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If a map induces the identity on homology, does it also on cohomology?

Let $X=S^3\vee\Sigma\mathbb{RP}^2$ and consider the following map $f:X\to X$. On the $S^3$ summand, $f$ is the identity. On the $\Sigma\mathbb{RP}^2$ summand, $f$ is the composition of the pinch map ...
Eric Wofsey's user avatar
8 votes
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What is the relation between homotopy groups and homology?

As commented above the Hurewicz theorem computes some homology groups from the knowledge of homotopy groups; it indicates that homotopy groups are harder to compute than homology groups. In degree 1 ...
plm's user avatar
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8 votes
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Question about isomorphism given by the Serre Spectral sequence

In the Serre spectral sequence the edge homomorphism $$H_p(X;M)\to E^\infty_{p,0}\subset E^2_{p,0}=H_p(Y;M)$$ is the map induced by $X\to Y$. If $F$ is acyclic then the Serre spectral sequence ...
Vincent Boelens's user avatar
7 votes

Degree of a map $\Bbb RP^n\to S^{n-1}\times S^1$

Let $n$ be odd and $n\ge 2$. We may consider the induced map $f^\ast: H^\ast(S^{n-1}\times S^1)\to H^\ast(\Bbb RP^n)$ on cohomology rings. If $a$ generates $H^1(S^1\times S^{n-1})$ and $b$ generates $...
Kevin.S's user avatar
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6 votes
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Abelian group as second cohomology group of a pair G,M

If $A$ is any abelian group, then $H^2(\mathbb{Z}\times\mathbb{Z},A)\cong A$, where $\mathbb{Z}\times\mathbb{Z}$ acts trivially on the coefficient group $A$. This follows from the universal ...
Eric Wofsey's user avatar
6 votes
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UCT gives different answer than explicitly calculating homology

Your computation of $H_1(X)$ is incorrect. You took a quotient by $(4\mathbb{Z},2\mathbb{Z},2\mathbb{Z})$ instead of $(4,2,2)\mathbb{Z}$. The correct computation gives $H_1(X)\cong \mathbb{Z}/2\mathbb{...
Sergey Guminov's user avatar
6 votes
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How simple are simple spaces in regards to twisted coefficients?

A simple space has the property that $\pi_1$ acts trivially on all homotopy groups, not just the higher ones, in particular $\pi_1$ is abelian. Otherwise there are very obvious counterexamples to your ...
Connor Malin's user avatar
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5 votes
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hodge number of complex surfaces

Here are a class of counterexamples. One can just consider the ruled surfaces $F_n$. These surfaces are not isomorphic for different $n$. However they have the same Hodge diamonds. One reference is ...
Invariance's user avatar
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5 votes
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If $M$ is compact connected non-orientable 3-manifold, then $H_1(M)$ is infinite.

This is simply false. Consider $M=RP^2 \times [0,1]$.
Moishe Kohan's user avatar
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5 votes
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Quotient of cohomology groups with different coefficents

Universal coefficients gives, for any abelian group $A$ and any $X$, an isomorphism $$H^1(X, A) \cong \text{Hom}(H_1(X), A)$$ (since $H_0(X)$ is free, so the Ext term vanishes). If in addition $H_1(X)$...
Qiaochu Yuan's user avatar
5 votes

Degree of a map $\Bbb RP^n\to S^{n-1}\times S^1$

Here's another way. Any such degree $k$ map gives a composition $$S^n\to \mathbb{R}P^n\to S^{n-1}\times S^1.$$ The first map has degree $2$ when $n$ is odd, so the composition has degree $2k$. ...
red whisker's user avatar
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A consequence of Hilbert 90

It turns out there really is only a $K^\times$'s worth of solutions: We can rewrite the key equation as $b=\sigma(b)c$. Assume that also $b' = \sigma(b')c$. Then $b' = bx$, for some $x \in L^\times$. ...
gimothytowers's user avatar
5 votes
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Twists of morphisms and torsors

To answer your direct question, it is useful to first abstract the situation slightly. That said, maybe the following will be helpful to help orient you to the rest of the answer. TL;DR: Implicitly ...
Alex Youcis's user avatar
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5 votes
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Proof of the "Whitehead theorem for homology"

As noted in Weak Homotopy Equivalence Induces Isomorphisms on Homology, we can replace $Y$ by the mapping cylinder of $f$; the effect is that we can assume that $f$ is an inclusion. Then we have long ...
John Palmieri's user avatar
5 votes

Question about isomorphism given by the Serre Spectral sequence

$\require{AMScd}$ I think you can use the fact that the Serre spectral sequence converges naturally; here is a rough argument: Consider the following commutative diagram of fibrations \begin{CD} F @&...
jasnee's user avatar
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5 votes
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excisive couples, sufficient conditions

Here are some observations (no claims of exhaustiveness): The map is an isomorphism if and only if $A$ is a union of path-components of $X$ or $B$ is a union of path-components of $Y$. (E.g. if $A=\...
Thorgott's user avatar
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5 votes
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Singular Homology is a special case of Simplicial Homology

Although the statements in your post are all true, none of them are correct interpretations of this passage from Hatcher's book. Instead, what Hatcher is saying is that for any topological space $X$ ...
Lee Mosher's user avatar
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5 votes
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Euler class of a principal $SO(2)$-bundle over a lens space

We may construct principal $S^1$-bundles over $L(p,q)$ as follows. We have a $\Bbb Z_p$ action on $S^3\times S^1$ generated by $((z_1,z_2),z)\mapsto ((e^{i2\pi/p}z_1,e^{i2\pi q/p}z_2),e^{i2\pi k/p}z)$,...
Kevin.S's user avatar
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5 votes
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What is the "higher cohomology" version of the Eudoxus reals?

Let $C_b^n(G, H)$ be the group of bounded functions $G^n \to H$, for discrete $H$ this is equivalent to functions with finite image, but this definition also makes sense for $H$ a normed $\mathbb{R}$-...
ronno's user avatar
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4 votes

Why is the orientable double cover orientable?

Let us go back to the roots and discuss the concept of local orientation and the local consistency condition for orientations. Let $M$ be an arbitrary manifold. $H_n(M \mid x) = H_n(M, M \setminus \{...
Paul Frost's user avatar
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4 votes
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Relationship between Cohomology groups/rings related to a topological group

By considering $G=S^1,$ so that $BG$ is the infinite-dimensional complex projective space, you can see immediately that there’s no easy relationship between your second two cohomologies. There is ...
Kevin Carlson's user avatar
4 votes
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$p$-torsion in fundamental group of Lie group.

First, the fundamental group of a connected Lie group is abelian and finitely generated. Secondly, any finitely generated abelian group is the fundamental group of a connected Lie group. To see the ...
Dietrich Burde's user avatar
4 votes
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Examples of CW-complexes wich are 1-acyclic but no simply conected.

The one general thing that can be said is simply to state the 1-dimensional Hurewicz theorem: $$H_1(X) \approx \pi_1(X) \, / \, [\pi_1(X),\pi_1(X)] $$ In words, $H_1(X)$ is isomorphic to the quotient ...
Lee Mosher's user avatar
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4 votes
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Vanishing of $r$-fold cup-products

Notice that the cup product defines a map $$H^*(X,A) \times H^*(X,B) \longmapsto H^*(X,A\cup B).$$ If $U$ is a set in your cover (you need only assume that the inclusion $U\to X$ is nullhomotopic, not ...
Pedro's user avatar
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4 votes
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Degree of a map $\Bbb RP^n\to S^{n-1}\times S^1$

Here's yet another way to see that the degree of a map $f: \mathbb{R}P^n \to S^{n-1} \times S^1$ is $0$ when $n \geq 2$. For $n \geq 2$, we have $\pi_1(\mathbb{R}P^n) \cong \mathbb{Z}_2$ and $$\pi_1( ...
Derived Cats's user avatar
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4 votes
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What if $X$ is not path connected?

Intuitively, it could not possibly be true for a non path-connected spaces in general. Notice one object possibly depends on $x_0$ (in this case) whereas the other object is invariant; it does not ...
FShrike's user avatar
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4 votes
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Defining Atiyah-Hirzebruch spectral sequence from exact couple

The $(-1,1)$ indicates the degree, i.e. $\alpha$ maps the summand $D^{p,q} = h^{p+q}(X_p)$ into $h^{p+q}(X_{p-1})=D^{p-1,q+1}.$ Analogously for the other maps.
Vincent Boelens's user avatar
4 votes
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Homology groups of $X \times S^1$

Let us understand $S^1$ as the set of complex numbers $z$ with $\lvert z \rvert = 1$. In Example 2.48 Hatcher derives an exact sequence "which is somewhat similar to the Mayer–Vietoris sequence ...
Paul Frost's user avatar
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4 votes

Proof of excission Axiom for singular Homology

In general, to say a chain $c\in\Delta_p(X)$ lies in a subspace $A\subseteq X$ just means it is contained in the subcomplex $\Delta_p(A)\subseteq\Delta_p(X)$, equivalently its support (the union of ...
Thorgott's user avatar
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4 votes
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Is homology dual to cohomology or is homotopy dual to cohomology?

Homotopy and cohomology are dual in a vague sense called "Eckmann-Hilton duality". As the linked page says, this is more a heuristic rather than a precise notion of duality. Certainly in the ...
John Palmieri's user avatar

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