Tag Info

1

For the sake of having an answer: no, the disk $D^4$ (mathematicians' notation for the hemisphere in $S^4$) is not the unique compact $4$-manifold with boundary $S^3$. Starting from any closed $4$-manifold $M$, you can cut a hole out of it to get a compact $4$-manifold with boundary $S^3$, and most of these are different (they can be distinguished e.g. by ...

1

After some months, I've finally arrived to a satisfactory answer to this question (I believe the answer given by Pedro Tamaroff is wrong...). First we are gonna need this lemma which is proven here surjective covering space If $p:\tilde X\to X$ is a covering space with $\tilde X$ nonempty and $X$ connected then $p$ is surjective. Note that Hatcher ...

0

On page 51 of Hatcher, it is assumed that we start with an orientable surface $M_g$ of genus $g$ and have its cell structure. The cell structure contains the information of not only the cells, but how they fit together - this data is contained in the gluing maps. Since the 1-skeleton is a wedge sum of $2g$ circles, almost all of the information is included ...

2

The discreteness assumption is meaningless since given a continuous group action $G\times X\to X$, it remains continuous if we equip $G$ with discrete topology. Now, a counter-example to the claim is the action of the group of the additive group of real numbers on itself $${\mathbb R}_{d} \times {\mathbb R} \to {\mathbb R}$$ via addition. Here the ...

2

This should be a comment, not an answer, but I'm new here. There are two typos that seem to be causing you problems. First, your formula for $\beta$ should read $$\beta(s) = \frac{(s_1-s)\alpha(s_0)+(s-s_0)\alpha(s_1)}{s_1-s_0} : [s_0, s_1] \to \mathbb{R}^n.$$ I've changed an $s_0$ to an $s$, which should fix your calculation for $\beta(s_0)$. Second, ...

1

It looks like you've got a few things backwards in question 1. You're not looking for a covering space of $S^2$ - there are no such things other than $S^2$ itself, since it's simply connected! Instead, you're looking for a covering space of $Q$. In fact, you're looking to show that $S^2$ is a covering space of $Q$. (The covering map should be $f$, so ...

9

Yes, you can reuse Brouwer's fixed point theorem. Let $\bar{\phi} : D^2 \to \mathbb{R}^2$ be the restriction of $\phi$ to $D^2 = \{ x \in \mathbb{R}^2 \mid \|x\| \le 1 \}$ (for whichever norm you want, usually the $L^2$ norm). Then since $D^2$ is compact, $\|\bar{\phi}\|$ reaches a maximum. Either $\max \|\bar{\phi}\| = 0$, in which case $\bar{\phi}(0) = 0 ... 0 A direct limit in topological spaces has as underlying set the corresponding direct limit of sets, and its topology is the "weak topology" or "final topology" for which a subset is open precisely if its restriction to all the stages (in the sequence that the direct limit is over) is open. Generally, the direct limit is a special case of a colimit of ... 0 This is a rather straightforward application of the van Kampen theorem. Write your space as the union of two half-cylinders. The intersection is$S^1$. What happens to the generator of$\pi_1(S^1)$if you include into the lower half? What about the upper half? Conclude that your space has fundamental group isomorphic to$\langle a,b |ab^{-r}\rangle$. 1 Your picture of two horns attached at a point is correct - but from there you can stretch the attached section until you have two small horns joined by a line, and from there deform it until it's$T \vee S^1$. I've attached a hastily-drawn picture. In answer to your "general construction" question: hopefully that's more obvious now that you've seen how the ... 0 It's important to distinguish between being equal to the constant map and being homotopic to the constant map. Eric Towers' answer gives you a map that induces the zero map on$\pi_1$but is not the constant map. However, it will be homotopic to a constant map. Doing the other way round: you seem to already be aware that$\pi_1(S^1, x_0) \cong \mathbb Z$, ... 1 Dave Glickenstein's notes give a short proof, see page 9. 0 If you've seen fundamental polygons before, one way for me to answer your question would be to say "allow repeated edges on your fundamental polygon!" and claim that the answer is$\langle a, b | aba^{-r}b^{-1}\rangle$(which it is). One way to see that this relation must be true is to take the loop$a = S^1 \times \{0\}$, drag it around the "torus" (that ... 2 If$f$omits any point of$S^1$, the image of$f$is contractible. In particular, if$f$is constant,$f(S^1)$is contractible and$f_*$is the trivial map. The other direction is false. Let$S^1$be represented by the set$\{(x,y) \in \Bbb{R}^2 \mid x^2 + y^2 = 1\}$. Let$f(x,y) = (|x|,y)$and$x_0 = (1,0)$. Then$f$is continuous, nonconstant, and ... 2 This follows from standard forklore in differential topology (maybe due to Thom or Poincare??): Let$M^n$be a smooth orientable$n$-manifold, then any homology class of codimension 1 or 2, that is, an element in$H_{n-1}(M^n)$or$H_{n-2}(M^n)$, can be represented by the fundamental class of a submanifold. 0 First, a point of terminology: a map is not said to be contractible (this is a property of spaces), it is said to be nullhomotopic. Now, when looking at relative homotopy groups, two maps$f,g : (D^n, S^{n-1}, s_0) \to (X,A,x_0)$are identified if they are homotopic relatively to$A$(in short, "rel$A$"). This means that there must exist a homotopy$H : ...

1

Call the lines $L_1, \dots, L_n$. Let $w$ be a direction vector for the lines, and let $P$ be the plane through the origin with normal vector $w$. Note, all the lines are orthogonal to $P$. Let $\{u, v\}$ be a basis for $P$, then $\{u, v, w\}$ is a basis for $\mathbb{R}^3$. Define the linear transformation $T : \mathbb{R}^3 \to \mathbb{R}^3$ by $T(u) = ... 1 In the accepted answer to the linked question$f(x)$was shown to be a monotone function which was strictly increasing on the open interval of those$x$such that$0<f(x)<|A|$. Remove connectedness and$f$may not be strictly increasing on a given open subset of such$x$. In particular, there can be infinitely many values with$f(x)=|A|/2$. A simple ... 2 This is a special case of the "Real Cancellation Theorem: suppose$X$is$d$-dimensional CW complex and$E\to X$an$n$-dimensional real vector bundle with$n>d$. Then, if$F$is another vector bundle and$E\oplus \mathbb{R}^k\simeq F\oplus \mathbb{R}^k$, then$E\simeq F$." By choosing$E=\mathbb{R^n}$you get the thesis. Another way of stating the ... 0 Well, almost any invariant of algebraic topology (since deformation retract is in particular a homotopy-equivalence). This includes for instance homology groups, the fundamental group etc... (This is why this notion is so useful in algebraic topology) 1 Edit: the question has been changed since this answer was written. I'm slightly confused as to where$\pm \sqrt{m+1}$has come from... Your function is$f(x) = x$outside the interval$[-T, T]$, so if you take$m+1$as your new regular value then your only preimage is$m+1$itself and your function has degree 1. In general a quadratic with non-degenerate ... 1 There are$2$closed surfaces with Euler characteristic$-2$. One is the standard genus$2$-torus, while the other is the connected sum of a regular torus with the Klein bottle, or equivalently, the connected sum of four projective planes. For any even, non-positive Euler characteristic, there are exactly two closed surfaces, up to homeomorphism: one ... 4 EDIT: Sigh... the torus is compact, but$T\backslash \{x_0\}$isn't.$\blacksquare$Nevertheless, I'm gonna keep the answer below. If$A \subset B$is a retract of$B$, then the induced map (by inclusion) on the fundamental group is injective. This is seen readily from functoriality. Note that the torus with a point removed has the free abelian group ... 8 Some points. 1) An oriented manifold$M$which occurs as the boundary of an oriented manifold$W$necessarily has$\chi(M)$even. This means that not only are all of your surfaces not going to work, but neither is anything that's the boundary of a manifold. 2) Every odd-dimensional manifold has$\chi(M) = 0$(even non-oriented ones). So you can't try that. ... 7 By the classification of surfaces, every closed oriented surface is a connected sum of tori, so you won't be able to find a$2$-dimensional example. By Poincare duality, any closed odd-dimensional manifold has Euler characteristic$0$, so the smallest dimension where you can find an example is$4$. To find a$4$-dimensional example, you can start with ... 1 $$\require{AMScd} \begin{CD} I\times X @>\pi_\sim>>(I\times X)/{\sim} \\ @V\exp\times 1_X VV @VVh V \\ S^1 \times X @>>\pi_\wedge> \operatorname{S}X \end{CD}$$ The map$\exp$is a perfect map, that is a closed map whose fibers are compact. For any space$X$, the map$\exp\times 1_X$is therefore a closed map. Now$\pi_\sim$identifies ... 0 Are you sure you read the question correctly? I don't think the spaces are homotopic. Have you tried to compute the$H_1(X,Z)$group for each space? Or the fundamental groups of each space? Do they agree? 2 In general if a topological group$G$acts on a topological space$X$then each$g\in G$gives a homeomorphism$\theta_g:X\to X$defined by$\theta_g(x)=g\cdot x$. Let$p:X'\to X$be a covering map. We say that the action of$G$lifts to$X'$and is compatible with the action on$X$if there exists a map$G\times X'\to X'$such that the following diagram ... 0 I think one can give definition of cubical maps of cubical complexes in a similar way to simplicial maps between simplicial sets. Cubical maps between cubical sets is a map of vertices which is compatible with face and degeneracy maps. 2 For (a) you want to prove that every$y\in Y$has an evenly covered neighborhood in$Y$. Let's start with what we are given. We are given a continuous map$f:Y\to X$. So let$x=f(y)$. Now we are also given a covering map$p:T\to X$. So this means$x$has an evenly covered neighborhood, call it$U$and let$p^{-1}(U)=\bigsqcup_\alpha U_\alpha$. Let ... 2 By simple computations, one checks$AB=BA$,$AC=CA$,$CB=BCA$is a presentation. So in the abelianization$A=1$,$BC=CB$the abelianized group is$\bf Z^2$. Note that$dy, dz$are invariant by$G$and form a base for the de Rham co-homology 0 Neither of these hold. Consider the standard covering of$S^1$by$\mathbb R$given by$x \mapsto (\cos(x),\sin(x))$. This is surjective, but on fundamental groups it is given by the inclusion of the trivial subgroup in$\mathbb Z$. For the other case, consider the inclusion$S^1 \to \mathbb R^2$; this is injective, but on fundamental groups this gives the ... 0 HINT: for the first question you can use the stereographic projection to see that a sphere minus a point is homeomorphic to a plane, in particular all homotopy groups are the same. For the second question: use the Van-Kampen theorem to see that$\pi_1(S)$is trivial. The conclusion follows. 2 This is just the pasting lemma again: you are pasting two continuous maps, one on$\Omega Y\times \{(s,t)\in I\times I:0\leq s\leq\frac{t+1}{2}\}$and one on$\Omega Y\times\{(s,t)\in I\times I:\frac{t+1}{2}\leq s\leq 1\}$. These are closed subsets of$\Omega Y\times I\times I$whose union is the whole space, and the continuous maps agree on the ... 6 Your idea is OK, using homotopy groups. Suppose$\mathbb{Z} = G \times G$for some (Abelian) group$G$, which must be infinite. Then note that$\{0\} \times G$and$G \times \{0\}$are infinite subgroups of$G \times G$that intersect only in the unit element$\{(0,0)\}$. By the isomorphism that supposedly exists, such subgroups should also exist in the ... 0 As Mambo mentioned, the group operation and$\pi$are continuous by definition. Therefore, the map$\phi: G\times G \to G/H, (x,y)\mapsto xyH$is continuous as a composition of the group operation an$\pi$. Hence, the preimage$\phi^{-1}(U)$of any open set$U\subseteq G/H$is open in$G\times G$. But$\pi$is an open map, so$\psi=(id,\pi): G\times G \to ...

0

As a counter example was already presented, I'll focus on the last part of your question, and conclude with a (a priori large) class of manifolds which will not be able to serve you with counter examples. Claim: The result holds if and only if $M$ is tame. One direction is immediate. For the other direction, let $M$ be tame and hence properly embeds ...

4

Two $R$-modules $M$ and $N$ are said to be "stably isomorphic" (see Stacks for example) if $M \oplus R^n$ and $N \oplus R^n$ are isomorphic for some $n \in \mathbb{N}$. It is possible for two non-isomorphic modules to be stably isomorphic. If $M'$ and $N'$ are such modules, let $n$ be the smallest (necessarily positive) integer such that $M' \oplus R^n \cong ... 0 So we have$u: A \rightarrow B$, continuous and$u(a) = b$and we have$[\gamma] \in \pi_(A,a)$. This means, as you say, that$\gamma: [0,1] \rightarrow A$is continuous and$\gamma(0) = \gamma(1) = a$. Then$u \circ \gamma$is continuous from$[0,1]$into$C$, as a composition of continuous functions, and $$(u \circ \gamma)(0) = u(\gamma(0)) = u(a) = b = ... 1 As homology with \mathbb Z coefficients consists of abelian groups, you can think of \mathbb Q-homology as measuring the free part and \mathbb Z/p-homology as measuring the p-torsion part. Hence we have: X is an integer homology point iff X is a rational homology point and a \mathbb Z/p-homology point forall p. Next note that we always have ... 0 For the first part, you want a path from a \in \mathbb{R}^{2n+1} to -a that doesn't go through zero. There are plenty of ways to do this; for example, let b be some point which is not colinear with 0 and a (this exists because 2n+1 \ge 2, draw a picture to convince yourself). Then concatenate the line segment from a to b and the line segment ... 2 To give an answer along the lines suggested by OP, one needs only remark that for a field F we have a vector space isomorphism$$Hom_F (H_n(X;F),F)\cong H^n(X;F)$$Now H_n(X;F) is a vector space over F and if Hom_F (H_n(X;F),F)=0 then H_n(X;F)=0. 1 Suppose X=A_0\cup A_1 and we have x_0\in A_0\cap A_1. Now the Siefert-VanKampen Theorem states$$\pi_1(X,x_0)\cong\pi_1(A_0,x_0)\ast_{\pi_1(A_0,x_0)\cap\pi_1(A_1,x_0)}\pi_1(A_1,x_0)$$is an amalgam provided several assumptions are satisfied: 1) We must have that A_0 and A_1 are open in X. 2) We must have that A_0 and A_1 and A_0\cap A_1 ... 1 Under certain conditions they do actually mean the existence of neighborhoods, containing the common point x_0 of the wedge, which do retract on x_0. With this condition we are able to choose suitable open path-connected sets A_i to compute the desired fundamental group. For instance if you take the n wedge of circles S^1 at x_0 you can ... 1 So we can actually build up the answer inductively here: let's label the simplices of \Delta_2 as v_0, v_1, v_2, and the corresponding vertices of the top and bottom faces of \Delta_2 \times I as t_0, t_1, t_2, b_0, b_1, b_2. So let's define P on C_0(\Delta_2). Here \partial P + P\partial = \partial P (as \partial vanishes on 0-simplices) ... 2 The long exact sequence of homology (which indeed comes from the snake lemma) reads$$H_i(X,\mathbb{Z}) \to H_i(X,\mathbb{Z})\to H_i(X,\mathbb{Z}/n\mathbb{Z}) \to H_{i-1}(X,\mathbb{Z})\to H_{i-1}(X,\mathbb{Z})$$where the maps$H_i(X,\mathbb{Z}) \to H_i(X,\mathbb{Z})$and$H_{i-1}(X,\mathbb{Z})\to H_{i-1}(X,\mathbb{Z})$are given by multiplication by$n$... 1 The reason that being$n$-simple is not enough is that, not only do we need to have a canonical identification of$\pi_n(F,*)$for every choise of basepoint (which is where the$n$-simplicity condition shows up), we also need a canonical identification of the homotopy groups of different fibers$\pi_n(F,*) \cong \pi_n(F',*)$. Put another way, there's an ... 1 I take "relative to$A$" to mean equivalence classes in$\pi_1(X, A, x_0)$, that is paths$\gamma: I \to X$with$\gamma(0) \in A$and$\gamma(1) = x_0$up to homotopy through other paths of the same form. The result then isn't too difficult to see. Let$i: \pi_1(X, x_0) \to \pi_1(X, A, x_0)$denote the inclusion. In one direction, if$[i(\gamma_1)] = ...

0

Reading the comments it seems your second and third questions have been answered. I address the first question. The hint suggests using Lemma 1.15, which reads: If a space $A$ is the union of a collection of path-connected open sets $A_\alpha$, each containing the basepoint $x_0\in A$ and if each intersection $A_\alpha\cap A_\beta$ is path-connected, ...

1

Question 1 has been answered in principle and 2. is quite a nice question! One way of doing it is to realize $S^2$ as the union of two hemispheres $E^2_+, E^2_-$ with intersection $S^1$. Then give $S^1$ the structure of a polygon with a set $X$ of $n$ vertices. Now each $E^2_+. E^2_-$ can be given a cell structure with $S^1$ as the $1$-skeleton, and one ...

Top 50 recent answers are included