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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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No, this is not true unless you restrict to quasicoherent sheaves. For instance, just as a topological space, $\mathbb{A}^1$ is homeomorphic to $\mathbb{P}^1$, and there are coherent sheaves on $\mathbb{P}^1$ with nontrivial $H^1$. In fact, there are even sheaves of $\mathcal{O}_{\mathbb{A}^n}$-modules on $\mathbb{A}^n$ which have nontrivial cohomology. ...

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Let $M$ be a connected non-compact manifold. Then $H_c^0(M)\cong 0$, and $H^0(M)\cong\mathbb{R}$. Let $f:M\rightarrow \mathbb{R}$ be a function with $df=0$. Then $f$ is constant (this uses the connectedness of $M$). If $f$ is assumed to be compactly supported, this constant must be zero. If $f$ is not assumed to be compactly supported, all constants occur. ...

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It helps to recall the formulation of the Kunneth theorem more fully. While the fact that $H_1(X \times Y) = H_1(X) \oplus H_1(Y)$ is true here, this is a particular fact for dimension $1$. The Kunneth isomorphism shows that $H^*(X \times Y) \simeq H^*(X) \otimes H^*(Y)$, with the isomorphism being one of graded rings. This means that $H^k(X \times Y) = ... 7 The generator of$H^2(S^2\times S^4)$is$\pi^*(\alpha)$where$\pi:S^2\times S^4\rightarrow S^2$is the projection on the first factor, and$\alpha$is a generator of$H^2(S^2)$. Then$\gamma:=\pi^*(\alpha)\smile\pi^*(\alpha)=\pi^*(\alpha\smile\alpha)$by functoriality. But$\alpha\smile\alpha=0$as$H^4(S^2)=0$, so$\gamma=0$in$H^4(S^2\times S^4)$. ... 0 The map is always trivial mod 2. Let me change notation a little bit. I'm going to write the coordinates of$S^{n+m}$as$(x_1,...,x_n, y_0,..y_m)$with involution given by negating the$y$coordinates. I'm going to view$S^{n+m}$as the$n$-fold suspension of$S^m$,$S^{n+m} = \Sigma^n S^m$, where the$S^m$is given by the$y$coordinates, and the ... 1 For arbitrary$i$the$i$-th factor$S^n/\mathbb Z_2$will be homeomorphic to the join$S^{i-1}\star\mathbb RP^{n-i}$, and action the covering map on the$n$-th homology will be actually zero because on the second component of the join the map is$2$-sheet. 1 Note that$E^k \setminus \{p\}$deformation retracts to$S^{k-1}$by$r : E^k \setminus \{p\} \to S^{k-1}, \; x \mapsto \frac x {\| x \|}$. Therefore,$\mathrm{id} \sqcup _f r : X \sqcup _f (E^k \setminus \{ p \}) = V \to X \sqcup _f S^{k-1} \simeq X$is a retract of$V$to$X$. To clarify why$X \sqcup _f S^{k-1} \simeq X$let us show that, in general, if ... 1 There is a more general notion of a mapping cone. That is, an adjunction space corresponding to$f:S^{k-1} \to X$is the same as a mapping cone$C_f$. For those you find plenty of answers which use the same strategy, hence you can build your intuition and see formulas by considering e.g. https://en.wikipedia.org/wiki/Mapping_cone_(topology), Homology of ... 0 The Klein bottle is the quotient of a square by the equivalence relation where you identify top and bottom, and left and right, but where one of those two identifications has a twist. If you look just at the boundary of the square, the identification gives you a wedge of two circles. So when you glue in the rest of the square you get an adjunction space ... 1 The key example is$f : S^n \to S^n$and you want to compute $$\mathbb{Z} \approx H_n(S^n;\mathbb{Z}) \xrightarrow{f_*} H_n(S^n;\mathbb{Z}) \approx \mathbb{Z}$$ This function is simply the homomorphism "multiply by$degree(f)$". And you can compute the degree very easily. For example: If$f$is an orientation preserving homeomorphism then$degree(f)=1$... 2 When taking the second boundary map, one can not remove the$ i $th term since it was already removed by the first boundary map. So$ j \neq i $in the sum, and so it is convenient to split up the sum as $$\sum_{\substack{ j \\ j \neq i } } \cdot = \sum_{\substack{ j \\ j < i } } \cdot + \sum_{\substack{ j \\ j > i } } \cdot$$ Now for the signs, in ... 0 For basic Mayer-Vietoris usage you need to thicken your link/knot a little in such way that you won't create new intersections in obtained in such way "link" of solid tori. Namely: let$K$be your link of$n$circles and$U_1 = K \times D^2_{3 \varepsilon}$, where$D_{3 \varepsilon}^2$is disk of "small" radius$3 \varepsilon\$ such that you can retract ...

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