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


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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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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) ...


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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 ...


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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 ...


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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 $f_\ast: \pi_k(X)\oplus \pi_k(X)\rightarrow \pi_k(X)$. ...


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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 $T$ ...


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We have that $X$ is a smooth projective algebraic variety and $X$ is also good reduction. Therefore, we can apply the Weil conjectures and have that$$\text{dim}_\mathbb{Q} H^m(X, \mathbb{Q}) = \deg(P_mn)),$$where$$Z(X/\mathbb{F}_q, n) = {{P_1(n)P_3(n) \dots}\over{P_0(n)P_2(n)\dots}}$$with$$P_m(n) = \prod_{j=1}^{b_m} (1 - \alpha_{mj}n),\text{ }|\alpha_{mj}| = ...


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This problem in Weibel actually appears to be a mistake. If $C$ and $H_*(C)$ are homotopy equivalent, $C$ is split. I was stuck on the same thing for a while, but resolved it by asking this question. See the question itself for a proof, and the accepted answer for a link to Weibel's website where the mistake is in a list of errata.


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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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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.


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As Sempliner says, it's a good idea to think of this question in De Rham language. Oriented area is just the differential $2$-form $dx\wedge dy$. Namely, it is given by$$S\mapsto\int_Sdx\wedge dy,$$where $S$ is a (regular enough) $2$-simplex. Yes, this $2$-form is closed and hence exact. As always, there are many primitives. For example, we have$$dx\wedge ...


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$\newcommand{\Ch}{ C_{\tilde h}} \newcommand{\CP}{\mathbb{CP}^2}$ 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 ...


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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 ...


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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 ...


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


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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 ...


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First, there's a mistake in in one of your formulas: $S_n(\amalg_{i\in I} X_i)\cong\bigoplus_{i\in I} S_n(X_i)$ and not disjoint union. Also, the two chain complexes you are interested in can't be exactly equal since the left hand side consists of linear combinations of chains, whereas the right hand side consists of $I$-tuplets of linear combinations of ...


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Okay, it actually is pretty simple. We want to argue that the map $H_k(A)\to H_k(S^n)$ is trivial. This follows because it factors through the contractible space $S^n\setminus\{p\}$ for any point $p\notin A$.


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It's important to keep in mind that there exist oriented and non-oriented grassmannians (depending on you have fixed orientation of subspace or not). For oriented grassmannian $\widetilde G(2,4)$ we can consider $S^1$-fibration $V(2,4)\to\widetilde G(2,4)$, where $V(2,4)$ is a Stiefel manifold. As it easy to see, $V(2,4)\simeq T_0S^3$ (unit tangent vectors ...


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Now I have some kind of answer. At first, note that short exact sequence $$0\to H^n(X,\mathbb Z)\otimes R\to H^n(X,R)\to \mathrm{Tor}(H^{n+1}(X,\mathbb Z), R)\to0$$ occurs for finitely-generated (as $\mathbb Z$-module) ring $R$, because of equality $C^*(X,R)=C^*(X,\mathbb Z)\otimes R$. Suggestions about being $X$ a compact or a $CW$-complex are ...


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Actually you cannot construct such $CW$-complex. Consider the map $d_k:C_k(X,R)\to C_{k-1}(X,R)$, the basis element ($k$-cell) $\alpha$ maps to $\sum [\alpha:\beta_i]\cdot\beta_i$, this is sum of $k-1$-cells with integer coefficients. So, for example, if $R=\mathbb Z[t]$, you cannot get $H_n(X,R)=\mathbb Z[t]/(t)$.


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This is a duplicate of this question, the answer to which is given in a comment. See also this website, which was the first thing I found when I searched for "cohomology of projective space".


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I think, $H^*(\mathbb RP^{2n+1})=\mathbb Z[\alpha,\beta]/(2\alpha,\alpha^{n+1},\alpha\beta,\beta^2)$, where $\alpha$ has degree $2$ and $\beta$ has degree $2n+1$. You can see it when you write down spectral sequence for $S^1$-fibration $\mathbb RP^{2n+1}\to\mathbb CP^n$ and look on multiplicative structure (as you doing for calculating ring $H^*(\mathbb ...


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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) = ...


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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 ...


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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.


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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 ...


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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 ...


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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 ...


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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 $degree(f)=1$ ...


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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 ...


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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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