# Tag Info

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You can make a "branch", the real line with doubled non-negative side. You can then have the two branches meet, and you have a "loop on a string". This is non-orientable. More specifically, we have the disjoint union of two copies of $(\infty, 1]$, divided by the relation $\sim$ given by $a\sim b$ iff both of the following are satisfied: $a$ and $b$ are in ...

5

The space $S^1\vee S^1\vee S^2$ is the wedge sum of two circles and one sphere. In particular this space is also known as mouse space. It is quite easy to see that the fundamental group of $S^1\vee S^1\vee S^2$ is $\mathbb{Z}*\mathbb{Z}\cong\langle a,b| \emptyset \rangle$ ( you can use the Seifert-VanKampen theorem to see it ); whereas the fundamental ...

3

This is not necessarily true. Consider the following bicomplex: $$K^{p,q} = \begin{cases} \mathbb{Z}, & q=-p \text{ or } q=-p+1; \\ 0, & \text{otherwise.} \end{cases}$$ The differentials are defined as follows: $d : K^{p,-p} \to K^{p,-p+1}$ (the vertical maps) is multiplication by $2$, while $\delta : K^{p,-p} \to K^{p+1,-p}$ (the horizontal maps) ...

3

You run into two technical problems with this attempt to describe homology, but they're not "obvious" problems. (In fact, your description is at least partially what motivated Poincare to define homology in the first place.) The first is the restriction to submanifolds. There are too many technical details about embedding submanifolds: how many submanifolds ...

3

Not necessarily. For instance, the shift functor is an automorphism of any triangulated category, but it usually does not preserve distinguished triangles (because the shift of a distinguished triangle need not be distinguished, only the shift with the signs of all the maps reversed is distinguished). For a general discussion of equivalences of ...

3

I'll explain an analytic approach to computing the de Rham cohomology, to complement Qiaochu's algebraic one. Put any Riemannian metric on the torus, then there is a natural map from the space $\mathcal{H}^k(T^n)$ of harmonic $k$-forms to the de Rham cohomology $H^k_{dR}(T^n,\mathbb{R})$, where we send a harmonic form to its de Rham class. In fact, the Hodge ...

3

Yes, and in fact you do not need $X$ to be an H-space: the coproduct $\psi$ of $H_*(X)$ exists anyway (if the ring of coefficients is a field), and the definition of a group-like element does not involve the product. So take $X$ to be any space, and define $S \subset H_*(X)$ as in your question (the set of classes in $H_*(X)$ satisfying $\psi(a) = a \otimes ... 3 I should warn you that I'm not an expert -- I similarly found the original papers unenlightening toward their motivation -- so I did a bit of guesswork, perhaps it will be helpful to you. As you said, the mystery lies in the motivation of the additional step: modding out the functions from$X^{k+1} \to R$by the subcomplex of functions which disappear on ... 3 Because to obtain our chain complex for a space$X$we look at arbitrary maps$\Delta^n \to X$which are not necessarily nice, i.e. singular. In contrast we look at nice embeddings when dealing with simplicial homology. 3 You could use that isomorphism$H_n(|K|,|L|)\cong H_n(|K|/|L|)$which holds for reduced homology, and then your analysis is correct. You could also interpret this more directly. Relative cycles are chains whose boundary lies in the subcomplex$L$. The two edges in$K$satisfy this property, so give two relative cycles, which generate the$\mathbb ...

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It sounds as if you're confused about the difference between singular/simplicial homology and cellular homology. If you look at the homology chapter of Hatcher's book again, you'll see that these subjects are covered in some detail. The relevant details are as follows: The homology of a space is a deep property of that space that is hidden behind a ...

2

The problem of understanding elements in $\mathcal{K}_0(\mathcal{D}^b(\mathcal{A}))$ can be somewhat reduced to that of understanding elements in $\mathcal{K}_0(\mathcal{A})$ via the isomorphism $$\mathcal{K}_0(\mathcal{D}^b(\mathcal{A})) \to \mathcal{K}_0(\mathcal{A}): [X] \mapsto \sum_{i=-\infty}^{\infty}(-1)^i [H^i(X)].$$ Thus if you know the cohomology ...

2

$\mathbb{R}^n$ is contractible, so this description tells you that the torus is a classifying space $B \mathbb{Z}^n$, or equivalently an Eilenberg-MacLane space $K(\mathbb{Z}^n, 1)$, and hence its cohomology can be identified with the group cohomology of $\mathbb{Z}^n$. You can also compute this using the Kunneth formula, but there are other ways. The ...

2

This is a supplement to Tyler's answer (it should be a comment, but it's too long). As Tyler mentions, the Hauptvertmutung (any two triangulations have a common refinement) is false for topological spaces with dimension greater than 2. This breaks the sentence "simplicial (co)homology is an approximation of oriented cobordism (co)homology theory in all ...

1

[Note: After the comments to this answer and the question, there have been major changes. Thanks to Najib Idrissi, the situation is much clearer now.] There are only slight differences in the definitions and strictly speaking, Hatcher's is more general. In addition to (equivalents of) Hatcher's axioms, May wants a Hopf algebra to be flat (a minor technical ...

1

The questioner is right to identify one of the inspirations for homology as coming from integration theory; this is emphasised in the article by S. Lefschetz in the book "History of Topology", edited I.M. James. It seems also that early articles on Betti nubers and torsion coefficients wanted to take "cycles modulo boundaries" but were not so clear about the ...

1

The comment I made above misunderstood the problem, and I'm sorry for that... To see they are the same, notice that $C_0(A)$ is not empty so the mapping $C_0(A)\stackrel{\epsilon}{\rightarrow}\mathbb{Z}$ is already surjective. Therefore $C_0(X,A)\stackrel{\epsilon}{\rightarrow}\mathbb{Z}$ is a zero mapping and its kernel is the same as $\ker \partial_0$.

1

The issue is that $a^*+b^*$ is not the generator of $H^1$, in fact it is not a cohomology class: $$(a^*+b^*)(dU) = (a^*+b^*)(a+b-c) = 2.$$ Instead, take $b^*+c^*$ as the generator, which is a cohomology class: $$(b^*+c^*)(dU) = (b^*+c^*)(a+b-c) = 0,$$ $$(b^*+c^*)(dV) = (b^*+c^*)(a+c-b) = 0.$$ Also, I believe that $\mathbb{Z}[x,y]/(x^3, 2x^2, xy, y^2)$ is not ...

1

When taking the second boundary map, one can not remove the $i$th term since it was already removed by the first boundary map. So $j \neq i$ in the sum, and so it is convenient to split up the sum as $$\sum_{\substack{ j \\ j \neq i } } \cdot = \sum_{\substack{ j \\ j < i } } \cdot + \sum_{\substack{ j \\ j > i } } \cdot$$ Now for the signs, in ...

1

I suggest looking up a few sources, which will allow you to piece together your own proof. The proof of Poincare duality in the book of Bott-Tu, with particular emphasis on the construction of a De Rham cocycle representing $[M]$. The construction of a cochain map representing the isomorphism between De Rham cohomology and singular cohomology with real ...

1

As a starting point, the book "Graphs, Surfaces and Homology" (3rd ed.) by Peter Giblin is an outstandingly readable and informative undergraduate introduction to homology. You probably should for now avoid MacLane's "Homology." There is no royal road, remember, and you are entering the gates to someplace where no road is a superhighway.

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$\newcommand{\Z}{\mathbb{Z}}$The Klein bottle has the following integral homology groups: $$H_*(K; \Z) = (\Z, \Z \oplus \Z/2\Z, 0, 0, \dots).$$ This gives an Euler characteristic $\chi_\Z = 1-1 = 0$. Over $\Z/2\Z$, the universal coefficient theorem yields: $$H_*(K; \Z/2\Z) = (\Z/2\Z, \Z/2\Z^2, \Z/2\Z, 0, 0, \dots)$$ and so the Euler characteristic is ...

1

Firstly, why are you writing $\hom_{F-modules}(H_n(X;F),F)$? UCT implies $$H^n(X;F)=\hom_{F}(H_n(X;F),F)$$. Now, every $F$-module is a $F$-vector space and dual of a (finite dimensional) vector space is isomorphic to itself. So, $H^n(X;F)=\hom_F(H_n(X;F),F)=H_n(X;F)$, whenever $H_n(X;F)$ has finite rank. For the infinite rank case see this SE post.

1

If $S^3/G$ was not orientable, then the orientation double cover would be non-trivial. From the correspondence between covers of $S^3/G$ and subgroups of $\pi_1(S^3/G)\cong G$, the cover would correspond to an index two subgroup of $G$ (call this subgroup $H$). But now $G/H \cong \Bbb Z/2\Bbb Z$ is a non-trivial abelian quotient of $G$, so $G$ cannot have ...

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