) \rightarrow U_\alpha x V$ with $\pi : E \rightarrow B$ the map in the vector bundle and $V$ the typical fibre.  Naturally, it is nice if we have this is problems because then we can use local coordinates along with a chosen basis on the typical fibre, to have local coordinates on open sets of the total space $E$. We have used this assumption often when treating connections. Another case is in proving the existence of a smooth positive definite inner product on the total space E. We use this along with partition of unity, which requires the open neighborhoods that are both charts and trivialising neighborhoods to cover B.  EDIT:  I think in fact this is trivial. By virtual of being a fibre bundle, for every $p\in B$ we have an open neighborhood $p \in U \subset B$ and a corresponding local trivialisation $\phi _U$. Likewise, since $B$ is a manifold then we have an open cover of coordinate charts $V_\alpha$. Since topology is closed under finite intersection, we need simply take the intersection $U_p \cap V_\alpha$ where $p \in V_\alpha$ and take the union of these for all $p \in B$. Now we have our open cover of $B$ consisting of coordinate charts that are trivialising neihborhoods.
121227
We know a vector  bundle as a manifold $E$ over a  manifold $M$ is defined using transition maps over a collection of open sets each giving  us a trivialization. Since $M$ is a manifold, it has an atlas. Can we intersect the two collections of open sets and obtain a chart for $E$ which also gives us the trivialization?
72468
Let $p : E \to B$ be a covering map. And also assume that $B$ is path-connected and locally path-connected. Then, given y in $B$, there exists a neighborhood $U$ of $y_1$ in $B$ that is path-connected and evenly covered by $p$. Now the inverse image of $U$ by $p$ consists of slices $\{V_a\}$. Take a point $x$ in $E$ such that $p(x)=y$ and let $V'$ be the slice containing $x$.   Now the questions is, if $N$ is an arbitrary neighborhood of $x$ in $X$, then how to replace $U$ by a smaller neighborhood of $y$ such that the slice $V'$ containing $x$ can be included in the $N$? It seems very nontrivial to me...In the text I read, there is no detailed explanation about this. Could anyone give me some rigorous proof?
66682
The derivative of a map $F$ between manifolds $M$ and $N$ is defined by $$F_*X(f)= X(f \circ F)$$ where $X \in T_P(M)$, the tanget space at the point $P$.  We know that $$\left\{\frac{\partial}{\partial x^i}\bigg|_P\right\}_i$$ is a basis for $T_P(M)$. How to show that $$\left\{dx^i\bigg|_P\right\}_i$$ is a basis for the cotangent space $(T_P(M))^*$?  First, by $dx^i$, I guess we mean the derivative of the map $x^i$ as defined above, right? Is this map $x^i$ just picking out the ith coordinate? Secondly, to show that it is a basis, we need to show that $dx^i\left(\frac{\partial}{\partial x^j}\bigg|_P\right) = \delta^i_j.$ Where to go from here: $$\underbrace{(dx^i)_P}_{(\Phi_*)_P}\underbrace{\left(\frac{\partial}{\partial x^j}\bigg|_P\right)}_{X}f = \left(\frac{\partial}{\partial x^j}\bigg|_P\right)(f\circ x^i)?$$  I can use the chain rule but I am not sure exactly. Please help.
71221
It's a well-known theorem (Corollary 8.10 in Lee Smooth) that given a smooth map of manifolds  $\phi:M\rightarrow N$ and a regular value $p\in N$ of $\phi$, the level set $\phi^{-1}(p)\subset M$ is a closed embedded submanifold. Is the converse true? That is, given an embedded submanifold $S\subset M$, is there necessarily a manifold $N$, smooth map $\phi:M\rightarrow N$, and regular value $p\in N$ of $\phi$ such that $S=\phi^{-1}(p)$?   Prop. 8.12 in Lee Smooth shows that this is true locally; specifically,     Let $S$ be a subset of a smooth $n$-manifold $M$. Then $S$ is an embedded $k$-submanifold of $M$ if and only if every point $p\in S$ has a neighborhood $U\subset M$ such that $U\cap S$ is a level set of a submersion $\phi:U\rightarrow\mathbb{R}^{n-k}$.   (and any level set of a submersion is of course the level set of a regular value). I feel like this is the kind of question where, if there is a counterexample, it probably is very simple, but I wasn't able to come up with one.
102933
It's well known that, in $\mathbb{R}^n$:     (1) Open and Connected $\Rightarrow$ Path-connected   The proof essentially goes through the fact that (2) Every path-connected component will be open. Using this fact, we arrive at a contradiction if we suppose there are more than one path-connected component.  Well, trying to see where (1) would keep being valid, I arrived at the following:     If $X$ is a locally convex topological vector space, then (1) is valid.   But, to prove this, I proved (2). But for that, you use the (rather strong imo) convexity of a local base (to essentially repeat the argument for $\mathbb{R}^n$). But I'm not satisfied with this, as I think that we are using a lot of strong conditions (using the existence of a "line segment" to prove the existence of a "curve" seems very bazooka-like to me).  Anyway, given the previous considerations, my questions are:  Are there more general examples of spaces where (1) holds? Is there a characterization of spaces that satisfy this property?  And bonus question, since (2) implies (1) if the set is open.:  Are there more general examples of spaces where open sets satisfy (2)? Is there a characterization spaces that satisfy this property?
26897
Let $X$ denote a topological space and suppose that $\mathcal{O}$ is an open cover of $X$. Assume $\emptyset \notin \mathcal{O}.$ (Thanks Niels!) Now make $\mathcal{O}$ into an (undirected) graph as follows:   Vertexes: elements of $\mathcal{O}$. Edges: we draw an arc between two vertexes iff they have a non-empty intersection when viewed as subsets of $X$.      Question. Is it true that if $X$ is connected (as a topological space), then $\mathcal{O}$ is connected (as a graph)?   The case where the open cover $\mathcal{O}$ has only two elements is easy, but the general case where $\mathcal{O}$ is allowed to have arbitrary cardinality is harder.
70030
Problem. Assume that $U$ is an open and connected subset of $\mathbb R^2$, and $\gamma :[0,1]\to U$ is a closed curve, which is not null-homotopic in $U$ and not necessarily simple closed. Show that there exists a simple closed curve, which is not null-homotopic in $U$.  I am sure that this is a standard problem, and I am looking for an elegant, if possible, solution or a reference. I have in mind the following sketch of proof:  a. Construct a polygonal closed curve $\tilde\gamma$ which is homotopic with $\gamma$ in $U$.  b. Construct $\tilde\gamma$ so that it intersects itself in finitely many points, and not whole segments.  c. Use induction on the number of self–intersections of $\tilde\gamma$, in order to show that $\tilde\gamma$ is written as a product (in the fundamental group's sense) of simple closed polygonal curves.
44338
Let $M$ be a manifold of dimension $n$ such that there exist two   charts $(U_a,\phi_a)$ and $(U_b,\phi_b)$ such that $U_a,U_b$ are   connected and $U_a\cap U_b\ne\emptyset$. Moreover the transition   function $\phi_{ab}$ neither preserves the orientation nor reverses the orientation. I have to show that $M$ is not orientable.   A manifold is orientable if it admits an oriented atlas, i.e. an atlas in which all the transition functions have Jacobian with positive determinant.
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.
112721
I have a question about a proof in John Lee's Introduction to Topological Manifolds (5.22).  Given CW-complex $X$ with skeletons $X_n$ and open cover $\left(U_\alpha\right)_{\alpha\in A}$, we inductively define partitions of unity $\left\{\psi_\alpha^n\right\}$ on $X_n$ subordinate to cover $\left(U_\alpha^n=U_\alpha \cap X_n\right)_{\alpha\in A}$, such that   $\psi_\alpha^k\mid_{X_{k-1}}=\psi_\alpha^{k-1}$ If $\psi_\alpha^{k-1}\equiv 0$ on an open subset $V \subset X_{k-1}$, then there is an open subset $V'\subset X_k$, containing $V$ on which $\psi_\alpha^k\equiv 0$   At the end of the induction step, prof. Lee proceeds to argue that supports of $\left\{\psi_\alpha^{n+1}\right\}$ form locally finite family.     If $x$ is in the interior of an $(n+1)$-cell, then that cell is a neighborhood of $x$ on which only finitely many of the functions $\psi_\alpha^{n+1}$ are nonzero by construction. On the other hand, if $x\in X_n$, because $\left\{\psi_\alpha^n\right\}$ is a partition of unity there is some neighborhood $V$ of $x$ in $X_n$ on which $\psi_\alpha^n\equiv 0$ except when $\alpha$ is one of finitely many indices, and then (2) shows that $\psi_\alpha^{n+1}\equiv 0$ on $V'$ except when $\alpha$ is one of the same indices.   I don't understand why does this imply local finiteness. Because of (2), there is a $V'$ for each such $\alpha$, but unless I'm missing something obvious, there can be infinitely many such alphas. How can we conclude there is a single neighborhood of $x$ in $X_n$ on which all these functions vanish?
11001
After googling this fact (which is used in Bott and Tu to show the existence of good covers of manifolds) I've gotten the impression this is somewhat difficult to prove.  But I also came across this homework problem (problem 0) with a hint:    which gives me the impression that it is maybe not so difficult as I thought.  If anyone has any thoughts about how to prove this result I'd be very appreciative. Thanks for your time.
22818
The problem:  Consider the smooth quadric $Q=V(X_{0}X_{1}+X_{2}X_{3}+X_{4}^{2})\subset\mathbb{P}^{4}$ and the line $L=V(X_{0},X_{2},X_{4})$ contained in it. Prove that there exists a vector bundle $F$ on $Q$ with a section vanishing exactly at $L$.  The hardest thing is to get the transition matrices. Any help is appreciated.
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.
94215
We have $A=\left\lbrace 2019,1010,\ldots,(n+2018)/n,\ldots \right\rbrace$ and $B=A \cup \left\lbrace 1 \right\rbrace$  How can I show directly (that is, not using the compactness criterion in $\mathbb{R}$) that every open cover $G$ of $B$ admits a finite subcover?  moreover, I know that $A$ is not compact and I am aware that I would need a single example to prove it, any hint on how to find an open cover of $A$ that does not contain a finite subcover?  My first thought was to choose the set of all neighborhoods with radius r centered around each point of A but apparently this is wrong and I do not understand why.
118489
On page $10$ of Hatcher's , he gives a proof that the Whitney sum of the trivial line bundle over $\mathbb{R}P^n$ and the tangent bundle is equal to the Whitney sum of copies of the tautological line bundle.  A summary of the proof is   1) The image of the tangent bundle of the sphere under the quotient is the tangent bundle of real projective space, and the image of the normal bundle is the trivial line bundle over real projective space.  2) He says the sum of the tangent bundle and normal bundle of the sphere is trivial.  3) He then proves that the trivial line bundle over the sphere is isomorphic to the normal bundle and proves that the normal bundle under the quotient is mapped to the tautological line bundle over real projective space.  But this doesn't make any sense because he claimed earlier that the image of the normal bundle is trivial and the tautological bundle is NOT trivial. Could someone please clarify what has happened? Is the image of the normal bundle of the sphere under the quotient the tautological bundle or the trivial bundle?
56066
I am reading Lee's book on Differential geometry. In Chapter 4, Lee has this exercise.  Let $M$, $N$ be smooth manifolds, and let $f:M\to N$ be a smooth map. Define $F : M\to M \times N$ by $F(x) = (x, f(x) )$. Show that for every smooth vector field $V$ of $M$ , there is a smooth vector field on $W$ on $M \times N$ that is $F$-related to $V$.  I can understand that we must have $W_{(x,f(x))}=V_x\oplus Df(x)V_p$ for all $x\in M.$ The set $\{(x,f(x)):x\in M\}$ is a closed set in $M\times N.$ Therefore, if we can show that for all $(p,f(p))\in M\times N$ there is a neighbourhood $U_p$ and a smooth vector field on $U_p$ extending $V_x\oplus Df(x)V_x$ we are done by partition of unity. But I cannot show that. Can someone help me out?
91569
Suppose that $f:M \to N$ is a smooth map of two manifolds and $E$ is a vector bundle over $N$. Given a connecton $\nabla$ on $E$ one can pull it back to $f^*N$ (which is a bundle over $M$) and obtain a connection $f^* \nabla$.   If $E=TN$ is a tangent bundle one can consider the Levi-Civita connection $\nabla$, with respect to the chosen metric $g$ on $N$. It is characterised uniquely as torsion free metric connection. Given a map $f:M \to N$ one can pullback the metric $g$ to $f^*g$. If it happens that $f^*g$ is again a metric (which is the case for example for immersions) one can consider the Levi Civita connection on $M$ with respect to $f^*g$. I wonder how these two constructions are related, in particular      Is there a way to pullback affine connections to get again affine connection (i.e. connections on tangent bundle)?     Additionally I would like to know:     What is the relation between Christoffels symbols for the connection and for its pullback?