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0

The proof of Owen for the first question is correct. Of course van Kampen works for the second question, but it is a slight overkill. I would prefer the following: $\pi_1(S^n)=0$ if and only if every loop based at $p$ is contractible through points based at $p$. Given an arbitrary path based at $p$, we can easily find a contraction by stereographically ...


2

The answer is yes; if $A \to B$ is any morphism of abelian groups then it induces a natural transformation $H_n(-, A) \to H_n(-, B)$, which implies that the map induced by $f$ on $H_n(-, \mathbb{Z}_2)$ is the reduction $\bmod 2$ of the map induced on $H_n(-, \mathbb{Z})$.


4

This is a copy of a part of the note found here (start at the bottom of the first page) In particular, if $M$ is a simply-connected closed 3 manifold, then $M$ has the same homology groups as the $3$ sphere $\mathbb S^3$. In fact $M$ is homotopy equivalent to $\mathbb S^3$. For by the Hurewicz theorem $\pi_2(M) = H_2(M) = 0$ and $\pi_3(M) = H_3(M) = ...


6

Let $\mathcal S$ be a compact surface, possibly with boundary. Let $\text{Homeo}^+(\mathcal S)$ refer to the group of orientation-preserving homeomorphisms that fix the boundary pointwise with the compact-open topology. Throwing an $n$ in there means we add $n$ marked points in the interior, which I prefer over deleting points. (I'd be worried about the ...


0

Generally, no. $\phi$ must be fibre-preserving to induce a smooth map of the base spaces, $$\pi_1(x) = \pi_1(y) \implies \pi_2(\phi(x)) = \pi_2(\phi(y)).$$ If a smooth map $\psi \colon M \to N$ of the covering spaces is fibre-preserving, it induces a smooth map of the base spaces. Locally, around each $x \in M_1$, we can define a smooth map via $\pi_2 ...


1

No. In fact, there is a more general result: Theorem: Let $M$ and $N$ be topological manifolds with boundaries and let $F:M\to N$ be a homeomorphism. Then, $F(\partial M)=\partial N$. The proof follows from the fact that a point in a topological manifold is either a boundary point or an interior point, but not both.


1

I think for differential topology, and geometry it is probably best to learn de Rham cohomology. Bott and Tu's book is the canonical reference.


2

Although Amazon says that Massey's Singular Homology Theory is a sequel to Massey's Algebraic Topology: An Introduction, the earlier book is not a logical prerequisite for Singular Homology Theory. There are plenty of online lecture notes that might be a right fit for you, for example: ...


0

Here is an example of a local homeomorphism which is not just a covering map with points from the domain removed (nor is it of the form $\bigsqcup U_i \to X$ where $U_i \subseteq X$ are open subspaces). Let $X= \Bbb{Z}_{\geq 0}$ with open sets $U_n = [n, \infty)\cap X$ for $n \in \Bbb{Z}$. The open sets are nested $X = U_0\supsetneq U_1 \supsetneq \cdots$, ...


1

Let $V$ be the covering space which is $X$ with a copy of the real line attached to each circle (not at the intersection point). At each integer point on the real lines, attach another copy of the real line. At each integer point on those real lines, attach another copy, and keep doing this (essentially creating something similar to the universal cover ...


2

Suppose the inclusion $V\to X$ induces isomorphisms in homology. Consider now the inclusion of pairs $(V,V\setminus\{p\})\to(X,X\setminus\{p\})$, fromwhich you can construct a big diagram: the two rows are the the long exact sequences of the pairs $(V,V\setminus\{p\})$ and $(X,X\setminus\{p\})$ and the maps going from the groups in one to the coresponding ...


0

You could use the fact that if $S \subseteq \mathbb{R}^n$ and every sequence $x_n \in S$ with $x_n \rightarrow x$ has $x\in S$, then $S$ is closed. If $S$ is finite, then the property clearly holds, therefore $S$ is closed.


1

Here's a way to prove that $\mathbb{R}^n \setminus\{x_1,\ldots,x_k\}$ is open. We first prove that $\mathbb{R}^n \setminus \{x_i\}$ is open: $$\mathbb{R}\setminus \{x_i\} = \bigcup_{c\neq x_i} B_{d(c,x_i)}(c)$$ So indeed $\mathbb{R}\setminus\{x_i\}$ can be expressed as an arbitrary union of open sets (open balls in this case) and therefore is open. Since ...


0

Let the points be $X=\{x_1, \dots, x_n \}$. Consider the function $f(x)=|x-x_1|^2\cdots|x-x_n|^2$, where $|x-y|$ is the Euclidean distance. Then $f: \mathbb R^n \to \mathbb R$ is continuous, since it is a polynomial function on the coordinates of $x$. Now, $X=f^{-1}(0)$ and so is closed.


1

A sigleton is closed because it has only one limit point (of course, the point itself), and a finite set is a finite union of singletons.


2

Let $x \notin \{x_1, \dots, x_n \}$ and $2\delta = \min_{i=1}^k d(x,x_i)$. Then $B(x,\delta) \subset \mathbb R^n \backslash \{x_1, \dots, x_n \}$ and therefore the complement of $\{x_1, \dots, x_n \}$ is open. Edit : as I said in my previous comment, it's even more simpler with the definition using covering : if $(U_i)$ cover $x_1, \dots, x_n$ then we can ...


1

You may just apply Borsuk-Ulam theorem directly. Define a function $f$ from $S^n$ to $\mathbb{R}^n$ as follows: If $x$ is a point on $S^n$, then there is a great $S^{n-1}$ that is orthogonal to the point $x$. The $S^{n-1}$ divides $S^n$ into two regions. Let's call them $U$ and $V$, where $U$ is the region containing $x$. If $\mu$ is the measure given, ...


0

Yes sure! There is the book by R. Brown, "Topology and Groupoids", which treats exactly this: http://pages.bangor.ac.uk/~mas010/topgpds.html Look also here: http://pages.bangor.ac.uk/~mas010/nonab-a-t.html On the nlab (if you didn't know the site, take a look!) there are also these articles that you may find interesting! ...


2

You're right that there are a very large number of edge symbols that you can get by pairwise identifying the edges of an octagon. However, many of them yield homeomorphic surfaces. Here are a few facts that might help you in your thinking: The classification theorem for surfaces states that any compact 2-manifold is homeomorphic to either a sphere, a ...


3

Let $X$ be a real $n$-manifold, $M$ a real $k$-manifold, $Z$ a submanifold of $X$ of codimension at least $k+2$. (Actually you can replace $Z$ by $f(Z)$, where $f: Z \to X$ is any smooth map, not just an embedding). Then the inclusion $X-Z \hookrightarrow X$ induces a bijection $[M,X-Z] \to [M,X]$. Invoke the transversality theorem: surjectivity follows ...


3

Simple connection is important in different areas of mathematics: $real$ $analysis:$ if you have an exact $1$-form $\omega$ and you need to integrate it along a closed path $\gamma$ contained in a simple connected space then $$\int_\gamma \omega =0$$ because $\gamma$ is homotopic to a point and the integral does not change over any path in the homotopy ...


1

You are missing a hypothesis they are assuming, which is that $H_*(M)[\pi^{-1}]$ (which is just $H_*(M)[m_1^{-1}]$ in this case) can be constructed by right fractions. This implies that $H_*(M_\infty)=H_*(M)[m_1^{-1}]$, as the colimit that computes $H_*(M_\infty)$ is exactly the right fractions for $H_*(M)[m_1^{-1}]$. Since element of $M$ is homotopic to ...


1

Your method is almost correct : you seem to have assumed somewhere that $H_2(\Bbb RP^2) = \Bbb Z$, which is wrong. $H_2(\Bbb RP^2)$ is actually trivial. That said, $H_2(X)$ should be $\Bbb Z$ instead of $\Bbb Z^2$. Here's another way of doing this : $\{U, V\}$ forms an open cover of $X$ where $U$ is the image of the open set consisting of a ...


1

In general, if $g\circ f$ is injective, for any functions $f, g$, then $f$ is injective. Proof: if $f(x)=f(y)$, then $g(f(x))=g(f(y))$. Since $g\circ f$ is injective, we conclude that $x=y$, so indeed $f$ is injective as claimed.


0

They don't have to be disjoint, actually. One can construct a counter example as follows. Take a closed curve $c$ on $S^n$, such that $c\neq-c$, but $c\cap-c\neq\emptyset$ (there's no problem doing that). Project $c$ (or $-c$) to the projective space, and you're done. Edit: If the initial curve $\alpha$ in $\mathbb{R}\mathbb{P}^n$ is simple, and its ...


2

See this blog post. It's true in general that the cohomology of a smooth hypersurface of degree $n$ in $\mathbb{CP}^d$ only depends on $d$ and $n$.


1

This question has been asked and answered on MathOverflow. I have replicated the accepted answer by Tom Goodwillie below. In most cases $\pi_{n+k}(S^k)$ is a finite group, so that the homotopy fiber of any map $S^{n+k}\to S^k$ is rationally equivalent to $\Omega S^k\times S^{n+k}$ and therefore has homology in arbitrarily high dimensions and cannot be a ...


1

Keep in mind the following picture: This is supposed to represent a homotopy of paths $[0,1] \to \Delta^2$ between the path that goes $0 \to 1 \to 2$ and the path that goes straight $0 \to 2$. If you want a precise definition, let: $$\tilde{H}(s,t) = \begin{cases} (1-s) \cdot (1-2t, 2t, 0) + s \cdot (1-t, 0, t), & 0 \le t \le 1/2, \\ (1-s) \cdot (0, ...


7

I've checked five of your claimed errors, none of which you're right about. There are other cases in which you've found obvious typos, which are admittedly annoying but which shouldn't cause any real trouble. It may be you've found some significant errors in the other points which are more complicated for me to check, but based on the following evidence it ...


1

For question 1 (per request): Suppose that $f : X \to Y$ is a weak homotopy equivalence. Put on $\mathsf{Top}_*$ the standard model structure where weak equivalences are weak homotopy equivalences, fibrations are Serre fibrations and cofibrations are determined via lifting properties (I believe they're retracts of generalized CW-complex inclusions). Then ...


2

This is a special case of the Brouwer plane translation theorem. See the paper of John Franks entitled "A new proof of the Brouwer plane translation theorem", Ergodic Theory Dynam. Systems 12 (1992), no. 2, 217–226.


2

No, there does not. This might not be the easiest argument, but it follows from Harold Bell, A fixed point theorem for plane homeomorphisms, Bull. AMS, 82 (5), (1976), 778-780. Bell's much more general result is that every homeomorphism of the plane leaving some continuum $M$ invariant has a fixed point in $T(M)$, where $T(M)$ is the "filled-in hull" of $M$, ...


1

Another possibility is the following paper: Samuel Eilenberg and Saunders MacLane. “Acyclic models”. In: Amer. J. Math. 75 (1953), pp. 189–199. ISSN: 0002-9327. JSTOR: 2372628. I read it a few months ago so I may not remember perfectly, but if I recall correctly, everything was done in detail and I didn't have to fill in any significant gap. The most ...


2

You should look at Barr, M. Acyclic models, CRM Monograph Series, Volume 17. American Mathematical Society, Providence, RI (2002), and see if that satisfies you for completeness. A theorem involving crossed complexes rather than chain complexes is in Section 10.4 of the book partially titled Nonabelian Algebraic Topology (2011). This version gives in ...


2

I think that for finite $G$, $M/G$ is a smooth manifold, regardless of any compactness assumptions. For a general discrete group $G$ acting on a manifold $M$, you need the action to be free and properly discontinuous in order to make $M/G$ into a smooth manifold (and $M\to M/G$ into a covering). The second condition means that for each $x\in M$ there is a ...


2

Suppose a diagram of simply connected pointed spaces has limit $X$ in $\mathbf{Top}_*$. If it has limit $\tilde{X}$ in $\mathbf{Top}_1$ then there is a canonical map $\tilde{X}\to X$ that is the universal map from a simply connected pointed space to $X$. The universal cover of $X$ (when it exists) comes close to this, but the failure of lifting theorems for ...


2

You have the right idea, but as AP says, you have to take the open cover $\{U, V\}$ given by the images of $\{z : 3/4 > |z| \geq 1/2\}$ and $\{z : 1 \geq |z| > 2/3\}$ under the quotient map. In that case, $U$ and $V$ are both homotopy equivalent to $S^1$, and $U \cap V$ deformation retracts to $S^1$. The inclusion induced maps $\pi_1(U \cap V) ...


0

Let $f\colon E\to X$ be any schematically dominant morphism of schemes, e.g., a surjective morphism with $X$ reduced. Let $U_{E/X}$ be the cokernel of the canonical morphism $f^{\flat}\colon\mathbb G_{m,X}\to f_{*}\mathbb G_{m,E}$ of etale sheaves on $X$ (this morphism is injective by the schematically dominant hypothesis. The sheaf just defined is, or ...


0

The tangent Bundle of sphere is another example. For arbitrary vector bundle E, it seems that $E\otimes \mathbb{C}$ has a trivial line bundle, as a summand. http://mathoverflow.net/questions/209247/almost-complex-structure-and-nontrivial-idempotents Moreover, I guess that the complexification of canonical line bundle over $\mathbb{R}P^{n}\;\;n>1$ ...


0

I believe I've found the answer myself. Let me now speak of $\omega$ on the level of forms, recalling that $E_{2} = H_{\delta} H_{d}(C^{*}(\pi^{-1} \mathfrak{U}, \Omega^{*}))$. By this I mean $\omega_{\alpha_0 \ldots \alpha_p} = \sum_{i = 1}^{n} a^{\alpha_0 \ldots \alpha_p}_{i} \sigma_i$ represents something in $H_{\delta} H_d$ cohomology, so $d\omega = 0$ ...


0

Rather than solving the problem for you, I collect here all relevant facts. Given these facts, actually solving it should be fairly straightforward. Convention. Write $T_0$ for the $2$-sphere. Classification Theorem. Suppose $X$ is a compact surface. If $X$ is orientable, then $X \cong T_n$ for some integer $n \geq 0$. If $X$ is non-orientable, then $X ...


0

I think all you need to know is that the sum of three projective planes is homeomorphic to the sum of a torus and a projective plane.


7

Yes, it is. Define $g:X\to S^n,\quad (p,q)\mapsto p.$ Then $g\circ f=id_{S^n}$, and we need to show that $f\circ g$ is homotopic to $id_X$. We have $f\circ g(p,q)=(p,p).$ The idea is that since $q\neq-p$, you can move $q$ continuously along the geodesic connecting it to $p$. Remark: The above $g:X\to S^n$ is a fibration with fiber ...


2

The restriction is not meant to avoid pathological behavior; I just put it there for convenience. For example, it would be perfectly reasonable to interpret $\langle a,b\mid aa^{-1}, bb^{-1}\rangle$ as a presentation of a disjoint union of two spheres, or $\langle a,b\mid ab, ab^{-1}\rangle$ as a presentation of a projective plane. But we don't need these ...


1

As Lee Mosher has pointed out in the comments, the correct representation for the fundamental group should be $F_3/\langle aabcb^{-1}c^{-1} =1\rangle$, or in other words, $\langle a, b, c|aabcb^{-1}c^{-1} = 1\rangle$. $H_1$ is precisely abelianization of this group, and the abelianization just attaches the relators $[a, b] = [b, c] = [a, c] = 1$ to the ...


8

The power of algebraic topology is that problems which seem to have little or nothing to do with algebra, or little or nothing to do with topology, can be converted into algebraic topology questions, and from there into purely algebraic computations, and thereby can be solved. Rather than trying to convince you on some philosophical level, let me list some ...


0

Another way to proceed, which is a bit different from polygonal presentations, is to think about connect sums. If $S$ and $T$ are surfaces, then we can form $S \# T$, a new surface, by cutting a disk out of each, and gluing together the resulting circle boundaries. For example, if $S$ and $T$ are both copies of $T^2$ (torus) then $S \# T$ is a copy of the ...


5

When I was a graduate student (in the late 1950s) I was advised to read the Introduction (and nothing else!) of Solomon Lefschetz's "Introduction to Topology". I think the point is that to understand what is going on in Algebraic Topology today you have really to understand the history, and how the subject arose. But this is usually neglected in the books, ...


1

Higher dimensional spaces! That's my answer. You can use intuition for lower-dimensional spaces, like the topological space of complex numbers, although even there, intuition often gives the wrong answer. But if you try to work out even plausible conjectures as to global topology in higher-dimensional spaces, intuition fails completely. If the word ...


4

Topology immediately poses a question: How can I characterize spaces, up to homeomorphism? More specifically, how can I tell if two spaces are homeomorphic? This happens because homeomorphisms are the structure-preserving maps of topological spaces, so we are interested in being able to determine when two spaces are "the same thing". Now, we have ...



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