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2

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

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 ...


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 ...


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 ...


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.


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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 $\ker \partial_0$.


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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 ...


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[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 ...


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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 ...


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 ...


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

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.


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As has been mentioned in the comments, you are correct. The beginning of the Mayer-Vietoris sequence is $$0 \to H^0(M) \xrightarrow{k^*\oplus l^*} H^0(U)\oplus H^0(V) \xrightarrow{i^*-j^*} H^0(U\cap V) \xrightarrow{\delta} H^1(M) \to \dots$$ and the end is $$\dots \to H^{n-1}(U\cap V) \xrightarrow{\delta} H^n(M) \xrightarrow{k^*\oplus l^*} H^n(U)\oplus ...


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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I would try to prove that the result is still false also in the case of a bounded double complex. Let's work in an abelian category $A$ with enough projectives, and let $F$ be a right-exact functor (for instance, you could think to the tensor functor over the category of modules). Let $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0 $ be a ses ...


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) ...


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 ...


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

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 ...


0

There is also a way showing the independence of $\chi (X)$ without using the universal coefficient theorem, but using cellular homology: Let $a_n$ denote the number of cells of $X$ in dimension $n$. Then we get the cellular chain complex $$\ldots \longrightarrow F^{a_k} \longrightarrow F^{a_{k-1}} \longrightarrow \ldots $$ Using that the Euler ...


1

$\newcommand{\Z}{\mathbb{Z}}$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 ...


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 ...


0

Choose a homomorphism $h:M\rightarrow N$. Let $x\in M$. There exists an open nighborhood $U\subseteq M$ of $x$ and a homomorphism $\psi_1:U\rightarrow B^m$, (where $B^m$ denotes an open ball in $\mathbb{R}^m$). Now, consider $h(x)\in N$. There exists an open set $V\subseteq N$ s.t. $h(x)\in V$ and a homomorphism $\psi_2:V\rightarrow B^n$. By restricting ...


0

Consider a neighbourhood $O_M$ in $M$. Since $M$ is a manifold, it's locally homeomorphic to $\mathbb{R}^m$ for some integer $m$. Let $f_1$ be the homeomorphic mapping from $\mathbb{R}^m$ to $O_M$. Now consider the homeomorphism $f_2$ from $M$ to $N$. Since $f_2$ is a homeomorphism, $f_2(O_M)$ is a neighbourhood in $N$, call it $O_N$. And finally, $O_N$ is ...


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Let $M,N$ be manifolds with dimension m, n respectively. Suppose $\Phi:M \to N$ is homeomorphism. Let $\psi_1: M \supset D \to U \subset \mathbb R^m$ be a map (that is homeomorphism of some open subset of $M$ and open subset of Euclidean space). $U$ is open so we can choose open some ball in it and restrict $\psi_1$ to subset that is mapped onto this ball. ...


9

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 ...


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

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 ...


2

$\newcommand{\RP}{\mathbb{RP}} \newcommand{\Z}{\mathbb{Z}}$ I will assume that $X$ is path-connected, because we can solve the lifting problem separately on each path component, and a map to $H^*(X;R)$ is zero iff it's zero when co-restricted to each path component. (1) The map $S^n \to \RP^n$ is a universal covering, so for a lift of $f : X \to \RP^n$ to ...


2

The answer is yes; if $A \to B$ is any morphism of abelian groups then it induces a natural transformation $H_n(-, A) \to H_n(-, B)$, which implies that the map induced by $f$ on $H_n(-, \mathbb{Z}_2)$ is the reduction $\bmod 2$ of the map induced on $H_n(-, \mathbb{Z})$.


1

I think for differential topology, and geometry it is probably best to learn de Rham cohomology. Bott and Tu's book is the canonical reference.


2

Although Amazon says that Massey's Singular Homology Theory is a sequel to Massey's Algebraic Topology: An Introduction, the earlier book is not a logical prerequisite for Singular Homology Theory. There are plenty of online lecture notes that might be a right fit for you, for example: ...


0

This is true for the dumbest of possible reasons. If $G$ is a compact, connected Lie group, then after inverting finitely many primes in the coefficient ring, $H^*(G) = \Lambda P$ is an exterior algebra. Over characteristics other than 2, exterior algebras are free commutative graded algebras (CGAs), so any surjection $H^*(E) \to H^*(G)$ splits. One simply ...


0

This is not a full answer, but maybe it can helps... As mt_ said, $\overline{f}$ and $\tilde{f}$ does not mean anything here. However, there is a map $H^2(G,\mathbb{Z}^\times)\rightarrow H^2(G,R^\times)$ (at least if $G$ acts trivialy on $-1\in R^\times$). So that if $f\in H^2(G,\mathbb{Z}^\times)$ there is a corresponding element in $f_R\in ...


1

You may just apply Borsuk-Ulam theorem directly. Define a function $f$ from $S^n$ to $\mathbb{R}^n$ as follows: If $x$ is a point on $S^n$, then there is a great $S^{n-1}$ that is orthogonal to the point $x$. The $S^{n-1}$ divides $S^n$ into two regions. Let's call them $U$ and $V$, where $U$ is the region containing $x$. If $\mu$ is the measure given, ...


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(Posting an answer to get this off the unanswered queue.) Ralph Mellish mentions that proof of the kind requested can be found as Cor 1.26 in Vick's Homology Theory. I note the same proof appears as theorem 2.28 in Chapter 2.2 of Hatcher's Algebraic Topology.



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