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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 $$\delta:C^p(\mathfrak{U},\Omega^k)\rightarrow C^{p+1}(\mathfrak{U},\Omega^k)$$ ...

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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 ...

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

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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 ...

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(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 ...

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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 ...

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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 ...

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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 ...

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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 ...

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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 ...

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