Tag Info

New answers tagged

0

If we denote inclusions $$I_{n_0 \dots n_p}^{n_i}:U_{n_0 \dots n_p}\hookrightarrow U_{n_0\dots \widehat{n_i} \dots n_p}\hspace{5pt}\text{and}\hspace{5pt}I_{n_0 \dots n_p}^{n_in_j}:U_{n_0 \dots n_p}\hookrightarrow U_{n_0 \dots \widehat{n_i} \dots \widehat{n_j} \dots n_p},$$ then $$\delta:C^p(\mathfrak{U},\Omega^k)\rightarrow C^{p+1}(\mathfrak{U},\Omega^k)$$ ...


3

General idea: Take the set $U=U_{\alpha_0\ldots\alpha_p}$. By performing $\delta$ twice on some form on $U$, you will get the set $V=U_{\alpha_0\ldots\alpha_p\beta_0\beta_1}$ twice, and the two forms on $V$ will cancel each other out. This is due to the $(-1)^p$ in the definition of $\delta$. To get the right feel, you may calculate $\delta$ explicitly for ...


3

Yes. $S^1$ is path connected, so every element of $S^1$ acts by a map homotopic to the identity.


3

Here's another (probably unnecessarily elaborate) proof that they are not diffeomorphic using Eliashberg's amazing theorem about which open $2$-handlebodies admit Stein structures and the adjunction inequality for Stein surfaces. I am sorry if this is not really accessible for the asker. $M$ is the 4-manifold given by gluing an open $2$-handle along the ...


6

(Let me answer the case of coefficients in $\mathbf Z$ (or $\mathbf Q$ or $\mathbf R$ or $\mathbf C$.) I think it should only take a bit of universal coefficient wizardry to deduce the answer in general, but it is too late here for that.) First, by the Lefschetz hyperplane theorem and Poincaré duality, we have an isomorphism $$ H_i(\mathbf P^n, \mathbf Z) ...


5

By the Lefschetz Hyperplane theorem, the inclusion $i : X \hookrightarrow \mathbb{CP}^n$ induces a map $i^* : H^q(\mathbb{CP}^n, \mathbb{Z}) \to H^q(X, \mathbb{Z})$ which is an isomorphism for $q \leq n - 2$ and injective for $q = n - 1$. Recall that $$b_q(\mathbb{CP}^n) = \dim H^q(\mathbb{CP}^n, \mathbb{Z}) = \begin{cases} 1 & q\ \text{even}, 0 \leq q ...


1

Your edit is correct. For abelian $G$, $H^1(X;G) = \text{Hom}(H_1(X);G) = \text{Hom}(\pi_1(X);G)$ because any map to an abelian group factors through the abelianization. This is frequently a good way of thinking about $H^1$. (For instance, if you care about characteristic classes: every real vector bundle $E$ over $M$ determines a cohomology class $w_1(E) ...


0

$P_1 : X \to X/H$ and $P_2: X/H \to X/G$ are covering maps such that $P_2oP_1$ is a covering map. Now $P_2$ is normal iff $P_{2*} (\pi_1(X/H))$ is a normal subgroup of $\pi_1(X/G)$. now $H= \pi_1(X/H) / P_{1*}(\pi_1(X)) \equiv P_{2*}(\pi_1(X/H)) / P_{2*}P_{1*}(\pi_1(X))$ which is a subgroup of $G= \pi_1(X/G) / P_{2*}P_{1*}(\pi_1(X))$ . Now $H$ is normal ...


0

This answer is beyond the scope or level of the question, but it may be of interest to some to note that there is a notion which reflects the geometry of the $3$-cube and $3$-torus better then the usual chain complex. This uses the crossed complex $\Pi X_*$ defined for a filtered space $X_*$, and in particular for a CW-complex $X$ with its skeletal ...


22

Such functions are called means. The earliest paper on the subject that I’ve seen is G. Aumann, Über Räume mit Mittelbildungen, Mathematische Annalen (1943), Vol. 19, 210-215. He shows inter alia that no $S^k$ has a mean; that the only $2$-dimensional manifold with a mean is the open disk; and that if $X$ has a mean, then so does every retract and every ...


19

Edit: This post has been edited to take Eric Wofsey's comments into account. The end result is the following theorem: Suppose $X$ is a closed manifold of positive dimension. Then $X$ does NOT admit an averaging function. If an averaging $f:X\times X\rightarrow X$ exists, it induces a map $f_\ast: \pi_k(X)\oplus \pi_k(X)\rightarrow \pi_k(X)$. ...


1

You've done half of the homotopy with the cone upside-down. Intuitively, the homotopy equivalence should shrink the cone to its vertex, so if your equivalence relation $\sim$ identifies $(x_1,s_1)$ with $(x_2,s_2)$ when $s_1=s_2=1$, then your homotopy should be between $\mathrm{id}_{\mathrm{Con}(X)}$ and $c_{[(x,1)]}$, not between ...


28

Here is a proof that no sphere (of dimension $>0$) admits an averaging function. Suppose there is an averaging function $f:S^n\times S^n\to S^n$. Let $T(x_0,x_1,x_2,\dots,x_n)=(-x_0,-x_1,x_2,\dots,x_n)$; then $T:S^n\to S^n$ is homotopic to the identity and satisfies $T(T(x))=x$. Now consider the map $g:S^n\to S^n$ given by $g(x)=f(x,T(x))$. Since $T$ ...


5

They are equivalent. Start with the left unknot. It has two upper strands: a left one and a right one. Grasp the right upper strand, and pull it towards the left, over the left upper strand. You will obtain a copy of the right unknot.


2

First of all, let me falsify $[\mathbb{CP}^1]+[\mathbb{CP}^1]=[\mathbb{CP}^1\#\mathbb{CP}^1]=[\mathbb{CP}^1]$ by a sledgehammer argument: The complex bordism ring $\Omega_*^U\cong \mathbb{Z}[x_{2i}|i\in\mathbb{N}]$ is an integral polynomial ring with one generator in each even dimension, so in particular has no torsion and $[X]+[X]=[X]$ can only hold for the ...


5

You can prove that they're not homeomorphic by employing the same strategy as in the answer here. If they were homeomorphic, so would be their one-point compactifications, and hence they would have the same cohomology. $TS^2$ has no embedded spheres with self-intersection number 1; the canonical line bundle does (the zero section!) Another way of phrasing ...


3

The second stiefel whitney class of your bundle $TS^2\oplus l$ is non-zero. For the chern classes you can see $$c(TS^2\oplus l)=(1+2a)(1-a)=1+a$$ where $a$ is the generator of $H^2(S^2;\mathbb{Z})$ (here I see $l$ with the complex structure still). Taking mod 2 reduction shows that the second stiefel whitney class is non-zero.


4

The NE and SW blocks both contain triangles with vertices 1,3,4. But these intersect in just the edge 13 and the vertex 4, while the intersection of two simplices in a triangulation must be a "face" of each (which might be the empty simplex), not a union of two or more simplices.


0

Imagine having two nails in 3d, positioned perpendicular to the face of the wall (i.e., nails are horizontal), but with the two nails in perpendicular directions to each other and one nail above the other nail. It should be much easier to come up with a solution for how to wrap the string in this visualization so that the string falls when you remove either ...


2

Here on this picture some simplexes are identified. When you give names to different vertices, it will look as follows


2

The component $D^2$ is contractible, so $H_*(D^2\times S^1)=H_*(S^1)$, because homology are homotopy invariant. It is well-known that $H_1(S^1)=\mathbb Z$.


1

If $\sigma$ is a two-dimensional simplex in $\mathbb R^2$, then $\sigma$ is a compact connected subset of $\mathbb R^2$, so a subset $S$ of $\sigma$ is open in the induced Euclidean topology if and only if for point $p$ in $\sigma$, you can find an open ball $B$ with center $p$ such that $B\cap S\subset \sigma\cap S$. Any nonempty proper face $\tau$ of ...


1

Let $\hat f: \Bbb R^n \to Y$ be the extension of $f$ and $i: X \to \Bbb R^n$ be the inclusion. By functorality of $\pi_1$, $f_*=(\hat f \circ i)_*=\hat f_* \circ i_*$. As $\Bbb R^n$ is contractible, both $\hat f_*$ and $i_*$ are zero (they are maps from and to the trivial group respectively), and therefore $f_*=0$.


2

If $f: X \subseteq \mathbb{R}^n \to Y$ is continuous it induces a homomorphism $f_*: \pi_1(X) \to \pi_1(Y)$. However if $F: \mathbb{R}^n \to Y$ extends $f$ then we have that $F_*: \{1\} \to \pi_1(Y)$ is certainly trivial because $\mathbb{R}^n$ is contractible. But this gives a nullhomotopy of $f(\gamma)$ for every $\gamma: I \to X$ where $H: I \times I \to ...


1

You need an understanding of the attaching map of the 3-cell in order to compute the differential $\mathbb{Z} \mapsto \mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z}$. Think of $T^3$ as the quotient of the cube $[-1,+1]^3$ by identifying $(x,y,-1) \sim (x,y,+1)$, $(x,-1,z)\sim(x,+1,z)$, and $(-1,y,z) \sim (+1,y,z)$. Let $q : [-1,+1]^3 \to T^3$ be the ...


3

The answer is no in general, because if $X$ is any nonempty set with the indiscrete topology (only $\emptyset$ and itself are open) then $\mathcal{O}_X \cong \mathbb{C}$, while two indiscrete spaces of different cardinality are not homeomorphic. When you restrict to compact Hausdorff spaces, this becomes true. (See A theorem due to Gelfand and Kolmogorov ...


0

This proof is indeed wrong. For a counterexample, consider the case where $\tilde X_1 = X_1 = \mathbb{R}$ and $\tilde X_2 = X_2 = \mathbb{R}$ and $p_1,p_2,p$ are all simply identity maps. In that case, you would be saying that for any open subset $U \subset \mathbb{R} \times \mathbb{R}$, letting $V = p^{-1}(U)=U$, the set $V$ can be written as a product $V ...


3

For $A\subset X$ you have long exact sequence $$ \dots\to H_n(A)\to H_n(X)\to H_n(X,A)\to\dots $$ The composition $r_*\circ i_*:H_n(A)\to H_n(X)\to H_n(A)$ is identity, so we see that $i_*:H_n(A)\to H_n(X)$ is inclusion for all $n$. Thus, we can write $$ 0\to H_n(A)\to H_n(X)\to H_n(X,A)\to0, $$ and $r_*$ gives us splitting of this short sequence.


1

Sam Nead has the correct suggestion here. In most instances of Mayer Vietoris you actually need to compute what the inclusion map induces or even, god forbid, what the snake homomorphism gives you. Here the boundary of a mobius strip includes into each mobius strip in such a way that it retracts onto the circle going around of the center of the mobius band. ...


0

You have to compute the homomorphisms $H_*(\partial M) \to H_*(M)$, induced by inclusion, where $M$ is the Mobius band. How to do this depends on what you know/understand about homology. Do you know what a deformation retraction does to homology groups? If you know about cellular homology, you can compute $C^{\rm cell}_*(\partial M) \to C^{\rm ...


0

I found a method in Dan Freed's notes on bordism. It's very, very slick, and I'll sketch it below for completeness. The main difficulty with the direct calculation is the use of "formal Chern roots" for the tangent bundle $T\mathbb{C}P^n$. If $T\mathbb{C}P^n$ were a sum of line bundles, then there would be no need to symmetrize the product and the ...


3

The contraction has to take place within $S^2$. As Hans Engler says in the comments, the continuous deformation is technically a continuous map $f:[0,1]\times S^2\to S^2$, specifically, one such that $f(0,x)=x$ for all $x\in S^2$, and there is a $p\in S^2$ such that $f(1,x)=p$ for all $x\in S^2$. For each $\alpha\in[0,1]$ define $f_\alpha:S^2\to S^2:x\mapsto ...


2

John has a good question. There is a tendency in algebraic topology to confuse a topological space and a topological space with a base point. Grothendieck wrote to me in 1983 in part: " both the choice of a base point, and the 0-connectedness assumption, however innocuous they may seem at first sight, seem to me of a very essential nature. To make an ...


0

If you have a continuous function between topological spaces (with base points) $$ f: (X, x_0) \to (Y, y_0) $$ then you have the induced function between the fundamental groups: $$ f_* : \pi_1(X, x_0) \to \pi_1(Y, y_0). $$ This function is defined by $$ f_*([\lambda]) = [f\circ \lambda]. $$ The $\circ$ is composition of functions. Recall that the ...


0

While I do not know whether it will help you much, you may view a dictionary between topological concepts and their equivalent algebraic concepts on pages 6 and 13 of "Very Basic Non-commutative Geometry" by Masoud Khalkhali.


1

$\newcommand{\Ch}{ C_{\tilde h}} \newcommand{\CP}{\mathbb{CP}^2}$ We will make use of natural inclusions $\alpha_i\colon S^2\hookrightarrow C_{\tilde h}$ and natural projections $\pi_j\colon \Ch\to \CP$. The composition $\pi_j\circ \alpha_j$ is the standard inclusion of $S^2$ in $\mathbb{CP}^2$ and the compositions $\pi_j\circ\alpha_i$ are the constant map ...


0

I don't want to do the entire question for you because I think it's an instructive exercise which is worth putting in the effort in order to build your intuition for similar questions. What I'll do is a calculation for a special case. Let $n=3$, $k,r=1$. I'll do part b. (which is actually the easier case) our space is $Y_b=S^3 - (S^1\vee S^1)$. b. To make ...


0

I'm not sure, but in the case of a double branched cover $W$ I was able to show the rational homology is of the form $$H_n(W;\mathbb Q)\cong\begin{cases}0&n\geq4\\ \mathbb Q^a&n=3\\ \mathbb Q^{2a}&n=2\\ \mathbb Q^{a}&n=1\\ \mathbb Q&n=0 \end{cases}$$ for some $a$. $W$ is constructed by gluing two homeomorphic $4$-manifolds $A$ and $B$ ...


2

As @BrianM.Scott noted, every space $X$ has a finite open cover, namely $\{X\}$. So this tells us nothing about $X$. One way to motivate the definition is to see that if $X = \mathbb{R}^d$ with the standard topology, it is satisfied by precisely the subsets which are closed and bounded. To illustrate this, I'll show this equivalence in one direction: if ...


0

It's not enough to demand that a space has a finite cover by open sets - every space has a finite cover by open sets, namely $\{$the whole space$\}$. This is a consequence of the fact that the definition of a topological space requires the whole space to be open. The idea behind compactness is that the open covers of the space are, in a sense, the only ...


0

Wouldn't it be sufficient to simply say that X has a finite, open cover? No. $\Bbb R$ has the finite open cover $\{(-\infty,1),(0,,\infty)\}$ but it is not compact.


2

A smooth manifold whose tangent bundle is trivializable is called parallelizable or frameable. On such a manifold, not only does the Euler characteristic vanish, but all characteristic classes vanish. Hence the nonvanishing of any characteristic class prevents a manifold from being parallelizable. In particular, the Klein bottle is non-orientable, so its ...


2

There are three great circles - if we write $$S^2=\{(x,y,z) : x^2+y^2+z^2=1\} $$ the great circles are when $x=0$, $y=0$, or $z=0$. Given any point $(x,y,z)\in S^2$ we must have at least one non-zero coordinate - if $x=y=z=0$ then the point is not in $S^2$. Then you can apply your covering map using the great circle for one of your non-zero entries. For ...


2

There is no general way of finding a CW-decomposition of an arbitrary space. In this specific example, note that $S^2$ with the prescribed identifications is simply $S^2$. Then in order to obtain our space $X$ from that we must glue in a 3-cell. Try to convince yourself that the attaching map of the $3$-cell is precisely the suspension of the degree $2$ map ...


2

Here's a hint: first find a CW decomposition of $S^2$ which is invariant under the $180^\circ$ rotation.


0

According to this text, branced covering $p:Y\to X$ is called irregular, iff the covering $(Y/\textrm{Aut}\,p)\to X$ is not homeomorphism. Here $\textrm{Aut}\,p$ is the group of all homeomorphisms $Y\to Y$ over $X$. In other words, automorphisms group acts non-transitive on the fiber of $p$.


1

This sheafification does not exist in general, at least if you want the target category to be the category of compact spaces. To construct the sheafification of a sheaf, you need the existence of certain limits and colimits, and the category of compact spaces does not have all limits and colimits (in particular, it does not have all equalizers and does not ...


1

Suppose $X$ is the open ball in $\mathbb{R}^n$. Denote by $\mathcal{C}$ (for compactification) the sheaf you defined, and $\mathcal{C}^s$ its sheafification. For a point $x \in X$ and balls $U_n$ centered at $x$ of radius $1/n$ we have the stalk of $C^s$ at $x$ is given by $\mathcal{C}^s_x = colim_{n \rightarrow \infty} \mathcal{C}(U_n)$. The colimit on ...


2

This is only a partial answer, but it's too long for a comment, so I'll post this as an answer and hope that nobody complains. For simplicity, let's call your presheaf $\mathcal{F}$. First, let's look at the stalk: The stalk $\mathcal{F}_x$ is just going to be the direct limit (with respect to the restriction you defined) of the spaces $U^*$ with $U$ an ...


2

Whether these chain complexes are literally equal depends on the precise set-theoretic definitions you have chosen for all the notation involved. For instance, a common definition of ${\bigoplus}_{i \in I} C^{sing}_n(X_i;R)$ is the set of all functions $f$ with domain $I$ such that $f(i)\in C^{sing}_n(X_i;R)$ for each $i\in I$ and $f(i)$ is the zero element ...



Top 50 recent answers are included