New answers tagged

1

You can view it as a toy example of classifying covering spaces of a given space. Every introduction begins with toy examples to understand basic concepts. In introductory group theory, subgroups of cyclic groups are classified for instance. There is a reason I use that analogy too: coverings of a space correspond to (conjugacy classes of) subgroups of a ...


5

As an addition to Najib's answer, I would like to provide a nice fact which follows from the fact that $S^1$ covers itself. First of all, to be honest to myself, I will point out that showing $S^1$ covers itself just for the sake of examples is enough a reason. Now, secondly, as a simple application, we have that, if a space $X$ is a finite CW-complex and ...


0

Let $x\in X$, consider a neigborhood $U$ of $x$ such that $V_x=\{g\in G:g(U)\cap U\neq \phi\}$ is finite. Write $V_x=\{y_1,...,y_n\}$ there exists $x\in U_i, y_i\in V_i$ such that $U_i\cap V_i$ is empty since $X$ is separated, set $W=U\cap_iU_i$, $W$ is an open subset containing $x$. Remark that if $g\in G$ and $g\neq I_d$ and $g(W)\cap W$ is not empty, then ...


4

$K(Z_n,4)$ is the forth Eilenberg McLane space, this implies that $\pi_4(K(Z_n,4))=Z_n$ and $\pi_n(K(Z_n,4))=1$ for $n\neq 4$. Now use the Hurewicz theorem, it implies that $\pi_5(K(Z_n,4))\rightarrow H_5(K(Z_n,4),Z)$ is surjective, since $\pi_5(K(Z_n,4))=1$, you deduce the result. https://en.wikipedia.org/wiki/Hurewicz_theorem


0

Edit: I'm rewriting this in light of the comment that $B$ is actually the intersection of $S^2$ with the positive orthant rather than the union as originally stated. In this case, you're correct that they are homeomorphic! To see this, note that these two sets can both be contracted down to a point and are the same dimension. That said, your functions ...


0

Of course the invariants you are asking for seem to be interesting and of course they are interesting in the sense that they contain information about the map $f$ but then you could just look at the element $f_*$ in the corresponding homomorphism spaces between the homology groups. This would be something like the finest invariant of this kind. The main ...


0

For manifolds with boundary, I think you need that $\partial X$ is mapped to $\partial Y$. Then everything works. For non orientable maps, one cheaply has the mod 2 degree : $H_n(X;\mathbb{Z}/2\mathbb{Z})\rightarrow H_n(Y;\mathbb{Z}/2\mathbb{Z})$. Another way of defining a integer valued degree is to use local coefficients. One should assume that the line ...


2

I assume what you're asking about is whether the Stiefel-Whitney class/numbers of a smooth manifold only depend on the homotopy type; similarly with the Pontryagin numbers of an oriented such manifold; similarly for the Chern classes of a complex manifold. SW: Yes. These can be defined entirely in terms of Steenrod squares, and indeed they only depend on ...


2

Classically Brown Representability says that if we have a cohomology theory $k^*$ satisfying the Meyer-Vietoris axiom and the wedge axiom that we can find a classifying spectrum $Y_n$ and element $u_n$ of $k^n(Y_n)$ where $B_{u_n}:H^n(-) \to [-:Y_n,y_*]$ is a natural equivalence. This is the case for reduced cohomology. To make this applicable to your ...


1

First one notes that the signature is a bordism invariant, additive, and multiplicative, hence defines $\sigma:\Omega \otimes \mathbb Q\to \mathbb Q$. Next you note that every ring morphism from the oriented bordism ring $\Omega \to \mathbb Q$ factors through $\Omega \otimes \mathbb Q$. But we know that on the latter space the Pontryagin classes are a ...


2

I've found the answer: the two are just the same things. A comment before made me be aware that $\mathsf A$ and $\mathsf B$ can be identical, and it's not hard to prove that this is exactly the case. First, let me make some convention about terminology: A topological space $X$ is a k-space when a subspace of $X$ is closed iff the preimage along any ...


0

Yes, an embedding is by definition injective. If you change the definition, then it's not what anyone calls an embedding anymore. It's like asking if the condition "every element has an inverse" is "really critical" in the definition of a group... Sure, you can consider "groups" where not all elements have inverses. Well, except nobody calls that a "group", ...


5

Ok, so lets start by identifying what group we're actually working with. We're acting with the action of a fractional linear transformation right? So we are looking at subgroups of $PSL(2, \mathbb{C})$. Ok, so with the restrictions given we know that we have to be in the matrix group generated by $$ \begin{bmatrix} a & b \\ 0 & 1 \end{bmatrix} $$ ...


2

If we consider $S^{n}$ as a subset of $\mathbb R^{n+1}$, then we get a continuous map $S^{n+1}\to\mathbb R^{n+1}$. By the Borsuk-Ulam Theorem, such a map must map two antipodal points in $S^{n+1}$ to the same point. In particular, these two antipodal points are not mapped to antipodal points in $S^n$. Alternatively, prove it directly - if $f\colon S^{...


2

Yes, this is correct and your argument is correct. (One can also think about this in terms of linear systems. If $D$ is a divisor of degree $0$, then $h^0(D)\le 1$, with equality holding precisely when $D$ is the trivial divisor.)


2

Following Mike Miller's suggestion, consider the cylinder $X =S^1 \times \mathbb{R}$ (as a Riemann surface, you may view it as either $\mathbb{C} \setminus \{0\}$ or $\mathbb{C}/\mathbb{Z}$). As this deformation retracts onto the base circle and homology is a homotopy invariant, we know that $H_2(X;\mathbb{C}) \cong H_2(S^1;\mathbb{C}) =0$. As $\mathbb{C}$ ...


1

With regard to your first question, the answer is no, we do not necessarily identify all vertices unless we expressly say so. Take for example the identification of the square into the cylinder. In this gluing, we identify one pair of opposing edges of the square, but we only identify the endpoints of these edges in pairs ($A$ with $A'$ and $B$ with $B'$). ...


1

$n$-cubes have the advantage that integrating over them (which is what Spivak wants to do) is easy, and another advantage (that Spivak doesn't really use) which is that the "cartesian product" of an $n$-cube and a $k$-cube is an $n+k$ cube. On the other hand, simplices have their own advantages, like simplicity in various contexts. Because every simplex ...


10

It is interesting because these coverings have different degrees. They correspond to different subgroups of the fundamental group of $S^1$: the map $z \mapsto z^n$ corresponds to $n\mathbb{Z} \subset \pi_1(S^1) = \mathbb{Z}$. This shows that for different values of $n$, these coverings are not isomorphic, i.e. there is no homeomorphism $f : S^1 \to S^1$ such ...


0

$\require{AMScd} \newcommand{\RP}{\mathbb{RP}}$ Note: The splitting is actually natural in $G$ contrary to what is written in many texts. What she is referring to is the following:$\newcommand{\z}{\mathbb{Z}}$. Lets fix $G=\z/2$. Thm: There is no possibility of a natural transformation from the functor $H_*(-,\z/2)$ to $(H_{*}(-,\z))^* \oplus Ext(H_{*-1}...


0

Let $f : S^2 \to T^2$ be a continuous map and let $p : \mathbb{R}^2 \to T^2$ be the universal covering map. As $S^2$ is simply connected, $f$ lifts to a map $\hat{f} : S^2 \to \mathbb{R}^2$ such that $f = p\circ\hat{f}$. As $\mathbb{R}^2$ is contractible, both $p$ and $\hat{f}$ are homotopic to constant maps, and therefore $f$ is homotopic to a constant map. ...


-2

as a non-specialist, I want in this answer, which definitely bring me a lot of -1, transmitted you what I think. functor classifying it seems discovery to respond to an algebraic matter of discrete groups, indeed for topological spaces is well embedded into universal topological space (topological cone) and this topological cone is homotopically zero (...


1

The problem in question (a) is stated (by your professor I guess) in a terrible way... The given data of the problem is the G-sets $X$ and $Y$ and the points $x \in X$ and $y\in Y$ : it is then asked to find a property $\mathsf P$ depending on those data such that $\mathsf P$ is equivalent to the property "$\exists f \colon X \to Y$ morphism of $G$-sets such ...


1

You are correct. Because each point in a manifold has a neighborhood homeomorphic to some Euclidean space, any manifold is locally contractible, which implies that is it both locally path connected and locally simply connected. Therefore if we restrict our attention to connected manifolds (which we usually do), we see that all manifolds admit universal ...


2

Instead of choosing lines going through the removed points, consider $n-1$ lines which separate the points into $n$ chambers. Each such punctured chamber deformation retracts onto its boundary circle. (Push along the dotted arrows in the following picture.) Then you are left with the 1-skeleton of this complex, which is precisely a wedge of $n+1$ circles. (...


2

All the constructions that you used to define the isomorphism are natural/functorial: Given a map $X \to Y$, you have a natural map that respect inclusions, which gives a starting point for all the applications of naturality to come: $$(\Sigma X, C_+ X, C_- X, X \times \{0\}) \to (\Sigma Y, C_+ Y, C_- Y, Y \times \{0\});$$ The long exact sequence of a pair ...


2

A split epi in $\mathbf{Top}$ need not even be a fiber bundle at all: its fibers don't have to all be homeomorphic. For instance, consider the map $p:[0,1]\to[0,1]$ given by $p(x)=\min(2x,1)$. This is a split epi (it is split by $x\mapsto x/2$), but $p^{-1}(\{1\})=[1/2,1]$ is an entire interval while the other fibers of $p$ are all single points.


1

OUTLINE: Let $H:X\times[0,1]\to X$ with $H|_{X\times\{0\}}=\mathrm{id}_X$ and $H|_{X\times\{1\}}=c_{(1,0)}$ the constant map to $\{(1,0)\}$ be continuous. Consider the sequence of points $p_n=(1,\frac{1}{n})$. Now prove (or recall) the following facts: There is a characterization of continuity from elementary analysis in terms of convergent sequences in ...


1

A $\Omega$-spectrum is not necessarily a connective one. In your argument, there is no reason "the spaces are more and [more] connected". For a concrete example, consider the $K$-theory spectrum $KU$. In its usual incarnation (with $BU \times \mathbb{Z}$ even and $\Omega(BU \times \mathbb{Z})$ odd), it is already a $\Omega$-spectrum by Bott periodicity. ...


1

In general if $X$ is contractible then every continuos map $f:X\rightarrow Y$ is contractible. A brief explication: if you have $f_1 :X\rightarrow pt$ be the constant map and $f_2 :pt\rightarrow X$ such that $f_1 \circ f_2 \sim Id_{pt}$ and $f_2 \circ f_1 \sim Id_{X}$ then clearly : $f\sim f\circ Id_X\sim f\circ f_2 \circ f_1$ but $f_1$ is the constant map ...


2

Let me put the whole argument together; hopefully you can fill in the gaps. Let $f:\mathbb{RP}^2 \to S^1$ be any continuous map. It induces a map on fundamental groups (after suitable choices of basepoint, which I omit) $f_\star:\pi_1(\mathbb{RP}^2) \to \pi_1(S^1)$. But the only such homomorhism is the zero map. (For this, you need to know what each ...


2

Let $A = [0,1]$ (which is compact). Let $a$ and $b$ be any points NOT on the equator of $S^2 = \{(x,y,z)\in \mathbb{R}^3: x^2 + y^2 + z^2 = 1\}$. Let $f:A\rightarrow S^2$ be defined by $f(t) = (\cos(2\pi t), \sin(2\pi t),0)$. Then $f$ is not injective since $f(0) = f(1)$. Further, $f$ is nullhomotopic because $[0,1]$ is contractible. Finally, $S^2\...


0

Taking complement, it suffices to show that $U^c$ lies in the unbounded component of $C$. Since $C$ is a simple closed curve, there is a homeomorphism $g:S^1\to C$ from $S^1$ to $C$. Now we use the assumption that $U$ is simply connected to conclude that the map $g:S^1\to U$ is path-homotopic (with range in $U$) to a constant map, hence nulhomotopic. Since $...


0

I'm not sure why you say the multiplication by $(-1)^a$ procedure is not being followed. In the case of removing the vertex $v_1$ from $[v_0,v_1,v_3]$, $a$ would be $2$ (since $v_1$ is the second vertex), so you would end up with $[v_0,v_3]$. So it seems that this is indeed the rule your book is following. (Beware, however, that this is not the convention ...


1

There is an extensive literature which studies topological algebras from the perspective of universal algebra. Among other things, it focuses on the question of how algebraic structure restricts the nature of compatible topologies on the underlying space (e.g. -- a T_0 topological group must be completely regular). Walter Taylor has studied this extensively....


2

There are indeed books on topological groups, Lie groups, topological rings, topological groupoids, Lie groupoids, .... , Lie double groupoids, ... A result I like, and the proof is accessible to an advnced undergraduate, is the classification of closed subgroups of the additive group $\mathbb R^n$. Any such is isomorphic to a product of copies of $\...


2

Hint: Apply the Serre exact sequence since the fibre $F$ is discrete, $\pi_i(F)=1, i\geq 1$, thus you have isomorphism between $\pi_i(P)$ and $\pi_i(B), i\geq 2$ where $P$ is the total space and $B$ the base space. https://en.wikipedia.org/wiki/Fibration#Long_exact_sequence_in_homotopy_groups


2

Morphisms of $G$-actions are the same as morphisms of $G$-sets. To clarify this, I need to introduce the necessary definitions. By right $G$-action (on set $X$) we usually mean a function $\cdot\,\colon X\times G\to X$ such that $x\cdot(gh) = (x\cdot g)\cdot h$ and $x\cdot e_G = x$. Morphism between $G$-actions on sets $X$ and $Y$ are defined as functions $...


2

This is just a standard abuse of notation that you are surely familiar with in other contexts. A group is not just a set $G$, but actually a pair $(G,\cdot)$ where $\cdot$ is a binary operation on $G$, but we nevertheless frequently refer to "$G$" alone as the group. Similarly, a right $G$-set is really a set $X$ together with an action $X\times G\to X$, ...


1

This is just a notational error. It should say $$l'(\tau_0)+l'(\tau_2) = l'(\tau_0+\tau_2) = l'(\tau_0\ast\tau_2) = l'(\tau_1).$$ To show that $l'(\tau_0+\tau_2)=l'(\tau_0\ast\tau_2)$, you just directly use the definition of $l'$: $$l'(\tau_0+\tau_2)=[u(x)\ast\tau_0\ast u(y)^{-1}\ast u(y)\ast\tau_2\ast u(z)^{-1}]=[u(x)\ast\tau_0\ast\tau_2\ast u(z)^{-1}]=l'(\...


0

I think I figured out a way to prove that cupping two degree $n$ generators in $X_k$ gives $k$ times a degree $2n$ generator. Let's do it in detail for $k=2$. I think $k>2$ can be treated in a similar fashion. The map $f_1:S^{2n-1}\to S^n\vee S^n$ is the image of $1\in \Bbb Z$ under the composition $$\Bbb Z\stackrel{deg}{\longleftarrow}\pi_{2n}(S^{2n})\...


1

I try to answer my question. I am pleased if you guys make some corrections. A mobius band can be think of a square $[0,1]×[0,1]$ identified in both ends ${0}×[0,1]$ and ${1}×[0,1]$ by $(0,x)$~$(1,1-x)$. After seeing this picture, we can intuitively shrink the band from time to time to get a circle. So there is a deformation retract $f_t:M → M, t∈I$, $M$ ...


2

Whenever you have a subspace $A\subseteq X$ and a deformation retraction $H:X\times [0,1]\to X$ onto $A$ (such that $H(x,0)=x$ and $H(x,1)\in A$ for all $x\in X$ and $H(a,1)=a$ for all $a\in A$), you get a homotopy equivalence as follows. Let $f:X\to A$ be given by $f(x)=H(x,1)$ and $g:A\to X$ be the inclusion map. Then $f\circ g=Id_A$, and $H$ is a ...


0

I would really appreciate feedback on the following answer, particularly the second part. Part 1. $H^i(K(\Bbb Z,3);\Bbb Z)$ contains no $\Bbb Z$ summand and it contains no $\Bbb Z_{2^k}$ summand for $k>1$, in the range $i\in\{4,\cdots, 9\}$ The Bockstein chain complex has the following form $$\underbrace{H^3(K(\Bbb Z,3); \Bbb Z_2)}_{\Bbb Z_2\cdot \...


1

In answer to a question that came up in comments under Najib's answer, let me point out that the category of pseudotopological spaces is a locally small cartesian closed category that contains $\text{Top}$ as a full subcategory. Actually, this category $\text{PsTop}$ has the even stronger property of being complete and locally cartesian closed and indeed is ...


2

If $\;x\in I\cap J\;$ , then $\;G_x\;$ appears as free factor in both $\;G_a,\,G_2\;$ , so it won't be true that $\;G_r\cap G_s=1\;$ always, as one of the possible intersections will be $\;G_x\cap G_x=G_x\;$ . The basic assumption is that the free products considered here are non-trivial, meaning: $\;G_i\neq 1\;$ forall the groups considered. Yes, as ...


2

Here is a comment from the author of the book you mention. I thought it might be relevant, however too long for a comment. The argument I had in mind was induction on the number of cells of X, but not explicitly using a cofiber sequence. Suppose X is obtained from a subcomplex X' by attaching an n-cell. Given a map f : X ---> Y, induction implies that ...


1

In topological spaces, a cell complex is usually the same thing as a CW-complex. The name "cell complex" comes from the fact that there exists generalizations to other categories, but if you're interested in topological spaces then for all intents and purposes "cell complex" = "CW-complex". A finite cell complex is a cell complex that has a finite number of ...


0

The first problem is that the homomorphism $\pi_1(Y, y_0) \rightarrow \mathrm{Sym}(F)$ could be too small. For instance, it could be the trivial homomorphism in a setting where the action of $\pi_1(Y,y_0)$ on $F$ is not always the identity permutation. A minimal example of this is the (helical) double-cover of $S^1$. Here, $F$ is two points and both the ...



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