# Tag Info

1

The key example is $f : S^n \to S^n$ and you want to compute $$\mathbb{Z} \approx H_n(S^n;\mathbb{Z}) \xrightarrow{f_*} H_n(S^n;\mathbb{Z}) \approx \mathbb{Z}$$ This function is simply the homomorphism "multiply by $degree(f)$". And you can compute the degree very easily. For example: If $f$ is an orientation preserving homeomorphism then $degree(f)=1$ ...

0

A bit long for a comment, but these classes might not be what you want. Note that for line bundles, the sw classes satisfy $w_1(L_1\otimes L_2)=w_1(L_1)+w_1(L_2)$. Lets call the proposed classes $f$, and that they satisfy similar axioms. For a real line bundle (over some (para)compact base say, so that we have metrics) a line bundle $L$ is isomorphic to its ...

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The first idea only works when $\phi$ is a reparametrized geodesic. Write $\dot \phi= \tfrac{d}{dt}\phi$ and denote parallel transport along $\phi$ by $P_{\phi}$. Then $P_{\phi}(\dot\phi(0))$ remains tangent to $\phi$ iff there is an $\alpha:(-\varepsilon,\varepsilon)\to \mathbb{R}$ s.t. $P_{\phi}(\dot \phi(0))(t)=\alpha(t)\dot \phi(t)$ and then ...

1

Say that $f$ and $g$ equivalent, if for two (holomorphic) homeomorphisms $h_1,h_2:\mathbb CP^1\to\mathbb CP^1$ satisfied $h_2\circ f\circ h_1=g$. Maybe, in this sense?

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After digging through tones of Internet material, I have answered my own question: currently, as of August 3, 2015, there is no resource on the Internet that contains all of the work "La longue marche à travers la théorie de Galois". Not scanned, not in TeX, not in PDF. There is only that which has been made available by Leila Schneps. "Long March Through ...

1

$\alpha_\gamma(t)$ just represents the angle made by the vector drawn from $\gamma(t)$ to $\psi(\gamma(t))$, which you could write down an explicit formula for if you wanted to. Although it is true that $\alpha_\gamma(t)\in[0,2\pi)$, if you define it this way, it won't be continuous since it will in general jump from $2\pi$ back to $0$. You could let ...

2

The answer is in your question. You say that collapsible⇒contractible is not invertible. Let $\Delta_1$ be a contractible but not collapsible simplicial complex. Let $\Delta_2$ be a collapsible complex. Then $\Delta_1$ and $\Delta_2$ are both homotopy equivalent to a point. However $\Delta_2$ is collapsible and $\Delta_1$ is not.

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When taking the second boundary map, one can not remove the $i$th term since it was already removed by the first boundary map. So $j \neq i$ in the sum, and so it is convenient to split up the sum as $$\sum_{\substack{ j \\ j \neq i } } \cdot = \sum_{\substack{ j \\ j < i } } \cdot + \sum_{\substack{ j \\ j > i } } \cdot$$ Now for the signs, in ...

1

Yes, as noted in comments, your argument works. If $K$ is a compact subset of $X$ and $U$ is open subset of $Y$, then the inclusion $A\subset Y$ maps $\{f\in C(X,A):f(K)\subset A\cap U\}$ to the $\{f\in C(X,A):f(K)\subset U\}\bigcap C(X,A)$. So the map $C(X,A)\to C(X,Y)$ identifies elements of subbase of own compact-open topology of $C(X,A)$ with elements ...

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Yes, if you think about $\bar U$ and $\bar V$ as about classes in $C_n(M)$, then $d(\bar U)=-d(\bar V)$ and $\bar U+\bar V$ equals the fundamental class of $M$. So you have $\delta :[M]\mapsto[\partial U]$, and fundamental class of $\partial U$ is the sum of fundamental classes of all components. For more details i can advice just ...

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I agree with programming/computation, probability/statistics, and linear algebra from Comments. Also, Optimization from Answer. Would add group theory. Find out what computer languages are currently in greatest use as the time graduation gets near, especially for managing and parsing large datasets; learn the basics of all. Unless the landscape changes ...

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I would say that the basic intuition is that a Riemannian metric defines an inner product on the space of differential forms, which can be used to define an adjoint operator to the exterior derivative. Nowconsider the adjoint $(d_{k-1})^*:\Omega^k\to\Omega^{k-1}$ of $d_{k-1}:\Omega^{k-1}\to\Omega^k$. If the spaces $\Omega^\ell$ were all finite dimensional, ...

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You can lift the sphere map $f: S^k \to S^1$ through the universal covering space $\pi: \mathbb R\to S^1$ since $1=f_*\pi_1(S^k)\leq\pi_1(S^1)=\mathbb Z$ and contract inside $\mathbb R$ for $k\geq 2$. I am not sure you can do HES through disconnected spaces.

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I think, "fattern manifold" is $k$-submanifold $M\subset\mathbb R^{n+k}$ with trivialized normal bundle, so you can identify neighborhood of $M$ with $M\times E^n$, $g$ is the inclusion $M\times E^n\subset \mathbb R^{n+k}$. And the map $\theta_g$ is constructed as follows: you should project $M\times E^n$ on the $E^n$, then cover $S^n$ without $\infty$ by ...

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For basic Mayer-Vietoris usage you need to thicken your link/knot a little in such way that you won't create new intersections in obtained in such way "link" of solid tori. Namely: let $K$ be your link of $n$ circles and $U_1 = K \times D^2_{3 \varepsilon}$, where $D_{3 \varepsilon}^2$ is disk of "small" radius $3 \varepsilon$ such that you can retract ...

1

Big Data is a very broad definition. If want to work in data-mining or machine learning my list would start with these Statistics/Measure theory Optimisation (in general and especially convex optimisation) Funcional Analysis

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Let $\alpha$ be the generator of $\pi_1(\mathbb RP^2)$ and $\beta$ be the generator of $\pi_1(S^1)$. Note that $f_*(a)$ can be written only two ways: $\alpha\beta^{k_1}\alpha\dots\alpha\beta^{k_n}\alpha\,\,$ or $\,\,\beta^{k_1}\alpha\beta^{k_2}\dots\alpha\beta^{k_n}\,\,$ for some $k_i\in\mathbb Z$. Then, as it easy to see, if $k_1\ne -k_n$, we cannot reduce ...

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Hodge theory is a vast subject, especially since Deligne and Griffiths revolutionized it and algebraized it around 1970 with the introduction of variations of Hodge structures and mixed Hodge structures, axiomatic approaches which found amazing applications in algebraic geometry. At the most elementary level I would say that Hodge theory is a refinement ...

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0) The excellent mathematician you evoke has as family name (=surname) tom Dieck and as first name Tammo: tom is part of his surname and has nothing to do with Tom, the endearing form of Thomas. 1) Your idea of "finding a book that starts with a formal treatment of the basics of category theory and moves to more advanced/specialized concepts in a ...

0

Of course, no. For example, let $Z\subset\mathbb R^2$ is unit open disc, $X\setminus Z = Y\setminus Z$ equal the boundary circle $S^1$, $Y\cap Z$ is all the disc interior without one point $(\frac13;0)$, and $X\cap Z$ is punctured disc with center $(\frac13;0)$ and radius $\frac23$. Note that both $X\cap Z$ and $Y\cap Z$ are homotopy equivalent (there ...

0

Let $X = \{0\}\times[-1, 1]$, $Y = S^1\cup X$, and $Z = \{(x, y) \in \mathbb{R}^2 \mid y < 1\}$. Then $X\cap Z \subset Y\cap Z$, they both strong deformation retract onto $\{(0, -1)\}$, and $X\setminus Z = \{(0, 1)\} = Y\setminus Z$. However, $X$ is contractible while $Y$ is not, so $X$ and $Y$ are not homotopy equivalent.

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It sounds as if you're confused about the difference between singular/simplicial homology and cellular homology. If you look at the homology chapter of Hatcher's book again, you'll see that these subjects are covered in some detail. The relevant details are as follows: The homology of a space is a deep property of that space that is hidden behind a ...

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Your claim will work fine. But you need to change the thing you're tensoring over back into a subalgebra! $F[\pi_0 X]$, the sub-$F$-algebra generated by $\pi_0(X)$, is the right choice. It might be easiest to justify by just showing that for $S$ a multiplicative subset of a commutative $F$-algebra $A$ and $G$ the free group on the monoid $S$, ...

3

Pick a basepoint $x_0 \in U\cap V$. Let $p:I \to X$ be some loop starting at $x_0$. Then we can write $p$ as a finite product of paths $p = f_1 \cdot \ldots \cdot f_n$, where each $f_i$ is entirely contained in either $U$ or $V$. WLOG we can assume that each $f_{2i}$ is entirely contained in $U$ and each $f_{2i+1}$ is entirely contained in $V$ (so they ...

3

Yes, and in fact you do not need $X$ to be an H-space: the coproduct $\psi$ of $H_*(X)$ exists anyway (if the ring of coefficients is a field), and the definition of a group-like element does not involve the product. So take $X$ to be any space, and define $S \subset H_*(X)$ as in your question (the set of classes in $H_*(X)$ satisfying $\psi(a) = a \otimes ... 3 The point is that if$p:X\rightarrow Y$and$q:Y\rightarrow Z$are covering maps in this sense, but some$z\in Z$has infinitely many$y_i\in Y$with$q(y_i)=z$, then each$y_i$may have some neighborhood$U_i\subseteq Y$"evenly covered" by$p$(I take this to mean the inverse image of$U_i$is a union of parts each mapped isomorphically to$U_i$) but as ... 2 Given a continuous map$f : X \to Y$, there is an induced chain map$f_\# : C_n(X) \to C_n(Y)$of singular chains. Since$C_n(X)$is a free abelian group whose basis is the set of singular simplices$\sigma : \Delta^n \to X$, in order to define$f_\#$all you need to do is to define it's value$f_\#(\sigma)$on cach basis element, and then$f_\#$is uniquely ... 1 Take$A, B$subspaces of$X$whose interiors cover$X$. There are three induced homomorphisms we need to understand:$$H_n(A\cap B)\xrightarrow{\quad\Phi\quad}H_n(A)\oplus H_n(B)\xrightarrow{\quad\Psi\quad}H_n(X)\xrightarrow{\quad\partial\quad}H_{n-1}(A\cap B)$$ The map$\Phi$comes from the inclusions of subspaces$\iota_A:A\cap B \hookrightarrow A$and ... 1 Firstly, why are you writing$\hom_{F-modules}(H_n(X;F),F)$? UCT implies $$H^n(X;F)=\hom_{F}(H_n(X;F),F)$$. Now, every$F$-module is a$F$-vector space and dual of a (finite dimensional) vector space is isomorphic to itself. So,$H^n(X;F)=\hom_F(H_n(X;F),F)=H_n(X;F)$, whenever$H_n(X;F)$has finite rank. For the infinite rank case see this SE post. 1 The comment I made above misunderstood the problem, and I'm sorry for that... To see they are the same, notice that$C_0(A)$is not empty so the mapping$C_0(A)\stackrel{\epsilon}{\rightarrow}\mathbb{Z}$is already surjective. Therefore$C_0(X,A)\stackrel{\epsilon}{\rightarrow}\mathbb{Z}$is a zero mapping and its kernel is the same as$\ker \partial_0$. 2 Sorry, this isn't a full answer, rather some algebraic hints at how you might relate the cross product to your intuition about the cup product (wrt intersections) and to linear algebra. It should really be a comment but it's too long. First, a quick review! The cross product,$\times$, is actually part of the cup product: $$x \smile y := \Delta^*(x \times ... 1 The issue is that a^*+b^* is not the generator of H^1, in fact it is not a cohomology class:$$(a^*+b^*)(dU) = (a^*+b^*)(a+b-c) = 2.$$Instead, take b^*+c^* as the generator, which is a cohomology class:$$(b^*+c^*)(dU) = (b^*+c^*)(a+b-c) = 0,(b^*+c^*)(dV) = (b^*+c^*)(a+c-b) = 0.$$Also, I believe that \mathbb{Z}[x,y]/(x^3, 2x^2, xy, y^2) is not ... 2 There is a homology theory that looks like this called (co)bordism, and it is very interesting. It comes in many flavors depending on what kind of extra structure you ask for on the manifolds. The basic problem with your proposal is functoriality: the image of a submanifold need not be a submanifold. The correct definition of bordism fixes this by allowing ... 1 [Note: After the comments to this answer and the question, there have been major changes. Thanks to Najib Idrissi, the situation is much clearer now.] There are only slight differences in the definitions and strictly speaking, Hatcher's is more general. In addition to (equivalents of) Hatcher's axioms, May wants a Hopf algebra to be flat (a minor technical ... 1 The questioner is right to identify one of the inspirations for homology as coming from integration theory; this is emphasised in the article by S. Lefschetz in the book "History of Topology", edited I.M. James. It seems also that early articles on Betti nubers and torsion coefficients wanted to take "cycles modulo boundaries" but were not so clear about the ... 1 I expect the restriction to open n-simplices is just to remove any potential ambiguity from the fact that closed n-simplices may overlap, while their interiors do not. 1 In Hatcher's Algebraic Topology, these are called characteristic maps, and the restriction \varphi\colon \partial\Delta^n\to \Delta to the boundary is referred to as the attaching map. In my experience, this usage is fairly standard. 2 This is a supplement to Tyler's answer (it should be a comment, but it's too long). As Tyler mentions, the Hauptvertmutung (any two triangulations have a common refinement) is false for topological spaces with dimension greater than 2. This breaks the sentence "simplicial (co)homology is an approximation of oriented cobordism (co)homology theory in all ... 3 You run into two technical problems with this attempt to describe homology, but they're not "obvious" problems. (In fact, your description is at least partially what motivated Poincare to define homology in the first place.) The first is the restriction to submanifolds. There are too many technical details about embedding submanifolds: how many submanifolds ... 5 The space S^1\vee S^1\vee S^2 is the wedge sum of two circles and one sphere. In particular this space is also known as mouse space. It is quite easy to see that the fundamental group of S^1\vee S^1\vee S^2 is \mathbb{Z}*\mathbb{Z}\cong\langle a,b| \emptyset \rangle ( you can use the Seifert-VanKampen theorem to see it ); whereas the fundamental ... 1 We will draw on the computation in Chapter 23, Section 3 of May's A Concise Course in Algebraic Topology that$$w(\mathbb{R}P^q) = \sum_{0 \leq i \leq q} \binom{q+1}{i} \alpha^i,$$where$$H^*(\mathbb{R}P^q ; \mathbb{Z}_2) \approx \mathbb{Z}_2[\alpha]/\alpha^{q+1}.$$Suppose that q = 2^k - 1. Then note that$$\binom{2^k}{i} \equiv 2 \text{ }(\text{mod }2)$$... 1 I am not sure If I understand the definition correctly, but if I do, then take a look at the maps$$ f : (0,1) \rightarrow [0,1], x \mapsto x$$and$$ g : [0,1] \to (0,1), x \mapsto \frac{1}{3} x + \frac{1}{3}.$$These are embeddings and composing these we get$$ f \circ g : [0,1] \rightarrow [0,1], x \mapsto \frac{1}{3} x + \frac{1}{3}$$and$$ g \circ f ... 3 Because to obtain our chain complex for a space$X$we look at arbitrary maps$\Delta^n \to X$which are not necessarily nice, i.e. singular. In contrast we look at nice embeddings when dealing with simplicial homology. 0 Here the answers to some of your questions. Since taking determinant and pulling back commute (and the latter is natural) we indeed have for$i:M-\nu N \hookrightarrow M$and$f:M \to RP^\infty$: $$detT(M-N) = det (i^*TM) = i^*detTM=i^*f^*\gamma_\infty=(fi)^*\gamma_\infty.$$ But you're right,$fi$factors through a disk and hence gives the trivial line ... 1 Here is a solution, actually not involving homotopy groups at all: In the real projective space, we want to follow the orientation given in a base point, going through the nontrivial closed loop until coming back to the base point. To do so, we take the loop in$S^n$which is the half-circle between two antipodal points; since$S^n$is orientable, what ... 4 Yes, being simply-connected is preserved under homeomorphisms. Suppose$f : X \to Y$is a homeomorphism and$X$is simply connected. First, path-connectedness is a topological invariant, so$Y$is path-connected too. Now, let$\gamma : S^1 \to Y$be any (continuous) loop. Then$f^{-1} \circ \gamma : S^1 \to X$is a continuous loop, and$X$is ... 0 I assume you are using cellular homology for the CW complexes$X$and$Y$. The problem you are posing comes up in multiple Algebraic Topology texts (it is problem 2.2.17 in the Hatcher text, for example). When we use cellular homology the problem reduces to naturality of the long exact sequence for singular homology, as the map$d_n:H_n(X^n,X^{n-1}) \to ...

0

Hint: Suppose $X$ is a non-empty connected open set, and that $f:X \to \mathbf{R}$ is locally constant. Pick an arbitrary point $x_{0}$ in $X$, and let $U = \{x \in X : f(x) = f(x_{0})\}$. Use local constancy of $f$ to prove $U$ is open and closed in $X$.

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A function $f : U \to \mathbb{R}$ is locally constant if for each $x \in U$, there is an open neighbourhood $V$ of $x$ such that $f|_V$ is constant. Note that for any $y \in \mathbb{R}$, $f^{-1}(y)$ is open, so for any $A\subseteq \mathbb{R}$, $f^{-1}(A) = \bigcup_{y\in A}f^{-1}(y)$ is open. In particular, $f^{-1}(A)$ is open for every open $A\subseteq ... 2 Consider$\mathbb{R} \setminus \mathbb{R} = \varnothing$, which does not have the same homotopy type as$\mathbb{R} \setminus \{0\} \sim S^0$even though$\mathbb{R}$deformation retracts onto a point. For more complicated examples (because this one could look "pathological"), you can remove a half-line from$\mathbb{R}\$ and get a contractible space (not ...

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