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If you've done any number theory, you are probably aware of the following classic formula: Let $L/K$ be an extension of number fields, $\mathfrak{p}\in\text{Spec}(\mathcal{O}_K)$ and $\mathfrak{p}\mathcal{O}_L=\mathfrak{P}_1^{e_1}\cdots\mathfrak{P}_n^{e_n}$. Define $f_i$ to be $\,[\mathcal{O}_L/\mathfrak{P}_i:\mathcal{O}_K/\mathfrak{p}]$. Then, $$\sum_i ... 5 I don't agree with the explanation on Wikipedia. They write down an isomorphism, which is an instance of a natural isomorphism, and then claim that it is not natural. This is not correct. But their reasoning is that they - secretly - change the domain categories. More specifically: The natural isomorphism \pi_1(X \times Y) \cong \pi_1(X) \times \pi_1(Y) ... 5 This only really works in the specific case of distinguishing 1 dimensional and n dimensional Eulcidean space. For higher dimensions you will need a more sophisticated approach. Normally this includes the use of techniques from algebraic topology. See this page for one approach to solving this problem. To comment specifically on why your approach ... 4 Here you only need to care about the topology coming from the neighborhood basis. Let B be a topological space. Assuming we can find local trivializations$$ f_{U}:U_{E}\cong U\times \mathbb{R}^{n} $$such that on U_{E} this is given by product topology of U\times \mathbb{R}^{n}. Then you "glue" two such neighborhoods together using ... 4 There are plenty of tests for seeing if a space is NOT a CW-complex, like checking to see if it fails to be normal, Hausdorff, locally contractible, etc. Usually, one only cares about a space being homotopy equivalent to a CW-complex. There is a statement in Hatcher's book (Proposition A.11) that says if Y is a space, X is a CW-complex, and there are ... 4 The comb space works although any point on the 'base line' can be deformation retracted to so depending on the interpretation of your question, this might not quite fit your criteria. Instead, we can take a space which is morally the comb space but where we quotient out by the subset of points which can be deformation retracted onto. I believe (though I have ... 3 Certainly there's no test for whether a topological space is equal to a CW complex. It's no more plausible to test whether it's homeomorphic to one, which may be what you meant by your phrasing "is a CW complex." Another answer has given a condition for a space to be homotopy equivalent to a CW complex, which is of much more significance. Another interesting ... 3 Here is a general fact that will yield the result you seek. If G is a Lie group and X is a manifold on which G acts freely and properly, then X/G has a natural manifold structure and the projection X\rightarrow X/G is a principal G-bundle. In particular, X\rightarrow X/G is a fibre bundle with fibre G. Since O(k) is compact, it ... 3 It's not at all clear to me that the two questions in Question 1 are asking the same thing, but in any case the answer to the first one is "definitely not". For example, think about a flat family of elliptic curves degenerating to a nodal rational curve. The general fibre is topologically a torus, so has H_1 \simeq \mathbf Z \oplus \mathbf Z, but the ... 3 The crucial fact here is that the fundamental group of a topological group must be abelian. As you point out in the comment, \pi_{1}(M_{f}) contains \pi_{1}(X) as a subgroup and hence if \pi_{1}(X) is not abelian, \pi_{1}(M_{f}) is neither and so M_{f} cannot be a topological group. The only closed surfaces with abelian fundamental group are the ... 2 Q1. No, it's not true that all fibers of a flat map are homotopy equivalent. For instance, the blow-up of \mathbf P^1 \times \mathbf P^1 at a point maps to \mathbf P^1. This map is flat and all fibers except for one are spheres; however, the remaining fiber is a wedge of two spheres. More generally, in a flat family of curves you can "collapse" a simple ... 2 Let \mathcal{F} be your space of functions (say continuous functions, or all functions etc.) on some space X to some space Y. The function e (evaluation) is defined on \mathcal{F} \times X to Y, by e(f,x) = f(x). A topology \mathcal{T} on \mathcal{F} is admissible iff e is a continuous map (in the respective topologies on Y and the ... 2 I'll stick with your question 2. By definition a fiber bundle E\to X is locally homeomorphic to U\times F where U is open in X and F is the fiber. So in fact if X is connected all the fibers of E are homeomorphic. On the other hand if X has distinct connected components the fibers over each one can differ arbitrarily, so that there's no ... 2 Try to use the the uniqueness (up to homotopy equivalence) of Eilenberg-Maclane spaces to show that if \pi_0 X = G and \pi_i X = 0 for i > n, then for some n onwards,you have that X_n is homotopy equivalent to an Eilenberg-Maclane space. With this you get maps defined on a cofinal subset, which induces homotopy equivalences of X_n with the ... 2 This is a difficult homotopy equivalence to visualize, but I've attempted to draw a picture of what's going on. First you poke a hole in your surface starting at +\infty and pushing in from the right. Similarly poke a hole from the left. You can see this is a homotopy equivalence by analogy with an infinite cylinder, which can be visualized as having two ... 1 "Unsigned" means just count up the number of crossings. "Signed" means count them with a sign. That is, some crossings count as +1 and others as -1. Loosely, the determination goes as follows. Let's say the curves C_1 and C_2 are parameterized. Define a crossing to be positive if when travelling along C_1 in the direction of the parameterization, ... 1 That's a tall order! I would wager that very few people understand Freedman's proof completely. I found this series of video lectures that he gave to be quite helpful. In particular, it will give you an idea about the sort of mathematics that is involved. Also, Freedman and Quinn's book on 4-manifold topology is great! Even if you aren't to the point ... 1 Say we want to contract it to (0, 0). Look at a point on the vertical interval. Any neighbourhood of that point contains some points of the sine-curve, but any point on the sine curve has to go all the way back through (1, 0) to get to the origin. Thus a contraction is impossible to do continuously. 1 The observations allow you to argue much in the same way you do for S^1. For the case of the fundamental group, you can construct a "universal cover" X for \mathbb{S} called the digital line. The space X can be viewed as the set of integers \mathbb{Z} with topology generated by the sets \{2n\} and \{2n,2n+1,2n+2\}. The "covering map" p \colon ... 1 I presume that you are computing \tilde{H}_{i}((\Delta^{n})^{k};\mathbb{Z}), i.e. you are taking coefficients in \mathbb{Z}. To say that this group is free of rank n \choose k+1 when i = k means that$$\tilde{H}_{k}((\Delta^{n})^{k};\mathbb{Z}) \cong \mathbb{Z}^{{n \choose k+1}} \cong \overbrace{\mathbb{Z} \oplus \cdots \oplus \mathbb{Z}}^{{n ...
Using Morse theory (I will let you fill in the details): Consider the Morse function $f:SU(n)\to\mathbb{R}$ given by $[z_{ij}]\mapsto\text{Re}(\sum_ic_iz_{ii})$ for fixed constants $1<c_1<<c_2<\ldots<c_n\in\mathbb{R}$. It has a critical point of index 0, and the next smallest index is 3 (figure out why we want $c_1<<c_2$). Thus $SU(n)$ ...