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0

Here is a diabolical argument which one can only love or hate: Put a complex structure on your torus and delete your point. Since the universal cover of this punctured torus is simply connected but not compact it is holomorphically isomorphic to $\mathbb C$ or to a disk by Riemann's uniformization theorem. In both cases it is thus homeomorphic to a ...


0

Note that $X\setminus B$ deformation retracts onto a lower dimensional subspace (assuming $X$ is connected) and so in particular $H^n(X\setminus B)=0$. Consider the long exact sequence in cohomology of the pair $(X,X\setminus B)$ given by $$\ldots\to H^n(X/(X\setminus B))\to H^n(X)\to H^n(X\setminus B)\to 0\to \ldots$$ As $H^n(X\setminus B)=0$ we can ...


0

Consider the map of pairs $(X,X-B)\to(S^n,*)$ with $S^n=X/(X-B)$ and $*$ the point resulting from collapsing $X-B$, and the diagram you get from it in the long exact sequences for cohomology. The map $H^n(S^n,*)\to H^n(S^n)$ is an isomorphism, and so is $H^n(S^n,*)\to H^n(X,X-B)$, so by the commutativity of one of the sqaures in the diagram you want to know ...


3

No, that's not true. Consider $p : \mathbb R \to [0, +\infty)$, $x \mapsto x^2$. Every fiber is a finite discrete set, and it is a quotient map: it's surjective, continuous, and $U \subset [0, +\infty)$ is open iff its preimage is open: Suppose $U$ is open, and let $x \in p^{-1}(U)$. Then $x^2 \in U$, so there's a small neighborhood of $x^2$ contained in ...


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If we take the basic approach of defining the sphere spectrum as a sequence $\mathbb{S}=(S^0,S^1,...)$ with a monoidal structure given by the smash product, this is only associative and commutative up to homotopy. From this point of view, the product $[f]\cdot [g]$ in $\pi_*^S$ is anticommutative in the sense that if we have representatives $f:S^{n+i}\to ...


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The proof you give above for the non-triviality can be extended quite elegantly to show what you want. Use the same triangulation, and try to find a cycle of $2$-simplices. Let $s_1,\ldots,s_n,$ denote the simplices, and suppose$$d(\sum a_is_i)=0.$$If $s_i$ and $s_j$ share a side, we must have $a_i=a_j$. If the surface is connected, it follows that all ...


1

You wrote already everything down precisely then! You know that a homomorphism between free abelian groups can be represented by a matrix. Write it down and you will see the kernel (i.e. $H_2$) which is $\langle (1,\cdots,1) \rangle$.


2

There's a proof of this fact (in the category of CW-complexes, at least) in section 4.G of Hatcher's "Algebraic Topology." The basic idea is to construct an explicit map $X\to N(\mathfrak{U})$, then use Whitehead's theorem to show that it's a homotopy equivalence. The proof uses paracompactness, but it's only used to construct a partition of unity; the local ...


2

Since the question itself has already been answered, this answer will add some references for further study. As Joseph Zambrano points out in the comments to his answer, "local rigidity" fails for two dimensional hyperbolic structures. This is a hint of the deep, beautiful theory of moduli spaces of geometric structures. Bill Goldman's survey paper ...


3

A big difference between topological spaces and varieties is that, for topological spaces, we can define $H^{\ast}(X, \mathbb{Z})$ where as, for varieties, there is no cohomology theory with coefficients in $\mathbb{Z}$. (I've blogged about this here and here; the examples I'm giving are originally due to Serre.) This is important because, in the topological ...


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Suppose $i_m^{-1}$ does not converge uniformly to $i^{-1}$. Then there are $\epsilon > 0$ and a sequence $x_m$ in $M$ such that $d(i_m^{-1}(x_m), i^{-1}(x_m)) > \epsilon$. This means there are $y_m$ and $z_m$ with $i_m(y_m) = x_m$, $i(z_m) = x_m$, and $d(y_m, z_m) > \epsilon$. Taking subsequences, we may assume $x_m \to x$, $y_m \to y$ and $z_m ...


0

There is always the two volume classic by Arnold, Guisein-Zade, and Varchenko. Here are the Springer pages for both: Volume 1 and Volume 2. I've only looked at volume 1 but it is very good. As far as I know, these books are "THE books" on the subject though I'm not an expert. There is also this very nice book by Greuel, Lossen, and Shustin that takes a ...


1

Ok, I'll elaborate on fixedp's answer which was sort of cryptic for me. You only have to observe the general fact that if there is only one $E_\infty^{p,q}$ not zero for $p+q=n$ given, then that one gives $H^{p+q}(X)$. This follows from the definition of convergence. This is a bit more tricky. I don't know how Davis & Kirk might have concluded the ...


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You're right. One does not want that $E_r$ as a whole becomes stationary but usually $E_r$ comes with a grading $E_r = \bigoplus_{p,q\in \mathbb Z} E^{pq}_r$ and then you want that $E^{pq}_r$ becomes stationary. $E_\infty$ is then the graded abelian group $\bigoplus_{p,q\in \mathbb Z} E^{pq}_\infty$. If this graded abelian group is the same as the ...


2

The fundamental theorem of algebra Every nonconstant polynomial with coefficients in $\mathbb{C}$ has a root in $\mathbb{C}$ can be shown using algebraic topology. See for example Hatcher (http://www.math.cornell.edu/~hatcher/AT/AT.pdf) Page 31. Theorem 1.8. Sketch of the proof: Take a polynomial: $p(z) = z^n +a_1 z^{n−1} +\cdots+a_n$ . Then ...


0

The reason is each simplex has a defined order given by the order inherited by the sub-simplices: A simplex always looks the same. In case of the projective plane in the diagram, they are just arranged in a convenient way, such that the gluing map of the boundaries exactly is the one needed to result in the desired space. In case of $\Delta^2$-simplices ...


1

Using Seifert-Van Kampen Theorem: Let $M$ be the graph you described two circles joined by an arc. Choose points $x_1 \in S_1$ , $x_2 \in S_2$ and $x_1,x_2$ are not the intersection points of circle and arc. It's easy to see $S_2$ is the deformation retract of $M-\{x_1\}:=M_1\Rightarrow \pi_1(M_1)=\mathbb Z$ And $S_1$ is the deformation retract of ...


2

Hindman's Finite Sums Theorem: If you partition $\mathbb N$ into finitely many classes, there is an infinite sequence $a_1\lt a_2\lt a_3\lt\cdots$ in $\mathbb N$ such that all of the finite sums $a_{i_1}+a_{i_2}+\cdots+a_{i_k}$, where $i_1\lt i_2\lt\cdots\lt i_k$ and $k\in\mathbb N$, belong to the same partition class. The proof uses some kind of topological ...


1

I was really surprised about the following theorem. (General Question, sharp result) Theorem: If a compact Lie Group $G$ acts freely on a sphere, then $G$ is either finite or isomorphic to $S^1$, $S^3$ or the Normalizer of $S^1$ in $S^3$. (As usual i write $S^1=SO(2)$ and $S^3=SU(2)$. The proof of this results is an easy (If you know a tiny bit of Lie ...


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The Kneser graph is a graph whose vertices are labeled by size $k$ subsets of $\{1, \ldots, n\}$. Two vertices are connected by an edge if their corresponding sets are disjoint. Kneser conjectured in 1955 that the independence number of the Kneser graph is $n-2k + 2$. In 1978 Lovasz proved the Kneser conjecture using (a variant of) the Borsuk-Ulam theorem: ...


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The Necklace Splitting problem in combinatorics has been beautifully solved by Alon and West using the Borsuk-Ulam theorem. http://en.wikipedia.org/wiki/Necklace_splitting_problem


8

The Nielsen–Schreier theorem (a subgroup of a free group is itself free) can be proven using methods of elementary algebraic topology. My personal favourite is the pretty deep result that every finite dimensional divison algebra over $\mathbf{R}$ has dimension $1,2,4$, or $8$. This result seems to be due Kervaire and Milnor; the proof uses methods of ...


5

The proof of the Cayley–Hamilton theorem in the case of different eigenvalues is very easy. The extension to general case in any field is possible using the Zariski topology.


0

So this is one quite interesting exercise in Milnor-Stasheff. The exercise is not too hard, so I will give you a hint but also motivation to solve it. Here we go: What the exercise means: So the exercise means that whenever you have a submanifold in a compact manifold, you can look at the fundamentalclass of the submanifold and the euler class of its normal ...


6

Francis Su described in 1999 ("Rental Harmony: Sperner's Lemma in Fair Division", Amer. Math. Monthly, 106, 1999, 930-42) how to apply Sperner's Lemma---which says that every so-called Sperner coloring of a triangulation of an $n$-simplex contains a cell colored with a complete set of colors---to produce a list of variously sized rents for rooms in a shared ...


2

The Borsuk-Ulam Theorem states that for any continuous map $f : \mathbb{S}^n \to \mathbb{R}^n$, there are two antipodal (opposite) points of $\mathbb{S}^n$ that $f$ maps to the same value. (This follows from the fact that every antipodes-preserving map from the sphere to itself has odd degree.) For example, in the case $n = 2$, we can conclude that there ...


9

How about Furstenberg's proof of the infinitude of prime numbers?


3

Are you familiar with the upper-half plane model of the hyperbolic plane? In this model, we consider the upper half plane (like $\mathbb{R}^2$ with y values greater than $0$); however, a geodesic ("line") between two points is either a circle centered on the x-axis or a vertical line (depending on whether the points have different x coordinates). You can ...


3

The beginning of your argument is correct. The first vector $v = v_1$ of a three-by-three orthogonal matrix is indeed a point of $S^2$, and every point of $S^2$ arises this way. It is also true that the set of vectors perpendicular to $v$, call it $P(v)$, is homeomorphic to $S^1$. However, after that your reasoning breaks down. While $P(v)$ is ...


1

There is a nice and, to my opinion, more natural way to motivate cohomology - a geometric one, rather than an analytical one. Please read carefully the following question and answer in math.stackexchange: Intuitive Approach to de Rham Cohomology


2

Intuitively: the points of $\tilde{G_n}(\mathbb{R}^{n+k})$ are pairs $(p,\varphi)$ of an $n$-plane and an orientation of it. The tautological bundle has fiber $p$ at point $(p,\varphi)$, which can naturally be given the orientation $\varphi$. A clean proof of orientability is as follows: the oriented Grassmannian is simply connected, as you can compute ...


1

Letting $M$ denote the Mobius band, there is a universal covering map $\mathbb{R}^2 \to M$ with infinite cyclic deck transformation group $\langle r \rangle$ generated by the "glide reflection" $$r : (x,y) \to (x+1,-y) $$ As with any universal covering map, $M$ is the orbit space of this action. One can verify this by noticing that the vertical strip $[0,1] ...


3

The space $E=\mathbb P^2(\mathbb R)\setminus\{0\}$ obtained by deleting $0=[0:0:1]$ from the $[x:y:z]$ plane $\mathbb P^2$ can be sen as the tautological line bundle on the line at infinity $z=0$ of the plane and since that line is homeomorphic to $\mathbb P^1(\mathbb R)\cong S^1$ we obtain the non trivial (=Möbius) bundle $p:E\to S^1.$ Pulling back ...


3

So it's a Moebius band, as you say. The Moebius band is a quotient of a cylinder, which is a quotient of the real plane.


0

Homotopy doesn't really tell you much about the space, except $n$-connectedness and things like that. Path homotopy is nothing but homotopy of paths. As for the second question, the fundamental group of $\mathbf{R}^n$ vanishes. Therefore it is simply connected and hence any two real-valued maps are homotopic. I suggest reading Hatcher, Chapter 1.


2

First let me correct a mistake in your question: is it WRONG to think of this as an astroid. It is just 4 arcs of circles, NOT an astroid, though it looks somewhat similar. Second. The conformal map is certainly NOT fractional-linear (which you call Mobius). This conformal map is written explicitly in the paper I already referred to on MO: arXiv:1110.2696. ...


3

Let $L$ be the union over all $i \neq j$ of the lines through $p_{i}$ and $p_{j}$; this is a finite union, so is not the entire plane. Let $q$ be an element of the complement of $L$. For each $i$, the line through $q$ and $p_{i}$ contains only one of your points (namely $p_{i}$), so the rays are distinct.


2

This seems an example where the natural solution uses the fundamental groupoid on two base points, and groupoid methods, see the books Topology and Groupoids (first edition, differently named, 1968) and Categories and Groupoids (first published, 1971). See also ...


0

you can use of local degree and the fact degree of reflection is $(-1)^{n+1}$ .or if you use $I^n$, $\partial_k=<a_1-(-1)^{k+1}a_1+...>$ because degree of reflection is $(-1)^{n+1}$


8

First, it is not quite true that there is "only one cohomology theory" for topological spaces. One should rather say that there are several such theories, but that for reasonable spaces, they give the same results. The reason why this is true in a category of reasonable topological spaces is that nice topological spaces can be "built up" from smaller ones ...


3

This makes sense when $Y$ is a subspace of $X$ (otherwise we cannot restrict $X$ to $Y$). Then it just means that $h: X \to Y$ is a homotopy equivalence satisfying $h(y)=y$ for all $y \in Y$. A homotopy equivalence is a map which admits an inverse up to homotopy.


4

For $n = 1$, the space $\mathbb{R} \setminus \{0\}$ is the disjoint union of two intervals, so both connected components are contractible and have trivial fundamental group. When $n = 2$, you can see that $\mathbb{R}^2 \setminus \{0\}$ deformation retracts onto $$S^1 = \{ (x,y) \in \mathbb{R}^2 : x^2 + y^2 = 1 \},$$ the retraction being $z \mapsto ...


0

You are right, in higher dimensions ($n>2$) the (first) fundamental group of $\mathbb{R}^n$ minus a point is trivial. This is, however, not true for the removal of an arbitrary subset. Think of removing a tubular shape or just a circle from $\mathbb{R}^3$. Other means of classification are so called higher homotopy groups, where you look at homotopy ...


2

The answer is affirmative if $X$ is compact and compactness is also necessary. A reference for the first statement can be found in the discussion after Corollary 5.1.4, the item (ii) (page 235) in the book Complex Geometry by Huybrechts. This is also known as the Gauss-Bonnet-Formula and this Wikipedia Link contains a counterexample for the noncompact case. ...


3

Yes, there's a fibration with fibre $BG$. And you don't need a semidirect product: any extension of groups will do. The special case of the Serre spectral sequence that this gives is often called the Lyndon-Hochschild-Serre spectral sequence, and can be constructed algebraically.


0

A generator of $\pi_1(S^1)$ is the class of the path $$ \gamma:I\rightarrow S^1, \qquad \gamma(t)=\cos(\pi t)+i\sin(\pi t). $$ Now $$ (p_n)_*[\gamma]=[p_n\circ\gamma] $$ and $p_n\circ\gamma=\cos(\pi nt)+i\sin(\pi nt)$. Thus visibly $(p_n)_*[\gamma]=[\gamma]^n$ (multiplicative notation). So under the isomorphism $\pi_1(S^1)\simeq\Bbb Z$ we have ...


1

The subgroup $n \mathbb{Z} < \mathbb{Z} \cong \pi_1(S^1)$ is generated by $$\underbrace{[\gamma] \cdot \cdots \cdot [\gamma]}_n,$$ so one can show this directly by writing down a representative $\gamma$ of a generator of $S^1$ (the identity will do) and then writing an explicit homotopy $$\underbrace{[\gamma] \cdot \cdots \cdot [\gamma]}_n \simeq ...


1

This is one of the places where generalizing makes structure more visible. If you are willing to consider $S_n$ for all $n$ at once, i.e., study $$ \coprod_n BS_n, $$ then the cohomology algebra has two products making it a Hopf ring (a ring object in the category of coalgebras). Giusti, Salvatore, and Sinha studied this and include explicit rules for ...


0

With can trivially define a simplicial structure on the wedge product of $X$ and $Y$, as follows : $n$-simplices are defined by taking all the $n$-simplices of $X$ and all the $n$-simplices of $Y$ for $n>0$. For $n=0$ you take the same, but you take the quotient by the equivalence relation generated by the identification of the base point of $X$ with the ...


2

Hint: Mayer-Vietoris Solution added post-checkmark: Let $x$ and $y$ be the basepoints of $X$ and $Y$. Let $U$ be a contractible neighborhood of $x$ in $X$ and $V$ a contractible neighborhood of $y$ in $Y$. (I am uncertain of the details but the fact that $X$ and $Y$ are triangulated should be enough to ensure this.) Let \begin{align*} A &= X \vee ...



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