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5

It is nice to see so many elaborate examples, but here is a trivial one... Let $X$ be a space with one point. There is no continuous function $f:X\to X$ such that the map $f:H_\bullet(X)\to H_\bullet(X)$ is multiplication by $2$. Later. The above example is for unreduced homology homology, but a similar sort of example (in that the obstruction comes simply ...


1

It's $H_1(G, \mathbb{Z})$ that's the abelianization of $G$. $H_1(G, \mathbb{R})$ is the tensor product of the abelianization of $G$ with $\mathbb{R}$, which is in particular a real vector space. It is much smaller than what you said.


4

Here are some aspherical examples. Suppose that $M$ is a closed hyperbolic (real or complex) manifold with 1-st Betti number $>1$ (such exist in abundance); I will assume that the real dimension of $M$ is $>2$. Then the group of outer automorphisms of $\pi_1(M)$ is finite (by the Mostow Rigidity Theorem), while the group of automorphisms of the 1st ...


5

No. There is an obstruction coming from the cup product, and so it will be easier to talk about cohomology. A simple family of examples when $X = Y$ is to take $X = Y = \mathbb{CP}^n$ for $n \ge 2$. As a graded ring with the cup product, $$H^{\bullet}(\mathbb{CP}^n) \cong \mathbb{Z}[x]/x^{n+1}$$ where $x$ is the generator of $H^2(\mathbb{CP}^n)$, living ...


0

The proof you give above for the non-triviality can be extended quite elegantly to show what you want. Use the same triangulation, and try to find a cycle of $2$-simplices. Let $s_1,\ldots,s_n,$ denote the simplices, and suppose$$d(\sum a_is_i)=0.$$If $s_i$ and $s_j$ share a side, we must have $a_i=a_j$. If the surface is connected, it follows that all ...


1

You wrote already everything down precisely then! You know that a homomorphism between free abelian groups can be represented by a matrix. Write it down and you will see the kernel (i.e. $H_2$) which is $\langle (1,\cdots,1) \rangle$.


4

A big difference between topological spaces and varieties is that, for topological spaces, we can define $H^{\ast}(X, \mathbb{Z})$ where as, for varieties, there is no cohomology theory with coefficients in $\mathbb{Z}$. (I've blogged about this here and here; the examples I'm giving are originally due to Serre.) This is important because, in the topological ...


1

Ok, I'll elaborate on fixedp's answer which was sort of cryptic for me. You only have to observe the general fact that if there is only one $E_\infty^{p,q}$ not zero for $p+q=n$ given, then that one gives $H^{p+q}(X)$. This follows from the definition of convergence. This is a bit more tricky. I don't know how Davis & Kirk might have concluded the ...


0

The reason is each simplex has a defined order given by the order inherited by the sub-simplices: A simplex always looks the same. In case of the projective plane in the diagram, they are just arranged in a convenient way, such that the gluing map of the boundaries exactly is the one needed to result in the desired space. In case of $\Delta^2$-simplices ...


1

There is a nice and, to my opinion, more natural way to motivate cohomology - a geometric one, rather than an analytical one. Please read carefully the following question and answer in math.stackexchange: Intuitive Approach to de Rham Cohomology


1

That all looks pretty decent to me. As an alternative, you could just say that given a 1-form $\alpha$ on the first factor and a function $f$ on the second factor, you can define a 1-form on the product, $$ \beta_p(v) = f(p) \alpha(D\pi_1(p)[v]) $$ where $\pi_1$ is the projection on the first factor. A natural name for $\beta$ is $\alpha \times f$. So the ...


0

To achieve a contradiction, show by some other means that the integral is nonzero. Since $d\theta \wedge d\phi$ is in fact the standard volume form on $T$, this shouldn't be too difficult.


8

First, it is not quite true that there is "only one cohomology theory" for topological spaces. One should rather say that there are several such theories, but that for reasonable spaces, they give the same results. The reason why this is true in a category of reasonable topological spaces is that nice topological spaces can be "built up" from smaller ones ...


3

Yes, there's a fibration with fibre $BG$. And you don't need a semidirect product: any extension of groups will do. The special case of the Serre spectral sequence that this gives is often called the Lyndon-Hochschild-Serre spectral sequence, and can be constructed algebraically.


1

This is one of the places where generalizing makes structure more visible. If you are willing to consider $S_n$ for all $n$ at once, i.e., study $$ \coprod_n BS_n, $$ then the cohomology algebra has two products making it a Hopf ring (a ring object in the category of coalgebras). Giusti, Salvatore, and Sinha studied this and include explicit rules for ...


5

To be (as requested) nitpicky and (I hope) meticulous: Step $2$ suggests that $d\theta$ is the differential of an angle function $S^1 \to (0, 2\pi)$, but this isn't true for two reasons: First, it is not defined at the point $(1, 0) \in S^1$. More seriously, being closed but not exact, $d\theta$ is not actually the differential of any real-valued function on ...


1

On a smooth manifold with a smooth chart $(U, \phi)$, the $1$-forms $dx^1, \ldots, dx^n$ form a local coframe of $TM$ defined on $U$, that is, for any $1$-form $\alpha$ on $M$, there are smooth functions $\alpha_i$ such that $\alpha\vert_U = \alpha_1 dx^1 + \cdots + \alpha_n dx^n$; this only requires knowing that for an $n$-manifold the rank of $TM$ is $n$, ...


2

You can define two forms which give you a basis of the cohomology, and show by hand thta every other 1-form is cohomologous to a linear combination of them. This is not hard to do.


0

Here are two ways of seeing that $H^1_{dR}(S^1\times S^1)$ is two-dimensional without using the Künneth Theorem. For any smooth connected manifold $X$, $H^1_{dR}(X) \cong \operatorname{Hom}(\pi_1(X), \mathbb{R})$ (see Lee's Introduction to Smooth Manifolds (second edition) Theorem $17.17$ and Problem $18$-$2$), so $$H^1_{\text{dR}}(S^1\times S^1) \cong ...


3

Every embedding $I^k\hookrightarrow S^n$ is indeed closed. Since $I^k$ is a compact space, its image $h(I^k)$ is also compact and hence closed in $S^n$. Now $S^n$ is Hausdorff and compact subspaces of Hausdorff spaces are closed. Thus, $A$ and $B$ are open subsets of $S^n$ and you can apply Mayer-Vietoris on $A\cup B$.


3

This is a way to explain the intuition behind de Rham cohomology: Cohomolgy comes up as a dual answer to homology. Homology identifies the shape of an object finding ‘holes’. More concretely, it looks for objects without boundary which are not the boundary of an object (and therefore the definition $H_k(M)=\text{ker}\partial_n/\text{im}\partial_{n+1}$). ...


0

You do need to decide what homology/cohomology theory you use and then what theorems you assume as tools. One result that is not that easy to prove, but intuitive is the following: a manifold of dimension $n$ can be covered by a family of open subsets so that any nonvoid intersection of these open sets is homeomorphic to a ball and at most $n+1$ of these ...


2

In Algebraic Topology by Allen Hatcher: Theorem 3.26. Let $M$ be a closed, connected $n$-manifold. Then: (a) If $M$ is $R$-orientable, the map $H_n(M;R) \to H_n(M | x; R) \approx R$ is an isomorphism for all $x \in M$. (b) If $M$ is not $R$-orientable, the map $H_n(M; R) \to H_n(M | x; R) \approx R$ is injective with image $\{ r \in R | 2r = 0 ...


2

Let $\sigma:S^3\times S^3\to S^3\times S^3$ such that $\sigma(a,b)=(ab,b^{-1})$. As $\sigma\circ\sigma$ is the identity map, $\sigma$ is a homeo of order $2$. The subset $X=S^3\times(S^3\setminus\{e\})$ is invariant under $\sigma$ and, if we call $G$ the group generated by $\sigma$ (which is cyclic of order two), you are asking for the cohomology of the ...


0

Since $\mathbb R^2 \times S^1$ is an open solid torus, if we remove one point, we have a space homotopy-equivalent to a torus with one meridional disk glued in: $(S^1 \times S^1) \cup (D^2 \times \{-1\})$. Pinching the disk to a point, this is a $S^2$ with two points identified. Expanding that point to an interval, yields the union of a sphere with an ...


1

2) is not true: $X\#Y \cup D \cong X\vee Y$ (the first union is along the boundary of the disc D along which we perform the connected sum) and from the Mayer-Vietoris sequence ($n$ is the dimension of the manifold): $$0\rightarrow H_n(X\#Y)\oplus H_n(D) \rightarrow H_n(X\vee Y) \rightarrow H_{n-1}(S^{n-1}) \rightarrow H_{n-1}(X\#Y) \oplus H_{n-1}(D) ...


0

As Grumpy Parsnip suggests we can take Alexander duality so that \begin{align*} \tilde H_i(S^n - S^k \vee S^\ell; \mathbb Z) &\cong \tilde H^{n-i-1}(S^k \vee S^\ell; \mathbb Z)\end{align*} which is $\mathbb Z$ for $i=n-1-k$ or $n-1-\ell$ and trivial otherwise.


2

Part of the confusion is in defining the trace of a group homomorphism. Once this is done, the proof is not too difficult since, from the comments in the question on how to prove this in the setting of vector spaces, it is easy to see how the proof might follow for finitely generated abelian groups. In general, let $A$ be a finitely generated abelian group ...


1

If you know a few formal properties of homology, you might argue as follows: Realize ${\mathbb S}^n$ as ${\mathbb D}^n / \partial{\mathbb D}^n$ with basepoint $\overline{\partial{\mathbb D}^n}$, you have a pinch map ${\mathbb S}^n \to {\mathbb S}^n\vee{\mathbb S^n}$ collapsing $\{x_n = \frac{1}{2}\}$ to a point, and which you can prove to induce the ...


6

As Steven Gubkin and Ryan Budney have already pointed out, cohomology is often used to measure how far a "locally consistent" object is from being "globally consistent". I thought I might describe another example of this in set theory, which it turns out contains a lot of open problems. As far as I know, this example hasn't been written down anywhere, so ...


14

I guess I will take a crack at this. First of all, it is probably worthwhile for you to learn some cohomology in its original home so that you have some intuition for it, and some knowledge of what theorems there are, how to compute it, etc. You do not give much indication in your answer as to how much knowledge of cohomology you have currently. Generally ...


2

First: Are there any other terms $E_2^{p,q}$ with $p+q=5$ which survives to $E_\infty?$ Similar considerations also answers the second part of this question. Second: You'll probably have to use the fact that $\pi_i(S^n)$ is finite for $i \not=n $, $n$ odd, which should help you calculate the Ext group. This fact follows from Serre's mod $\mathcal{C}$ ...


0

There are several different ways of computing the induced map on $H_2$. For example, one can give $\mathbb{T}^2$ a CW- complex structure. It's a general theorem that every map of CW complexes is homotopic to a CW-map (one which maps the $k$-skeleton to the $k$-skeleton), and that homotopic maps induce the same map on homology. One your map is CW, it's ...


0

First of all, isn't Borel-Moore Homology defined for manifolds, not for varieties (i.e. the analytic structure is used, which is not available on an algebraic variety unless you perform some sort of analytification). But with this caveat: Borel-Mooore Homology coincides with singular homology for compact spaces, so in particular the Kunneth Formula you've ...


4

The answer to question a is not always, and the answer to question b is that there may be missing terms. What you want is the universal coefficients theorem for homology. The general statement is that for any dimension $n$ (in your case $n=1$) and any coefficient group $Q$ (in your case $Q = \mathbb{Z}/N\mathbb{Z}$), and any topological space $\Sigma$ there ...


1

There is no relation whatsoever, basically. Every manifold embeds in a contractible space ($\mathbb{R}^N$ for big enough $N$), and every nonempty manifold contains a contractible subspace. Of course manifolds can have all sorts of Betti numbers, while all the Betti numbers of a contractible are zero (except for $b_0 = 1$). Any nonempty space always ...


5

You ask a lot of different questions... $\newcommand\Z{\mathbb{Z}}\newcommand\Q{\mathbb{Q}}$Question 1: For starters, the rank of an abelian group is the maximal cardinality of a linearly independent subset. Not the number of generators (see the comments). So for example the rank of $\Z/2\Z$ is zero. Equivalently, the rank $r$ of $A$ is the maximal ...


1

To simplify notation, write $\hat H^*$ for the Tate cohomology group $\hat H^*\left(\mathbb{Z}_p, \mathbb{Z}_p\right)$. Then $\hat H^*$ has period $2$; that is, the map $x\to x\cup u$ is an isomorphism $\hat H^* \to \hat H^{*+2}$ for some $u\in \hat H^2$. To prove this, consider the periodic complete resolution of $G$ via its action on $S^1$, for example, or ...


3

In this answer I'll assume that $\mathbb{Z}_p$ means $\mathbb{Z}/p\mathbb{Z}$. For 2), more generally, let $G$ be a finite group and let $q$ be a prime not dividing $|G|$. Then $H^n(BG, \mathbb{F}_q)$ vanishes (for $n \ge 1$). One of many ways to see this is that under the given hypotheses $\mathbb{F}_q[G]$ is semisimple and so the category of ...


4

The short answer is yes. For example, $\text{Spec }H^{\bullet}(\mathbb{CP}^{\infty})$ (here the cohomology is concentrated in even degree so there is no issue with applying $\text{Spec}$ in the classical sense) can be identified with the formal affine line. The group structure on $\mathbb{CP}^{\infty}$ given by tensoring line bundles induces a group scheme ...


0

The Dold-Kan correspondence tells you that connective chain complexes of abelian groups are the same thing as simplicial abelian groups. In particular, under the Dold-Kan correspondence, singular chains arises from linearizing the singular simplicial set of a space, and taking singular homology corresponds to taking the homotopy groups of the corresponding ...


4

First, recall that $\mathbb{R}^{\times} \cong \mathbb{R} \times \mathbb{Z}_2$, and hence $$H^2(G, \mathbb{R}^{\times}) \cong H^2(G, \mathbb{R}) \times H^2(G, \mathbb{Z}_2).$$ Second, since $\mathbb{R}$ is a $\mathbb{Q}$-vector space and $G$ is finite, $H^2(G, \mathbb{R}) = 0$. So the problem reduces to the computation of $H^2(G, \mathbb{Z}_2)$. At this ...


1

I will do only the case of the sphere (the case of the torus is very similar). If $n \ge 4$ (and not $3$: you need $H_{n-1}(S^2)$ to vanish) then there's not problem and the long exact sequence gives you $$H_n(S^2, A) \cong H_{n-1}(A) = 0.$$ Let's write down the LES in low dimensions: $$\require{cancel} \begin{align} & \cancel{H_3(S^2)} \to H_3(S^2,A) ...


2

I disagree that index theorem is "one of the most important results in twentieth century's mathematics". This kind of superlatives usually does not mean anything, unless one can draw a fine line between "important", "more important", "most important", "of greatest importance", etc. I think it is better to be humble about it, even though my own research field ...


1

From the Wikipedia page on the index theorem: One can also define the topological index using only K theory (and this alternative definition is compatible in a certain sense with the Chern-character construction above). If $X$ is a compact submanifold of a manifold $Y$ then there is a pushforward (or "shriek") map from $K(TX)$ to $K(TY)$. The topological ...


2

$\require{cancel} \newcommand\R{\mathbb{R}} \newcommand\Q{\mathbb{Q}} \newcommand\Z{\mathbb{Z}}$ Write down the long exact sequence in homology: $$\cancel{H_1(\R)} \to H_1(\R,\Q) \to H_0(\Q) \to H_0(\R)$$ The space $\R$ is contractible so $H_0(\R) = \Z$ and $H_1(\R) = 0$. It's known that $H_0$ counts path components, and each point of $\Q$ is its own path ...


3

Still new at this so grains of salt are still advised. The second one looks like it may be aided by the long exact sequence of the pair: $$H_1(A)\to H_1(X)\to H_1(X,A)\to H_0(A)\to H_0(X)$$ The fact that $X$ contains at most one path-component of $A$ per path component means that the last arrow is injective. We are given that the first arrow must be ...


1

John's answer is good. I would just like to add that randomly creating differential forms and testing to see if they are closed and exact is not the best way to build a relationship with cohomology. As a grad student, I tried that approach and it was not very fruitful. Since we're in a low degree, you might understand what's going on by dualizing the ...


2

The first one is correct, since $S^n\setminus (S^n\setminus U_i)=U_i$ and so on. For the second one, you're definitely on the right track, but you need to look at another place in the sequence: $$H_n(S^n\setminus\{x_i\})\to H_n(S^n) \to H_n(S^n, S^n\setminus \{x_i\})\to H_{n-1}(S^n\setminus\{x_i\})$$ where the first and last terms are zero, and therefore the ...


1

I'm super new to homology so take this answer with a grain of salt. You definitely have the first part right. The third one looks like you apply excision, the fact that the homology of the disjoint union is a product of homologies, the second result, and the knowledge that $H_n(S^n)=\Bbb Z$. For the second part, I would encourage you to use a different ...



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