New answers tagged

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 ...



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