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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 ...


6

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. ...


5

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 ...


4

We can use the same approach that works for the covering $\Bbb R \to \Bbb S^1$. By definition, a deck transformation $\phi : \Bbb S^1 \to \Bbb S^1$ must preserve the covering map $\pi : z \mapsto z^n$, that is $\pi = \pi \circ \phi$. Substituting gives $$z^n = \phi(z)^n ,$$ and rearranging gives $(z^{-1} \phi(z))^n = 1$, and hence for all $n$ $$\phi(z) = ...


4

I think that when he says $f$ is continuous from $E$ to $B$. he intends to mean that the important part which was justified was: (..) to $B$. and assumes continuity is a trivial matter (which is, since it is division by $\frac{1}{1+\Vert x \Vert }$. Since the norm is continuous, multiplication by scalar is continuous, inversion is continuous in ...


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 ...


4

There's a little abuse of notation. First, $M$ has to be compact for all this to make sense; it's probably written somewhere before in the paper. When $M$ is compact, orientable, and connected, $H^{2d}(M;\mathbb{R})$ is isomorphic to $\mathbb{R}$, and the choice of an orientation basically amounts to the choice of an isomorphism $H^{2d}(M;\mathbb{R}) \cong ...


3

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 ...


3

As an alternative to Travis' nice answer (+1), you can also note that firstly all rotations by multiples of $2\pi/n$ are Deck transformations, and that secondly they together act transitively on each fiber. Since a Deck transformation is determined by its action on a single point, it follows that they are already all.


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 $n$ ...


2

So $f\circ g \equiv Id_Y$ i.e $H:Y\times I\to Y$ is a continuous map and $H(y,0)=f\circ g (y)$ and $H(y,1)=y$. $Img(f\circ g) \subset U$ (lets say). Then consider the path $\gamma (t)=H(v,t)$. This connect $U$ with a point in $V$. But $U,V$ are disjoint open sets so they cannot be connected by a path. (Contradiction).


2

Every reasonable path connected subspace of $\mathbb{R}^2$ has fundamental group a free group, which is either trivial or infinite. "Reasonable" means homotopy equivalent to a tubular neighborhood of it in $\mathbb{R}^2$, and in particular means homotopy equivalent to an open subset of $\mathbb{R}^2$. Now, an open subset of $\mathbb{R}^2$ is a noncompact ...


2

Your question was answered in affirmative by George Lowther here: The Fundamental group of every subset of $\mathbb{R^2}$ is torsion free? More precisely, he proves that if $X$ is a subset of any closed surface $S$, for instance, $S=S^2$, and $x\in X$, then $\pi_1(X,x)$ is torsion free. Note that no assumption is made about the nature of the subset $X$, ...


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 $H_n(X;F)=0$.


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


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

Here are some details on how to get the fibration in Qiaochu Yuan's answer:$\newcommand{\Z}{\mathbb{Z}}$ Let $H \curvearrowright E_H$ act freely on a weakly contractible space $E_H$, and let the exact sequence of groups $1 \to G \to G \rtimes H \xrightarrow{\phi} H \to 1$ be given. Associated to the map $\phi : G \rtimes H \to H$, there is an action $G ...


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 ...


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)] = ...


1

The zero locus of a monic (in $x$) polynomial $P(x,y)=0$ in $C^2$ is always path-connected. The multiset of roots varies continuously in $y$ for any suitable distance between multisets, such as edit distance based on a metric in $C$. If you want to trace the path of a particular $x$ as a function of $y$ this is not hard to extract from a given path of all ...


1

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 ...


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) ...


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 ...


1

Consider $X=S^1\vee S^3$ and its double cover $X_2$ i.e, attach two copy of $S^3$ one in north pole and one in south pole of $S^1$. Then $\pi_1(X) =\mathbb{Z} = \pi_1(X_2)$. And covering map induced isomorphism in $\pi_n$ for all $n\geq 2$. But they are not homotopically equivalent/ they have different homology groups since their Eular Characteristics are ...


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

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

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 ...


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

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 ...



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