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  1. R

    Showing the Fundamental Group of S^1 is isomorphic to the integers

    There are basically still 3 issues with your homotopy: 1) You forgot an extra coefficient of 2 in front of n in your H for s >= 1/2. 2) It doesn't line up on s=1/2, i.e. the two expressions you gave are different when s=1/2 3) It isn't a homotopy of loops (i.e. not a based homotopy). 1 is...
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    Showing the Fundamental Group of S^1 is isomorphic to the integers

    Your idea seems correct. To separate into cases depending on whether s is smaller or larger than 1/2 is a good idea in this problem (as you have done). However try to think about what you are doing. f_{m+n} loops with a speed of (m+n) for the whole interval [0,1]. f_n * f_m instead first...
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    A Question about de Rham's Theorem

    I'm a bit rusty on de Rham cohomology so excuse my overly long answer which was mostly just to convince myself, but now I feel I might as well leave the details in there. It seems like (for smooth manifolds at least) it follows from the universal coefficient theorem and deRham's theorem as you...
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    Infinite set of points on a line.

    What does it mean to do a thing to infinity? Until you are clear on that your question makes no sense. There is an accepted way to define what you are trying to express which is the intersection of all the line segments. Let us consider the case where we start with the line segment from 0...
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    Finding inverse of linear mapping

    That is sufficient. Since f is a homeomorphism it has a continuous inverse g : Y -> X. You can show fairly easily that g is linear. g is linear and continuous and therefore bounded. Whether this is explicit enough or not I do not know, but I doubt you will find a much more explicit...
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    Finding inverse of linear mapping

    By a mapping do you mean a bounded linear map? And by inverse do you require your inverse to be bounded linear as well? Maybe you are only working with isometries? If you do not impose more conditions on f, X or Y than that f is bounded linear, then the problem is impossible in general. Just...
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    Fundamental Group

    Just construct it in the obvious way. Given two loops f_1 : I \to X and f_2 : I \to Y you can construct a path I \to X \times Y where t \mapsto (f_1(t),f_2(t)). Conversely given a path f : I \to X \times Y you can construct paths I \to X and I \to Y by projecting onto X and Y. You can show that...
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    Fundamental Group

    Yes. Just consider a loop f : [0,1] \to X. This is homotopic to the constant map c(t)=p by the usual straight-line homotopy (t,i) \mapsto (1-i)f(t) + ip. You can verify yourself that this works. The intuitive idea when dealing with fairly nice subsets of \mathbb{R}^2 is that the fundamental...
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    Square root loop

    You probably know this, but let me just re-iterate one important result. Given a topological space X, and an equivalence relation ~ on X we can construct the quotient space X/~. This quotient space posses the property: - Let g : X \to Y be a continuous function with the property that g(x)=g(y)...
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    Square root loop

    That would be the case if \sigma sent x to [x], but it does not. To see why this doesn't give any problems suppose \sigma(\cos(2\pi t_1)+i\sin(2\pi t_1)) = \sigma(\cos(2\pi t_2) + i\sin(2\pi t_2)) with t_1,t_2 \in [0,1) and show that you must get t_1=t_2. To conceptually visualize why this...
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    Square root loop

    I'm assuming [x,y] refers to the equivalence class of the point (x,y) under the antipodal equivalence relation. Why don't you think it's a bijection? Are you doubting it's injective or surjective? The map is indeed a bijection so it would be helpful if you could point out why you doubt this...
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    Sphere as an embedded submanifold

    This is correct. I'm sorry if I somehow conveyed with my post that it was a non-trivial fact. jem05 was asking for an embedding from the n-sphere into some larger topological space, so I just provided the simplest one I could think of.
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    A sequence or a single point?

    When you're working with (A,T), then a point is not a real number, but a sequence of real numbers (i.e. an element of A). So you take two points and see if you can separate them by open neighborhoods, but a point is a sequence. Thus you pick two sequences satisfying your given condition and try...
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    Sphere as an embedded submanifold

    Well in that case you know the n-sphere is a subset of the (n+1)-dimensional Euclidean space \mathbb{S}^n \subseteq \mathbb{R}^{n+1}. \mathbb{R}^{n+1} is of course a manifold so we just define \phi : \mathbb{S}^n \to \mathbb{R}^{n+1} to be the inclusion \phi(x)=x. This is an embedding.
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    Sphere as an embedded submanifold

    Submanifold of what? It makes no sense to speak of a submanifold without referring to what manifold it's a submanifold of. For instance in your theorem you assert that \phi(M) is a submanifold of N. If you don't care what manifold it's a submanifold of you can just use the rather trivial...
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    What is a nontrivial ultranet?

    The definition of an ultranet is given as definition 11.10. Just before 11.11 the author notes that for any directed set \Lambda and fixed element x \in X, the map P : \Lambda \to X defined by P(\lambda)=x is an ultranet. Such ultranets are called trivial ultranets. Thus a non-trivial ultranet...
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    Homeomorphism (defining a chart)

    Maybe I'm misunderstanding what you mean by basis-independent, but couldn't you just choose some basis e_1,\ldots,e_n \in V beforehand and define the homeomorphism in terms of that in the obvious way, and then show that it is actually independent of the basis? For instance in the...
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    Filling different shaped volumes

    Remember that a square is two-dimensional surface (and in this case the cross-section) not the whole figure itself. Similarly a circle is two-dimensional while the 3-dimensional equivalent is a ball. If every cross-section is a circle of the same radius r, then we get a cylinder, but imagine...
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    Filling different shaped volumes

    I'm not sure what you mean by all opposite sides parallel. You could also construct a cylinder with radius \sqrt{1/\pi} and height 1. That would fill at the same rate as the two containers I mentioned. You could also have something that transforms from being circular at the bottom to square at...
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    Filling different shaped volumes

    No it's not possible. What you know is the surface area at all points (actually its derivative, but since you know it at all points it's the same). Consider for instance a 1x1x1 cube. And instead consider a box with bottom 2 x 0.5 and height 1. These two containers are different, but they...
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    Simple Topology problem (Munkres)

    That's correct. This is also correct though I'm not sure why you present 1 and 2. The observations made in these are good to make if you want to find a counterexample, but they are not a necessary part of the proof that the statement is false. For that you just need to find a counterexample...
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    Simple Topology problem (Munkres)

    This example is not correct, but the idea is the right one. The problem is that T1 and T2 are not topologies since \emptyset and X are not open in either, but they should be open in both. Try to modify T1 and T2 to include these and see if your idea still works. For the former consider a family...
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    Number of monomials of degree d in finite field F[x]

    Ok so the difficulty is counting the number of (n+1)-tuples (d_1,\ldots,d_{n+1}) of non-negative integers. I think I actually gave the necessary details in the previous post, but it may be hard to visualize so let me provide an example. To visualize the strategy imagine you have n=3 and d=6...
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    Number of monomials of degree d in finite field F[x]

    I don't think Phyisab**** meant to be offensive, but rather to put out that it's kind of crazy to get such an answer from a professor. Personally I don't think that response (from the professor) was appropriate unless you're in an extremely advanced graduate class (he could at least have kindly...
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    Number of monomials of degree d in finite field F[x]

    I must admit I'm not sure if I totally understand the OP, but as he hasn't gotten any answers to his question I'll try. I usually take F[x] to mean the polynomial ring in ONE variable, and in that case there is one monomial if you require it to be monic (namely x^d) and |F|-1 if not (the...
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    Sphere Intersects with Plane

    A point (x,y,z) is in both the sphere and the yz-plane iff: (x-3)^2 + (y-2)^2 + (z+5)^2 = 36 x=0 So just substitute x=0 in the sphere equation to get a circle equation. For the intersection with the z-axis we have: (x-3)^2 + (y-2)^2 + (z+5)^2 = 36 x=0 y=0
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    What's the questions asking me to show?

    That the topologies are equal. A topological space is a set X with a topology T. In your case you are probably given a set S1 with a topology T1 (maybe product topology, maybe subspace topology, maybe discrete topology, maybe some other topology) and you are given a quotient map q : Y \to S1...
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    Connected vs. Path Connected Sets

    Sorry about stating that A is open. I switched back to an alternative definition of a separation that says that A and B are disjoint and both open which is equivalent to your definition. Anyway you don't need it by the argument you presented as you show that the limit of the sequence is in A and...
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    Connected vs. Path Connected Sets

    If it was disconnected you could find a separation A, B of it. Let A be the component containing (0,0). Clearly the ray y=0,x<=0 is connected so it must be in A. The graph of (x,sin(1/x)) for x>0 is also connected and so it must be B (since if it were in A, B would be empty). Now define a...
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    Connected vs. Path Connected Sets

    Those definitions seem somewhat non-standard (especially 2). What is a line segment? If it has the elementary meaning consider the unitcircle \{(x,y) | x^2+y^2 =1\} which doesn't seem to be path connected according to your definition 2 (but is path connected according to the standard definition...
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