Use this tag if your question involves some type of (co)homology, including (but not limited to) simplicial, singular or group (co)homology. Consider the tag (homological-algebra) for more abstract aspects of (co)homology theory.

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How could someone conclude $\check{H}^i (M, \mathbb{R}) = 0$ for arbitrary $M$?

sorry if this is a very stupid question and I'm missing something very trivial, though I could not solve it after thinking for a while. In page 18-19 of ...
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1answer
6 views

How does one compute the Hurewicz homomorphism for a (symplectic) nilmanifold?

I have a symplectic six-dimensional nilmanifold $X:=G/\Gamma$ in hand, characterized by the sextuple $(0,0,12,13,14+23,24+15)$, which records the exterior derivatives of a basis of $\Gamma$-invariant ...
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14 views

Finding the chern classes of line bundles over projective space using homotopy classes of clutching functions

I have just started to learn about characteristic classes and before learning more about the ways to compute them it would be nice to compute some examples using tools I already know. I only started ...
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1answer
34 views

There are no nonzero cocycles on $U$ vanishing on a def. retract of $U$

Playing around with cochains, I think I showed the following: Proposition: Suppose $U$ deformation retracts onto its subspace $A$, and suppose $\varphi \in C^k(U)$ is a singular cocycle which ...
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1answer
14 views

Why is $\ker(id\otimes \cdot b:R/(a)\otimes_R R \to R/(a)\otimes_R R)=R/(d)$?

Let $R$ be a commutative ring with unit $1_R$, $M$ a $R$-module. Let $a,b\in R\setminus \{0\}$ and $\gcd(a,b)=d$. I want to prove: $$\operatorname{Tor}_1^R(R/(a),R/(b))=R/(d).$$By definition, it is ...
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1answer
37 views

Property of second Steifel-Whitney class of tangent bundle?

Let $M$ be a closed, smooth, simply-connected $4$-manifold. Is $w_2$ the unique class in $H^2(M, \mathbb{Z}_2)$ such that $w_2 \cup x = x \cup x$ for all $x \in H^2(M, \mathbb{Z}_2)$ or not? I ...
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1answer
19 views

Is the transfer homomorphism in cohomology surjective?

Let $p:\widetilde{X} \to X$ be an $n$-sheeted covering space. Consider a singular simplex $c:\triangle^k \to X$, because the simplex is simply-connected, there exist $n$ different lifts ...
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1answer
11 views

Regarding embeddings and homological/cohomological injectivity

Possibly low-level question here: if one has an embedding $M\hookrightarrow X$, call it $\iota$, then this is of course an immersion, so the differential maps $\iota_{\ast}:T_{p}M\rightarrow ...
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1answer
19 views

strange implication of addivity axiom in homology theory

Let $H$ be a homology theory satisfying Eilenberg-Steenrod axioms and $X$ an arbitrary topological space. We can write $X$ as a disjoint union of its points $$X= \coprod_{x \in X}{\{x\}}$$ Now the ...
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2answers
66 views

Errata in Prof. Rotman AIHA book about projectives in the chain complex category

EDIT After thinking carefully with the help of the clear answer of ZhenLin, I think I will reformulate my question the following way. The text of my original question is kept below. When I look at ...
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2answers
59 views

Long exact sequence of a triple: working through the geometry

Suppose $X$ is a topological space with subspaces $X \supset U \supset A$ such that $U$ deformation retracts onto $A$. We know that $H^*(X,U) \cong H^*(X,A)$--one way to see this is to take the long ...
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1answer
16 views

Induced map in cohomology of a covering [on hold]

Is it true that if $p: E \to B$ is a $2$-fold covering, the map $p^*$ induced in cohomology is surjective?
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1answer
20 views

Definition of the Fundamental Class for $K(A,0)$

I'm having a little doubts on the definition of the fundamental class for the Eilenberg-MacLane space $K(A,0)$. Recall that a fundamental class $\imath_{A,n}$ for a polarized $K(A,n)$ is the element ...
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1answer
24 views

$f_*$ induces an isomorphism in homology iff the mapping cone of $f_*$ is contractible.

Let $f_*:C_*\to D_*$ be a chain map. I'm stuck in the proof of the following statement: $f_*$ induces an isomorphism in homology iff the mapping cone of $f_*$, cone($f_*$), is contractible. (For ...
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1answer
76 views

Injectivity in the zero homology

I'm struggling with following step in an excercises about Mayer-Vietoris sequences: In one step the solution says this map is injective since $A \cap B$ is path-connected: $$ H_0(A \cap B) ...
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2answers
56 views

Long exact sequence of $(I \times Y, \partial I \times Y)$.

There's a section in chapter 3 in Hatcher's Algebraic Topology where he talks about the long exact sequence of the pair $(Y \times I, Y \times \partial I)$, where $Y$ is any topological space. The ...
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1answer
49 views

How to use the Universal Coefficient Theorem to determine $H^i(M; \mathbb{Z}_p)$ from $H^i(M; \mathbb{Z})$? [on hold]

Let $M$ be a path-connected finite $CW$-complex. Suppose $$ H^2(M;\mathbb{Z})=\mathbb{Z}_{2k}, \text{ } k\geq 3; $$ $$ H^3(M;\mathbb{Z})=\mathbb{Z}\times\mathbb{Z}_{2}; $$ $$ ...
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0answers
23 views

universal coefficient theorem for mod p cohomology

In the book Algebraic Topology, Allen Hatcher, p. 266, Corollary 3A.6 (b): Question: I want to rewrite the above statement into a cohomology version. If I replace all homologies with cohomologies, ...
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1answer
44 views

Can binary ring for homology make life easier?

Do you know of a proof which uses homology to demonstrate a property about a topological space which is made easier (or even possible) because they work over $\mathbb{Z}/2\mathbb{Z}$ instead of ...
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31 views

$H^*(X,A;R)\cong H^*(X',A';R)\; \Rightarrow H^*(X\times Y ,A\times Y;R)\cong H^*(X'\times Y,A'\times Y;R)?$

I have a quastion about product spaces in singular cohomology. I only know a formula for sinugular homology for product spaces from lecture, the universal coefficient theorem. Let $R$ be a ...
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31 views

Simple modules and homological algebra

Let $A$ be an k-algebra, and $M$ a $A-$module. If $Ext^{1}(M,S)=0$ for every simple $A-$module $S$, then $M$ is projective. I know that this is true if $A$ is finite-dimensinal, but if $A$ is ...
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1answer
41 views

If induced map on homology is surjective, is induced map on cohomology injective?

Suppose I have topological spaces $X, Y$ and a continuous map $f: X \to Y$. Let $\mathbb{k}$ be a field, and $i \ge 1$ an integer. If the induced linear map on homology $f_* : H_i ( X, \mathbb{k}) ...
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0answers
20 views

When does the first cohomology group commute with inverse limit?

Let $M_i,i\in\mathbb{N}$ be an inverse system of continous, discrete G-modules and let $M=\varprojlim M_i$. Under what conditions on $M$ and $M_i$ do we have $\varprojlim H^1(G, M_i) \cong H^1(G, M)$? ...
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1answer
47 views

Is $[N]^\#([N])$ congruent to $w_n(\nu_N)([N])$ mod $2$, where $\nu_N$ is the normal bundle of the embedding of $N$ in $M$?

Let $M$ be a closed, smooth, orientable $2n$-manifold, and let $N$ be a closed, smooth, orientable $n$-submanifold. Let $[N]^\#$ denote the cohomology class (Poincaré) dual to the homology class ...
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1answer
45 views

$T^2-D$ does not retract to the boundary $\partial D$

First of all: yes, there is already a post about it, but I missread retract as strong deformation retract and wanted to know if this solution is right if we really do assume the stronger assumption of ...
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1answer
43 views

Why is $H^*(S^1\times X,\{\text{pt}\}\times X;R)\cong H^*(D^1\times X,\partial D^1\times X;R)$?

Let $X$ be a topological space, $R$ is a commutative, unital ring. In a proof from lecture there is claimed that $H^*(S^1\times X,\{\text{pt}\}\times X;R)\cong H^*(D^1\times X,\partial D^1\times X;R)$ ...
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35 views

Poincaré duality isomorphism maps cohomology to homology here?

Let $M = M^n$ and $A = A^p$ be compact oriented manifolds with smooth embedding $i: M \to A$. Let $k = p - n$. Does the Poincaré duality isomorphism$$\bigcap \mu_A: H^k(A) \to H_n(A)$$map the ...
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1answer
20 views

Do these non-homotopic maps induce the same map in reduced homology?

Consider two maps $f, g: X\to Y$, where $X=Y=\{ 0, 1 \}$ with discrete topology, $f$ is the identity and $g$ maps everything to 0. Then it's clear that $\widetilde{H}_0(X;\mathbb{Z})\cong \mathbb{Z}$ ...
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1answer
12 views

On the assumptions of cocyle conditions in a Lie algebra

To define the Cohomology (with values in $\mathbb{C}$) on a lie algebra $L$, we define a coboundary map $\delta:\Lambda^n(L)\to \Lambda^{n+1}(L)$. There is a general formula for the coboundary map but ...
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27 views

What is an integral differential form and how do we recognize it as such?

I am reading about embedding theorems of various types of manifolds (Kodaira's embedding theorem being one of them), and one condition that repeats in all of them is that the manifolds should be ...
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23 views

Cohomology of geometric realization of a simplicial topological space

Let $X$ be a simplicial topological space. We can consider to notions of cohomology of $X$. Denote by $|X|$ geometric realization of $X$. Then we can take just $H^*(|X| )$ (i.e. usual cohomology of ...
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1answer
36 views

A remark about the map $\partial^*:H^{*}(A;V)\to H^{*+1}(X,A)$ of the l.e.S. in cohomology

My question is about a remark from lecture about the connecting-homomorphism of the long exact sequence in homology of a pair $(X,A)$. Let $(X,A)$ a pair of topological spaces, $V$ be an abelian ...
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1answer
45 views

counterexample for $f_*:C_*\to D_*$ be a chain map such that $f_*$ induces an isomorphism in homology. Then $f_*$ is a chain homotopy equivalence

I want to understand a counterexample for: Let $f_*:C_*\to D_*$ be a chain map such that $f_*$ induces an isomorphism in homology. Then $f_*$ is a chain homotopy equivalence, because the statement ...
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44 views

Characterisation of Stable Cohomological Operations: $\Sigma (\tau_n (\imath_{A,n}))=\rho_{n+1}^*(\tau_{n+1}(\imath_{A,n+1}))$

I've started studying (stable) cohomological operations on my lecture notes, and I was given that an equivalent definition for a family of cohomological operations to be a stable cohomological ...
3
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1answer
40 views

Absolute Galois group $\text{Gal}(\overline{K}/K)$ of any number field $K$ has a non-open subgroup of any prime index $p$?

Let $K$ be a number field, and let $p$ be a prime number. Does $G = \text{Gal}(\overline{K}/K)$ necessarily have a subgroup of index $p$ that is not open?
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49 views

Cohomology algebra generated by Steifel-Whitney classes and dual classes subject to defining relations? [closed]

Is the cohomology algebra $H^*(G_n(\mathbb{R}^{n+k}))$ over $\mathbb{Z}/2$ generated by the Steifel-Whitney classes $w_1, \dots, w_n$ of $\gamma^n$ and the dual classes $\overline{w}_1, \dots, ...
3
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1answer
28 views

Left vs right projective resolutions and homology of monoids

Let me use the ad hoc notation $\mathbb Z^l$ and $\mathbb Z^r$ to distinguish between left and right modules. These are trivial modules. The homology of a (discrete) finite monoid $M$ with ...
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40 views

Killing vector field for Witten complex?

I am reading a classical paper by Atiyah Bott "The moment map and equivariant cohomology". In paragraph "Relation with Witten complex" (at the very beginning of this paragraph) they claims that ...
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1answer
22 views

Relation between symplectic blow-up of a compact manifold and fibre bundles over same manifold

The symplectic blow-up of a compact symplectic manifold $(X,\omega)$ along a compact symplectically embedded submanifold $(M,\sigma)$ results in another compact manifold $(\tilde{X},\tilde{\omega})$ ...
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1answer
62 views

Compatibility between Unreduced Suspension Iso and Reduced Suspension Iso

I need some clarifications on these two "basic" things because I realised I was using them carelessly and now I want to know once and for all the relation between the two. Let us assume working with ...
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0answers
23 views

How to find a counter-example that the centralizer of differential graded algebras does not preserve quasi-isomorphism?

Let $A^{\bullet}$, $B^{\bullet}$ be two differential graded algebras (dga) and $f: A^{\bullet}\to B^{\bullet}$ be a differential graded homomorphism between them. Now let $R$ be another algebra ($R$ ...
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1answer
46 views

Homology with local coefficients as a functor from pointed, path-connected spaces and $\pi_1$-modules.

A local system of coefficients on a space $X$ is a functor $F\colon \Pi(X)\rightarrow Ab$ from the fundamental groupoid to the category of abelian groups. From this, one can define the homology groups ...
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1answer
49 views

$n \times n$ invertible matrix defining diffeomorphism

I was reading the proof of the Hairy Ball Theorem in Madsen and Tornehave's book "From calculus to cohomology", and at some point they refer to the Lemma 6.14, which says the following: An invertible ...
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25 views

Relation between compactly supported cohomology and locally finite homology

Building up on a previous question, I am currently investigating in the properties of locally finite homology. Suppose that $X$ is a reasonably well-behaved space. I want to find out whether there is ...
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1answer
40 views

Basic question on cohomology ring

To show (1) $S^2\vee S^1\vee S^1$ is not homotopy equivalent to $S^1\times S^1$ (2) $S^1\vee S^2\vee S^3$ is not homotopy equivalent to $S^1\times S^2$ I use the same method: For (1) the ...
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1answer
32 views

On the boundary map of a locally finite chain complex

I am just learning about locally finite homology and I'm having a bit trouble understanding some of its concepts. There doesn't seem to be a whole lot of (non-advanced) literature on this topic, so I ...
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2answers
42 views

Is there any expression to calculate the homology groups of a quotient space?

Let $B \subset A$ where $A$ is a topological space and $A/B$ the space obtained from $A$ via collapsing $B$ to a single point. I was wondering if there is any expression for $H_k(A/B)$ in terms of ...
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28 views

Homology of SO(3)

In Tohru Eguchi, Peter B. Gilkey, and Andrew J. Hanson's review paper "Gravitation, gauge theory and differential geometry," I came across the following claim about the Homology of SO(3): I cannot ...
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1answer
35 views

Cohomology of Eilenberg Maclane space

In a book on spectral sequences that I am reading, it is stated, without proof, that $H^i(K(\mathbb{Z},2);H^0(K(\mathbb{Z},1);\mathbb{Z}))$ is isomorphic to $\mathbb{Z}$ for even $i$ and $0$ for odd ...
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4answers
1k views

Why Cohomology Groups?

Why do we need cohomology groups? Homology groups are easier to compute and given two topological spaces, there is an isomorphism in homology groups if and only if there is an isomorphism in ...