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0

Let's start with $\Bbb RP^2$ since the rest are analogous. Description 1: The points of $\Bbb RP^2$ are the 1-dimensional subspaces of $\Bbb R^3$. A line in $\Bbb RP^2$ is a set of such points such that the 1-d subspaces all lie in a single 2-d subpace of $\Bbb R^3$. A point is on a line if the 1-d subspace for the point is contained in the 2-d subspace for ...


2

Homeomorphism is the ultimate topological equivalence. Two homeomorphic spaces are identical, topologically. That is, you can continuously turn one to the other, and back again, so continuous functions seem them as "identical". Homotopy equivalence says roughly that two spaces have the same "shape". For example, $\mathbb{R^n}$ is homotopy equivalent for all ...


1

$g$ does have degree $-1$. Degree only describes the action of a map on top homology, i.e. $H_1$ in this case. Every map between connected spaces acts as the identity on zeroth homology: for any chain $\alpha=\sum a_i \sigma_i, g_*(\alpha)=\sum a_ig(\sigma_i)$. In particular, for a 0-chain $x$, $g_*(x)=g(x)$, which is a generator of $H_0$. The confusion here ...


0

Would all these new diagonals would be completely new edges? Yes. They do not seem to be part of your description of an $n$-gon. Notice you will need more vertices, too, if you want the resulting object to be a simplicial complex. (Your construction is not well-defined for exactly this reason. How do you add new edges? Where? How are you ensuring that ...


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The problem is uniqueness: there may be multiple lifts (and in fact there is always an infinite number of lifts if $Y \neq \emptyset$). For example if $Y$ is a point and $I \to S^1$ is a constant map, every point in the fiber over the image (something that looks like $\mathbb{Z}$ if the covering map is $t \mapsto e^{2\pi it}$) is a possible lift.


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The proof is correct; however, remember to add the assumption that both $X$ and $Y$ are manifolds. As Idrissi says in the comments, this proof also applies to homology manifolds.


3

$\require{AMScd}$ You have a commutative diagram $$\begin{CD} 0 @>>> ker(\varepsilon) @>i>> C_0(X) @>\varepsilon>> \Bbb Z @>>> 0\\ @. @VVV @VVV @| \\ 0 @>>> \tilde H_0(X) @>\bar i>> H_0(X) @>\bar\varepsilon>> \Bbb Z @>>> 0 ...


1

@Vrouvrou: You have asked many questions on the subject of Morse theory. I have the feeling any book on the subject must discuss this on the first page. The basic observation is (under suitable compactness assumptions) the following. Let $f:M\rightarrow \mathbb{R}$ be a smooth function. And $c-\epsilon$ and $c+\epsilon$ are regular values such that the ...


0

You can just do the same calculations as with integer coefficients. Using the same notations as in this answer to your previous question, you have $H_1(K;G)=\langle d\rangle/\langle 2d\rangle\oplus\langle c\rangle=G\oplus G/2G$. The boundary of a chain $g_1U+g_2L$ with coefficients $g_1,g_2\in G$ is $g_1(a+b-c)+g_2(a-b+c)=(g_1+g_2)a+(g_1-g_2)b+(g_2-g_1)c$ ...


1

Since $\gamma$ is a loop, $\gamma(\alpha) = \gamma(\beta)$, and so $$\theta(\gamma(\beta)) - \theta(\gamma(\alpha)) = 0 \in \mathbb{R} / 2 \pi \mathbb{Z}.$$ Thus, for any lift $\tilde{\theta}$ of $\theta$, $$\tilde{\theta}(\gamma(\beta)) - \tilde{\theta}(\gamma(\alpha)) \in 2 \pi \mathbb{Z},$$ and so $$n(\gamma, a) \in \mathbb{Z}.$$


3

First, do not take the preimage of an arbitrary point. Instead, take the preimage of a regular value $p \in S^1$. The inverse function theorem then guarantees that $N=f^{-1}(p)$ is a submanifold of codimension 1. Next, show that $N$ has orientable normal bundle. What this means is that $T(M)\mid N$, the restriction to $N$ of the tangent bundle $T(M)$, has a ...


0

You may look at [Poset Topology] and the related lecture notes by Michelle L. Wachs.


1

This question is relevant to the question of the universal cover of a topological group, $G$, if $G$ is not connected. We would like the answer to be a surjective covering map $p: U \to G$ such that $p$ is a universal cover on each component, $U$ has the structure of topological group and $p$ is a morphism of topological groups. The obstruction to this ...


3

I don't think there's a widely accepted definition of what the universal cover of a disconnected space is; the standard definition, as the maximal connected cover, only applies to (sufficiently nice) connected spaces, since a disconnected space has no connected covers. One candidate is "the disjoint union of the universal covers of its connected ...


2

Let me summarize my comments in an answer. I don't have answers to all the questions but I don't expect (myself) to really come up with any more, so here goes. First, I am mostly thinking about this space as the algebraic variety over $\mathbb{R}$ (complex conjugation is not $\mathbb{C}$-scalar) cut out by the $n^2$ (in the $\mathbb{R}$ case) or $2n^2$ (in ...


0

Since your groups are most likely abelian and finitely generated, you can factor $G$ into the direct sum of its torsion group and a free group. This reduces any such problem to a problem about finite abelian groups.


1

I would start by thinking of the torus as a square, with two pairs of sides identified. If the exercise were to find an equivalent for the once punctured torus, I would think of a punctured square, which is obviously equivalent to a circle. But considering the identifications, the punctured torus is homotopy equivalent to two circles that share a point. ...


1

The topology on a CW complex $X$ is the final topology on $X$ with respect to the inclusions $X^n \to X$, ie. it's the finest topology on $X$ such that these inclusions are continuous. On the other hand, as you say the weak topology is the initial topology wrt continuous linear functionals. These two notions are dual (initial and final topologies). I doubt ...


4

The crucial point is how $RP^{k-1}/RP^{k-2}$ is identified with $S^{k-1}$. $RP^{k-1}$ is the quotient of the northern hemisphere $N$ that is obtained by identifying opposite points on its boundary $S^{k-2}$. If $RP^{k-2}$ is divided out, this is the same as building the quotient $N/S^{k-2}$, which is the same as $S^{k-1}$ since it is the one point ...


6

I think it's simpler to see that if $f : X \to Y$ is a constant map, then it factors through as $X \to * \to Y$ where $*$ is a singleton. Therefore $f_*$ factors through $H_n(X) \to H_n(*) \to H_n(Y)$. But $H_n(*) = 0$ for $n > 0$, from which you conclude that $f_*$ is the zero map. In fact this proof shows that any cycle in $C_n(X)$ (I guess it's ...


3

For instance, the mountain pass lemma, says that if you have a connected "manifold" with a Morse function with two local minima then there is a path between them passing through an index one critical point. I placed manifold in parentheses because it might be infinite dimensional with appropriate conditions on the Morse function. This has been used to prove ...


0

N.Owad's figures are very nice and much better than I could do, but it may also be helpful to show the paths at a crossing, as follows, and which also includes the graph case. See also the part on knots of the presentation on Out of Line: Motion. I have had a pentoil made of copper tubing and given demos of the fundamental group of (the complement of) ...


2

I'll add that this means the group is isomorphic to $\mathbb{Z} \times F$, where $F$ is a free (nonabelian) group on two generators. In particular, before modding out, there are two pairs of generators, say $\{x,y\}$ for the first $\mathbb{Z}\times \mathbb{Z}$ and $\{a,b\}$ for the other. Each pair commutes with itself, but the two pairs do not commute. ...


4

Here are two tori with a copy of $S^1$ identified. The picture is clearer when we move $x_0$ to the inner rim of one torus, and to the outer rim on the other torus. The two images below show a path on one of the tori that belongs to the homotopy class of the latter generator. Basically Van Kampen says that these two become homotopic in $\pi_1$ of ...


2

$\tilde{X}\times_{\pi_1(X)}G$ is an example of the associated bundle construction: Suppose you have a principle $H$-bundle $E\rightarrow B$ and a left action of $H$ on some other space $F$, then you can form the associated bundle $E\times_H F$ as described in wikipedia in the section "Fiber bundle associated to a principal bundle". This is a fiber bundle ...


1

Ok, let me answer these questions in the order you asked them. About almost flat, I could not understand the 'U' turns part of your question, but to see what Hatcher means by almost flat, you should think about this picture: So, the only time our knot is not in the $x,y$ plane is at the crossing and the overarc just is bumped up (in the positive $z$ ...


2

The $n$-skeleton $X^n$ is obtained from the $(n-1)$-skeleton by gluing a union of $n$-balls along their boundaries to that $X^{n-1}$. This gives the first of the two pushout squares in the diagram. $\require{AMScd}$ $$\begin{CD} \coprod\nolimits_\Lambda S^{n-1}_\alpha @>\varphi>> X^{n-1} @>>> \{*\} \\ @VVV @VVV @VVV\\ ...


1

Use Yoneda lemma in the pointed category $Top_*$ and a bit of fancy: $Top_*(X^{(n)}/X^{(n-1)}, Y) \simeq \{ f:X^{(n)} \to Y \ s.t. f_{| X^{(n-1)}} = constant \} \simeq \{f_{\alpha}: D^n \to Y \ s.t. f_{\alpha| S^{n-1}} = * \}_{\alpha \in \Lambda _n} \simeq Top_*(S^n,Y)^{\Lambda _n} \simeq Top_*(\bigvee _{\alpha \in \Lambda _n} S^n, Y) $ .


1

Let $X_1, X_2, \dots X_n$ be the closed convex sets, and let $f \colon [0,1] \to X$ be a path. We first prove that given any $\alpha \in [0,1)$, there is some $\epsilon > 0$ such that for all $\eta \leq \epsilon$, the restriction of $f$ to $[\alpha,\alpha + \eta]$ is homotopic in $X$ to a linear path via a homotopy that is constant for $t = \alpha$ and ...


1

Yes, it deformation retracts to a wedge of four circles. Additionally, $\pi_1(\bigvee^4 S^1)=\ast^4\pi_1(S^1)$, where $\ast$ is the free product of groups.


3

As Hakim pointed out in a comment, there is no natural distance between two points in your concept. Of course, you could take a classic 0-1 metric (where the distance between any two distinct points is 1) but that is unlikely to lead anywhere. It would be more natural to use a graph. A graph is an extremely simple structure. It has a set of points (usually ...


1

This quotient is a sphere bundle over $RP^n$: You can think of this construction as a special case of an associated fiber bundle, associated with the representation $$\pi_1(RP^n)\to O(n+1)=Isom(S^n)$$ sending the generator to the matrix $-I$. This bundle is non-orientable if $n$ is even, hence, the bundle is nontrivial for even $n$. The bundle is orientable ...


1

Yes, they are the same. For a quick explanation, I'll paraphrase Hatcher's discussion from his book Algebraic Topology. In terms of CW complexes, a graph is a space $X$ obtained from a discrete set $X^0$ be adding a collection of 1-cells $e_\alpha$. In non-CW speak, this means $X$ is formed from the disjoint union of $X^0$ with closed intervals $I_\alpha$ ...


0

The space is homeomorphic an open ball in $\mathbb{R^3}$. I would look first at the analogous case : $X$ closed ball in $\mathbb{R^2}$ imbedded in the sphere $S^2$ by a stereographic projection from the North pole.


2

Let $X$ be a closed disk in $\mathbb R^2$. Think about the family of maps where you first crush the top half of the disk down, so that disk becomes just the lower semi-disk. Now grab hold of the "vertices" of the semi-disk (i.e. the points where the lower semi-circle meets the horizontal diameter), or rather small neighborhoods of these vertices, and start ...


1

We may take $X$ to be $\{x\in\mathbb R^3:\|x\|\le 1\}$. The mapping $f:X^\circ\to B$ given by $$ f(x) = \begin{cases} x/\|x\|^2 & \text{if }\|x\|>0, \\[6pt] \text{the compactification point} & \text{if }x=0, \end{cases} $$ is a homeomorphism from $X^\circ$ to $B$ (where $X^\circ$ is the interior of $X$). Therefore if $X^\circ$ is simply ...


1

The space $B$ is homotopy equivalent to $S^3$ minus a point, which by Seifert-van Kampen has trivial fundamental group (or notice that $S^3$ minus a point is again homotopy equivalent to $\mathbb{R}^3$).


2

The answer is No, even if $X$ and $Y$ are compact. Let $X = S^1$, $M$ be $S^1\setminus\{p\}$, that is , $M$ is $S^1$ with a point removed. Let $Y =[0,1]$ and $N = (0,1)$. Obviously, $M$ and $N$ are homeomorphic, so let $f:M\rightarrow N$ be such a homeomorphism. Then there is no continuous extension $F:X\rightarrow Y$. This follows because $F(p)$ is ...


1

The Morse complex on a manifold (without orientation assumptions) can be defined using the following data: A Morse function $f$, a metric $g$ and a choice of orientation $\mathfrak{o}$ of the unstable manifolds. Then the space $W(x,y)=W^u(x)\cap W^s(y)$ is given an orientation by the following exact sequence $$ 0\rightarrow TW(x,y)\rightarrow ...


0

I hope it is allowed to still add my bit. Since by your reference the action of $H$ is properly discontinuous, your problem is a special case of corollary 4.11 describing fundamental groups of orbit spaces. Just in case you are interested in some theory behind the question. Yet another way I think you can even solve it is by using the fact the $G ...


2

Hi, I think you need to work on your homotopy theory, particularly the notion of homotopy fibre. The following reference might be helpful, once you have done that! Gilbert, N.D. "On the fundamental $\rm cat^ n$-group of an $n$-cube of spaces". In "Algebraic topology, Barcelona, 1986", Lecture Notes in Math., Volume 1298. Springer, Berlin (1987), ...


0

Your second question has been answered by Lee Mosher. Here is the answer to your first question: Consider two smooth paths $$t\mapsto z_0(t), \quad t\mapsto z_1(t)\qquad(0\leq t\leq 1)$$ which both avoid the origin and are homotopic with respect to $\dot{\mathbb C}$. This means that there is a continuous map $$F: \quad [0,1]^2\to\dot{\mathbb C},\qquad ...


1

$\require{AMScd}$ The diagram $$\small{\begin{CD} A @>i>> X \\ @VVV @VVV\\ A/B @>j>> X/B \end{CD}}$$ is a pushout. It is a general fact in category theory that if a map in a diagram $\bullet\leftarrow\bullet\to\bullet$ has the LLP (left lifting property) w.r.t. a class of maps, then the parallel map in the induced pushout diagram has the ...


1

To answer 2, analytic maps of $\mathbb{C}$ preserve orientation, but continuous maps such as $f(z) = \overline z$ may reverse orientation.


0

In Algebraic Topology, one of the most fundamental questions is whether or not $X$ and $Y$ are homotopy equivalent spaces. It turns out that this question is sometimes easier to answer by algebraic means. Proving homotopy inequivalence is difficult because it requires demonstrating that no maps exist which satisfy the requirement. But if we know that ...


2

Sure: $w(\xi_1) \cdots w(\xi_n) \overline{w}(\xi_1) \cdots \overline{w}(\xi_n) = w(\xi_1) \overline{w}(\xi_1) \cdots w(\xi_n)\overline{w}(\xi_n) = 1\in H^*(M, \mathbb{Z}_2)$.


2

The signature is $0$. One way to see this is that $\Sigma_1$ is a boundary, so $\Sigma_1 \times \Sigma_2$ is also a boundary, and the signature vanishes on boundaries. You can also directly compute the intersection form using the Kunneth formula.


3

The statement is wrong, of course. The correct one is that a covering is regular if and only if for every loop in the base one lift is closed iff all lifts are closed. (One still has to define properly what a closed lift means.)


2

The resulting object will be (assuming everything's compact, say...) an orientable manifold, but not an oriented one. If you want it to be oriented, with the orientation matching that of $M_1$, say, then the map $f$ should be orientation reversing (which may not be possible without reversing the orientation on $M_2$, and hence the induced orientation on ...


0

being just a fiber bundle with discrete fibers - yes. Look at some proof of pullbacks of fiber bundles, which are used more often in general than in basic covering space theory. But good question!



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