# Tag Info

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First, $M(X,Y)$ equals $Y^X=\prod_{x\in X} Y$, the set of all functions from $X$ to $Y$ since ever function on a discrete space is continuous. Now the subbasis for the product topology on $Y^X$ consists of all $$e_x^{-1}(U)=\{(f(x))_{x\in X}\mid e_x(f)=f(x)\in U\}$$ ranging over the open subsets $U$ of $X$ and the elements $x$ of $X$. The subbasis for the ...

1

This is a great question! I know no references for the following facts and I am pretty sure one can make more general statements, but these are the ones that I have seen used in practice. 1) Yes, at least when $R$ is a ring. Indeed, $H_{*}(X, R)$ can be defined as homology of $C_{*}(X, R)$, the chain complex which in degree $n$ is the free $R$-module ...

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This is a special case of the main result of this paper: Morton Brown, "Locally flat imbeddings of topological manifolds", Annals of Mathematics, Vol. 75 (1962), p. 331-341.

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E. H. Connell, in this 1967 paper.

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I think a possible solution is to let $X_n$ be the $n$th space in the Postnikov tower of $X$, and let $Y^n$ be an $n$-connected CW model for $(Y, y_0)$, as constructed in Proposition 4.13. Let $Z_n$ be the disjoint union of $X_n$ and $Y^n$. Let $X \to Z_n$ be the inclusion of $X$ in $X_n$ and $Z_n \to Y$ be the disjoint union of $Y^n \to Y$ (from the CW ...

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There is a very nice proof of the existence of a maximal torus in a compact Lie group involving Lefschetz fixed point theorem, given in the third chapter of "J.F.Adams, Lectures on Lie groups, 1983".

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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 ...

1

A simply connected set is by definition path-connected (= any two points in it can be connected by a path contained in the set). And a path-connected set is connected (means: not a union of two open sets that have no points in common), so a simply connected set is connected.

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Hint: do the obvious thing, use the van Kampen theorem with respect to the open covering of your space by the two open sets $U$ and $V$ such that $U$ is the leftmost circle and a little bit more, and $V$ is all the rest of the circles and a bit more.

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Maybe not what you want but this may help: Surgery on the Hopf link is the same as gluing two solid tori to the two sides of $S^1 \times S^1 \times [0,1]$ by the gluing maps determined by the framings. Let $T_1,T_2$ be solid tori where $T_1$ is glued to $S^1 \times S^1 \times \{0\}$ with framing $p$, and $T_2$ is glued to $S^1 \times S^1 \times \{1\}$ with ...

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 ...

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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 ... -1 As Martin writes, one has to be clear about the categories one is dealing with There is a functor \pi_1: Top \to Gpd, giving the fundamental groupoid of a topological space. This functor preserves products. (6.4.4 of Topology and Groupoids). A consequence (6.5.10) is a results on the effect on the morphisms of fundamental groupoids of homotopies of ... 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) ... 0 yes, it is true (more genrally for free actions of compact lie groups): Let [x]_1 denote the equivalence class of the action of G_1 via h.x = (h,e_2)x, where e_i is the identity element of G_i. First the action of G_2 on X/G_1 via g.[x]_1 = [(e_1,g).x]_1 is well defined and free, what is easy to check. Let its equivalence classes be denoted ... 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 ... 0 Every set in the discrete topology is open, so the inverse image of an open set in the range is obviously an open set, since every set is open. Thus every function out of a discrete space is continuous. 0 So, I am answering my own question whose answer I found some time ago. One appropriate term to use to search the literature about spaces with fundamental group G is K(G,1). These are special cases of the Eilenberg-MacLane spaces. Quoting Wikipedia: Let G be a group and n a positive integer. A connected topological space X is called an ... 0 You're forgetting about all the other morphisms in \Pi(X). Once you've selected a path representative p_y:x\rightarrow y for each y\in X, you have to worry about all the morphisms y\rightarrow x. By choosing p_y, concatenation with any class of paths y\rightarrow x will give you a morphism x\rightarrow x. This morphism will map, under F, ... 0 I like Topology and Geometry by Glen E. Bredon, especially as a reference for cohomology and homology products. I second Weibel's book on homological algebra, especially for the universal coefficient theorem. 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 ... 2 The easiest argument is to observe that the 1st Cech cohomology of this space is nontrivial. You can see this by observing that it separates the plane in 2 components and applying Alexander duality. 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 Start by reading Milnor's "Lectures on h-cobordism theorem", since Freedman's proof is a very difficult variation on Smale's proof in higher dimensions. 0 As I noted in the comments above, what you originally asked is not true - the standard genus one splitting of S^1 \times S^2 does not have an intersection between its \alpha and \beta curve. However, it does hold for rational homology spheres. We take a Heegard splitting for M with \alpha and \beta curves (\alpha_i)_{i=1}^g, ... 0 If a space X is the union of two path connected open sets U,V whose intersection W=U \cap V has n path components, then the natural thing is to choose a set A consisting of one point in each path component of W; the form of the Seifert-van Kampen Theorem given in Topology and Groupoids, Section 6.7, determines the fundamental groupoid ... 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 ... 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 ... 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 ... 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 ... 0 I like to put what is in some ways the opposite point of view, and ask: how does one specify a space? One way is to give constructions from other spaces, e.g. functions spaces. But how does one give the other spaces? The history of CW-complexes is that Whitehead generalised them from other constructions in his highly original papers published 1939-1941 ... 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 ... 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} $$such that on U_{E} this is given by product topology of U\times \mathbb{R}^{n}. Then you "glue" two such neighborhoods together using ... 3 Certainly there's no test for whether a topological space is equal to a CW complex. It's no more plausible to test whether it's homeomorphic to one, which may be what you meant by your phrasing "is a CW complex." Another answer has given a condition for a space to be homotopy equivalent to a CW complex, which is of much more significance. Another interesting ... 4 There are plenty of tests for seeing if a space is NOT a CW-complex, like checking to see if it fails to be normal, Hausdorff, locally contractible, etc. Usually, one only cares about a space being homotopy equivalent to a CW-complex. There is a statement in Hatcher's book (Proposition A.11) that says if Y is a space, X is a CW-complex, and there are ... 1 "Unsigned" means just count up the number of crossings. "Signed" means count them with a sign. That is, some crossings count as +1 and others as -1. Loosely, the determination goes as follows. Let's say the curves C_1 and C_2 are parameterized. Define a crossing to be positive if when travelling along C_1 in the direction of the parameterization, ... 2 The higher homotopy groups are all abelian, so for finite indices you have the canonical isomorphism:$$ \prod^{n}_{i=1}A_{i}=\bigoplus^{n}_{i=1}A_{i} $$by mapping (a_{1}\cdots a_{n}) to \sum_{i=1}^{n}a_{i}. 2 Finite direct sums and finite direct products are the same thing on groups. The homotopy theory part has nothing to do with it. (Infinite direct sums and products are different, though, so be careful about this.) 1 Say we want to contract it to (0, 0). Look at a point on the vertical interval. Any neighbourhood of that point contains some points of the sine-curve, but any point on the sine curve has to go all the way back through (1, 0) to get to the origin. Thus a contraction is impossible to do continuously. 3 The crucial fact here is that the fundamental group of a topological group must be abelian. As you point out in the comment, \pi_{1}(M_{f}) contains \pi_{1}(X) as a subgroup and hence if \pi_{1}(X) is not abelian, \pi_{1}(M_{f}) is neither and so M_{f} cannot be a topological group. The only closed surfaces with abelian fundamental group are the ... 1 I presume that you are computing \tilde{H}_{i}((\Delta^{n})^{k};\mathbb{Z}), i.e. you are taking coefficients in \mathbb{Z}. To say that this group is free of rank n \choose k+1 when i = k means that$$\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 ...

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A free abelian group of finite rank $n$ is isomorphic to the group $\mathbb{Z}^n$.

1

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 $SU(n)$ ...

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 ...

0

Break up the space $X$ into infinitely countably many open pieces $\{X_i\}$, where each piece is homotopy equivalent to a torus minus two disjoint disks, and the intersection of each two consecutive pieces is homotopy equivalent to a circle. Pick a straight line $L$ that extends through all pieces, and pick a base point $y_0$ on this line. Let $Y_i$ be the ...

1

Let $X$ be the infinite-holed torus and $G$ its fundamental group. You can write $X$ as an increasing union $X= \bigcup\limits_{n \geq 2} X_n$, where $X_n$ is a $4$-punctured $n$-holed torus; let $G_n$ denote its fundamental group, which is a free group of rank $2n+2$. In the figure below, $X_3$ is represented with a free basis of $G_3$ in red. Now, you ...

4

$$O \to Ext(H_{n-1}(M),\mathbb{Z}) \to H^n(M) \to Hom(H_n(M),\mathbb{Z}) \to 0$$ As the latter arrow is an isomorphism when M is closed, connected and orientable, it follows that $Ext(H_{n-1}(M),\mathbb{Z})=0$. You just need to understand why it implies your $n-1$ torsion group $T_{n-1}$ is $0$...

1

The connected surface that results from cutting along a loop on a Mobius strip is homeomorphic to a cylinder. But cutting along a loop on a cylinder will always produce a disconnected space (either a disk and a cylinder minus a disk, or two cylinders).

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The Zariski topology is just too coarse. For example, the Zariski topology on any algebraic curve is just the cofinite topology. In fact, as I recently learned on MO, any topological space with a generic point (in particular, the spectrum of any integral domain) is contractible.

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