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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 $T$ ...


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


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 $H^4(S^2\times S^4)$. ...


6

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 $p\notin A$.


3

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}| = ...


3

You need one additional tool, which is the Hurewicz theorem. Let me write $B^n A$ for $K(A, n)$. Then the Hurewicz theorem tells you that $$A \cong \pi_n B^n A \cong H_n(B^n A, \mathbb{Z})$$ and then universal coefficients tells you that $$H^n(B^n A, C) \cong \text{Hom}(H_n(B^n A, \mathbb{Z}), C) \cong \text{Hom}(A, C)$$ as desired. (Note that all we ...


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

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.


3

Chain homotopies are an abstract way to define homotopy in the category $\mathsf{Ch}_\bullet$ of chain complexes. But indeed, there is a geometric interpretation of this. $f, g : X \to Y$ be two homotopic maps, with homotopy given by $F : X \times [0, 1] \to Y$. $f_{\#}, g_{\#}$ be the induced maps $C_\bullet(X) \to C_\bullet(Y)$ between the singular chain ...


2

In general, $[X, K(G, n)]$ is a group whenever $G$ is abelian. This comes from the homotopy commutative, homotopy associative $H$-space structure on $K(G, n)$ which has inverses modulo homotopy. This structure can be easily seen to come from the map $$K(G \times G, n) \simeq K(G, n) \times K(G, n) \to K(G, n)$$ induced from the group addition map $G \times ...


2

With field coefficients $\Bbbk$ and a finite type space $X$ (i.e. all the homology groups are finite-dimensional), the homology coalgebra $H_*(X)$ is dual to the cohomology algebra $H^*(X)$ equipped with the cup product, see e.g. this answer by Ben Webster on MO. I'll try to write down all the details. In the case of the sphere, if $1 \in H^0(S^m) \cong ...


2

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 ...


2

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) ...


2

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. ...


2

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 ...


2

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) = ...


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 ...


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.


1

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. ...


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

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 ...


1

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".


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 ...


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 ...


1

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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