Hot answers tagged

6

To elaborate on Dmitri Pavlov's comment, your questions have been answered in a recent paper by Yehonatan Sella. Sella give an example (Example 0.3) of a locally contractible space $X$ for which $\mathcal{S}^k(X) \xrightarrow{\mathrm{shf}_X} \tilde{\mathcal{S}}^k(X)$ is not surjective. Sella's space has $5$ points, and the argument that $\mathrm{shf}_X$ is ...


4

I think the function you are looking for is $f(x)=\int_{\gamma_x} \alpha$ where $\gamma_x$ is a path from some chosen base point $x_0$ to $x$. But then for this to be well defined you have to say that all paths between $x$ and $x_0$ are homotopic, i.e. that your sphere is simply connected, so the whole argument gets a bit circular... P.S. Traying to write ...


4

You used it when you wrote this equation: $$\frac{\ker \phi \oplus H}{\operatorname{im} \partial_1} = \frac{\ker \phi}{\operatorname{im} \partial_1} \oplus H$$ You're implicitly using the fact that $\operatorname{im} \partial_1 \subset \ker \phi \iff \phi \circ \partial_1 = 0$. Otherwise taking the quotient $\ker \phi / \operatorname{im} \partial_1$ doesn't ...


4

In terms of the logical process of inducing maps, you can think in the following way: A continuous map $f : X\rightarrow Y$ induces chain maps $f_\# : C_k(X) \rightarrow C_k(Y)$ (for each $k$) which induces maps on homology $f_* : H_k(X) \rightarrow H_k(Y)$. It's important to note that $f_*$ exists only because $f_\#$ sends homologous cycles in $X$ to ...


4

Consider the space $K$ given by the disjoint union of two circles. If $\phi$ is the map interchanging the two components (preserving the orientations), then we see that $K_{\phi}$ is a torus. However, $K\times I$ is the disjoint union of two tori, hence the Betti numbers are obviously different: $$(b_0(K_\phi),b_1(K_\phi),\ldots)=(1,2,1,0,\ldots)$$ ...


4

If $f:M\to M$ is a diffeo, then there is a linear map $f^p:H^pM\to H^pM$ and then we can consider the spaces $$H^p(M)_f=H^p(M)/(1-f^p)H^p(M)$$ and $$H^p(M)^f=\ker(1-f^p).$$ One can then show that there are exact sequences $$0\to H^{p-1}(M)_f\to H^p(S_fM)\to H^p(M)^f\to 0$$ with $SM$ the mapping torus of $f$. One speedy way of getting this, by the way, is to ...


4

Let me try to outline the proof, and you can fill in the details. We take an $n$-dimensional smooth quadric hypersurface $Q$ sitting in $\mathbf P^{n+1}$. By the Lefschetz hyperplane theorem, for $k \neq n$, the restriction maps $$H^k(\mathbf P^n, \mathbf Z) \rightarrow H^k(Q,\mathbf Z)$$ are isomorphisms, and so the cohomology groups of $Q$ in these ...


3

I'm going to use singular cohomology. Recall that if $C_i(M)$ denotes the collection of singular $i$-chains, then for an abelian group $G$, $C^i(M; G) = \operatorname{Hom}(C_i(M), G)$ is the collection of singular $i$-cochains with values in $G$. Note, if $H$ is a subgroup of $G$, then $C^i(M; H) \subseteq C^i(M; G)$. There is a coboundary operator ...


3

Note that the restriction of $p$ to $S\setminus\delta S$ is a homeomorphism onto $X$. So $X\cap Y$, being a subset of $X$, is homeomorphic to a subset of the open unit square. Which subset is that? Well, $p^{-1}(Y)$ is just a small neighborhood of $\delta S$, so $Y\cap X$ will look like a small open ring along the edge of the square, but not actually ...


3

First, let us note that a map induces isomorphisms on cohomology with coefficients in all rings iff it induces isomorphisms on cohomology with coefficients in all abelian groups. Indeed, if $G$ is any abelian group, consider the ring $R=\mathbb{Z}\oplus G$, with multiplication defined by $(m,g)\cdot(n,h)=(mn,mh+ng)$. Then there is an isomorphism ...


3

$\require{AMScd}$I'd rather think about the negation of your first condition. So call a space bad if there exists a decomposition $A \cup B$ into connected open sets with disconnected intersection. Call a space semi-bad if there exists a decomposition into open sets $A \cup B$ such that the induced map on relative homology $\tilde H_0(A \cap B) \to \tilde ...


2

Your De Rham cohomology algebra is $H^*(\mathbb P^3_\mathbb C)=\mathbb R[w]/(w^4)$, where $w=[\omega]$ is the class of some closed $2$-form $\omega$. We then have $0\neq w^2=[\omega \wedge \omega]$, so that $\omega\wedge \omega$ is not exact and a fortiori $\neq 0$.


2

Since $\alpha = (1,0)$ and $\beta = (0,1)$ generate $\Bbb{Z}^2$, let's see where your operator takes them. Let $T = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$. Then $T\alpha = (1,3)$ and $T\beta = (2,4)$. Then for arbitrary, $a,b \in \Bbb{Z}$, $T(a \alpha + b \beta) = (a+2b, 3a+4b)$. As $a,b$ independently range through all of the integers, you ...


2

The standard answer is to observe $S^1$ is an Eilenberg-MacLane spaces, specifically that $S^1$ is a model for $K(\Bbb Z, 1)$. (This relies on showing that all of the higher homotopy groups of $S^1$ vanish.) In general, there is a natural isomorphism $H^n(X;G) \cong [X, K(G, n)]$. This is referred to as the representability of cohomology by ...


2

$H_n(M,M-x;R)= H_n(M,M-x)\otimes R$. For each $r\in R$ determines a covering space $M_r$ of $M$ consisting of points $\pm \mu_x\otimes r \in H_n(M,M-x;R)$ where $\mu_x\in H_n(M,M-x)$ is a generator. If $r=-r$ i.e order $2$, then $M_r$ is just a copy of $M$, otherwise it is isomorphic with the oriented $2-$sheeted cover $\bar{M}$. So observe that a ...


2

Let $M$ be an orientable (connected, closed) $n$-manifold. Then classically $H_n(M;F) \cong F$. But moreover, $H_{n-1}(M;\mathbb{Z})$ is torsion-free. This is Corollary 3.28 in Hatcher's book. The proof basically goes like this: if $H_{n-1}(M;\mathbb{Z})$ weren't torsion free, by the UCT then $H_n(M; \mathbb{F}_p)$ would be bigger than just $\mathbb{F}_p = ...


2

There is a homotopy $H_t: C_r \to \Bbb R^2 - 0$ given by $H_t(z) = f(tz)$. At time $1$ this is just $f(z)$ and at time $0$ it is the constant map $f(0)$. Note that $f(tz) \neq 0$ by our assumption on $f(z)$. Therefore $f$ induces the same homomorphism on homology as the constant map does, and the latter clearly induces the zero homomorphism. Edit: When I ...


2

Write $S^n$ as the union of two open sets $U= S^n-N$, $V=S^n-S $where $N,S$ are the north and south pole. Note that the intersection $U\cap V$ is connected if $n\geq 2$. $U,V$ are contracile (diffeomorphic to $R^,$, and on each of these sets the form $\alpha$ is exact and andmits primitive $f_U, f_V$. On the intersection $d(f_U-f_V)=0$ and by connexity ...


2

It is hard to know what will give you an intuition without knowing what you want and what background you already have. If you are asking roughly what kind of geometric or topological information Cech cohomology reveals about a space, then answer is the same as for all cohomology theories: it reveals the global connectivity of spaces with relatively simple ...


2

I'm confused. $H_1(\mathbb{R}P^3)=\mathbb{Z}/{2\mathbb{Z}}$, but is orientable (on edit, this answers your question 2). EDIT: Maybe it refers to the fact that $H^{m-1}(M)$ can only be torsion if the manifold $M$ is non-orientable. It is Corollary 3.28 in Hatcher.


2

All of the homology groups of $\mathbb{S}^n$ are trivial, except of top and bottom one. The induced map $H_0(\mathbb{S}^n) \to H_0(\mathbb{R}\mathbb{P}^n)$ will always be isomorphism (this is very easy to calulate). The top homology group $H_n(\mathbb{R}\mathbb{P}^n)$ is either trivial or $\mathbb{Z}$, depending on whether $\mathbb{R}\mathbb{P}^n$ is ...


2

No, having a CW complex structure is absolutely not enough to be a manifold. Being a CW complex is very easy, being a manifold is hard. Take the wedge sum of two circles $S^1 \vee S^1$ for example: it's a CW complex, but not a manifold. If a space $M$ is a compact manifold without boundary, then you can read its dimension using singular homology: it is the ...


2

In applying the Mayer-Vietoris sequence, not only do you need to know how to evaluate homology groups of spaces, but you also need to evaluate induced homology homomorphisms of continuous maps. In this example, you have correctly evaluated the homology groups $H_1(K) \approx \mathbb{Z}$ and $H_1(K_0) \approx \mathbb{Z} \oplus \mathbb{Z}$. The additional ...


1

Let $\pi \colon M \to S^1$ denote the usual projection and let $C$ denote a non-zero homologous simple closed curve with no self intersections, i.e. an injective map $c$ from $S^1$ to $M$, which induces something non-zero on homology. We can isotop $c$ s.th. $\pi \circ c$ is a covering of $S^1$ (This is not that hard, you just make the map locally ...


1

As OP correctly remarked in the comments, my answer was wrong! And his proof was right (I guess). Well done, there's just a trivial mistake you've made: $H_2(Y)=0$ would imply that the map $H_2(Y)\to H_1(A\cap B)$ is the zero map. On the other hand, the image of this map is the kernel of the map $\phi$, which you correctly described; the kernel of this map ...


1

Recall that the $H_0$ of a path-connected space $X$ is isomorphic to $\mathbb{Z}$ via the augmentation map, which counts the sum of the coefficients of a given (class of a) $0$-chain, i.e., a formal sum of points of $X$. Therefore the generator $1 \in \mathbb{Z}$ corresponds to the class of any point $x \in X$; alternatively, if you pick $x,y \in X$, then ...


1

Why not directly prove it is isomorphic to the projective line? For example, changing variables to $(x_0+ix_1, x_0-ix_1,ix_2)$, this curve $M$ is same as the one given by $x_0x_1-x_2^2=0$ and you have a map $\mathbb{P}^1\to M$, by $(u,v)\mapsto (u^2, v^2, uv)$, which you can easily see to be an isomorphism.


1

This is the fundamental lemma in homological algebra (see here theorem $2.22$ page $36$ and the following corollary.) If you take the identity the thesis of the theorem says in particular that you obtain a chain homotopy


1

From Universal Coefficient Theorem we have the following natural s.e.s. for homology we have $$ 0 \to H_1(X;\mathbb{Z})\otimes \mathbb{Z}_2 \to H_1(X;\mathbb{Z}_2) \to \text{Tor}(H_0(X;\mathbb{Z});\mathbb{Z}_2)\to 0$$ since $H_0(X;\mathbb{Z})$ is $\mathbb{Z}$ the s.e.s becomes $$ 0 \to H_1(X;\mathbb{Z})\otimes \mathbb{Z}_2 \cong H_1(X;\mathbb{Z}_2) \to 0 $$ ...


1

It is enough to show that $H_n(g):H_n(\Bbb RP^n)\to H_n(\Bbb RP^n)$ is multiplication by an odd integer because $2\deg(f)=\deg(p\circ f)=\deg(g\circ p)=2\deg(g)$. By the UCT, this is equivalent to showing that $$H_n(g;\Bbb Z/2):H_n(\Bbb RP^n;\Bbb Z/2)\to H_n(\Bbb RP^n; \Bbb Z/2)$$ is an isomorphism, which in turn is equivalent to the analogous statement for ...



Only top voted, non community-wiki answers of a minimum length are eligible