# Tag Info

5

The first thing to note is that the only homology group that will change is $H_0$. With that in mind, I think reduced homology actually better captures the intuition of "counts the holes in the space." Thinking that way, $H_0$ should count the 0-dimensional holes, which are just "gaps" between path-connected components. A point has no such "gaps," so its ...

5

There is the long exact sequence in homology of the pair $(X, X \setminus p)$ (everything is with $\mathbb{Z}$-coefficients unless indicated otherwise): $$0 \to H_n(X \setminus p) \to H_n(X) \to H_n(X, X \setminus p) \to H_{n-1}(X \setminus p) \to H_{n-1}(X) \to 0$$ (recall that the local homology groups satisfy, by excision, $H_n(X, X \setminus p) = ... 5 You can take homology$H_{\bullet}(X, A)$with coefficients in any abelian group$A$. Integral homology is the special case where$A = \mathbb{Z}$. "Homology" with no qualifier usually refers to this. 5 Here is a visualization for the regular singular homology theory with coefficients in$\mathbb{R}$. The generator for$H_2$is going to be the cycle that is "the entire manifold" (this is called the fundamental class). Since our manifold is closed, it has no boundary, and it has no$3$-cells, so it generates$\mathbb{Z}$(try contrasting this with ... 4 Two things that come to mind are It's the dual (in the sense of universal coefficients) to the fundamental class, as Mike Miller says, and It corresponds to any point in the torus by Poincaré duality, as Daniel Valenzuela says. The visualization via Poincaré duality might seem unsatisfying but in fact it works, suitably reinterpreted (augmented with ... 4 As mentioned in the comments, such a thing needs to be a rational homology sphere, and in particular a$\Bbb Z[1/d]$-homology sphere, where$d$is the degree. It also needs to have fundamental group of order coprime to$d$. As in the comments, we may as well assume the manifold is simply connected by passing to the universal cover. First, there are no ... 4 It often helps to visualize cohomology classes as the Poincaré dual homology classes (or more specific their representants). Let$M$be a closed oriented$n$-manifold. For$x\in H^nM \cong Hom(H_nM,\mathbb Z)$we have a dual class$P.D.(x)\in H_0M$. Let$m\in M$be any point representing this homology class (if$M$is not connected choose one for each ... 4 Merely by transport of structure, there is a natural map (covariant in$f$)$H^i(X,F)\to H^i(X,f_*F)$; equivalently, replacing$f$by its inverse, there is a natural map$H^i(X,f^*F)\to H^i(X,F)$contravariant in$f$(note that$f_*$and$f^*$are inverse functors when$f$is a homeomorphism). However, there is no reason there should be a natural map ... 4 An orientable manifold is$R$-orientable for any$R$. Now if$M$is closed connected$n$dimensional$R$-orientable manifold then$H_n(M;R)=R$. Again universal co-efficient theorem for homology says that , If$C$is a chain complex of free abelian groups, then there are natural short exact sequence$0\to H_m(C)\otimes G\to H_m(C;G)\to Tor(H_{m-1}(C);G)\to ...

4

The construction is not completely immediate because the map on cochains defining the cup-product $\smile : C^k(X) \otimes C^l(X) \to C^{k+l}(X)$ doesn't directly map $C^k(X,A) \otimes C^l(X,B)$ to $C^{k+l}(X, A \cup B)$. However if you look at the definition: $$(\varphi \smile \psi)(\sigma) = \varphi(\sigma_{\mid[v_0, \dots, v_k]}) \cdot ... 4 Intuitively, the isomorphism preserves products because just as the cup product comes from pulling back Künneth via the diagonal, the product seems to come from pulling back, i.e. dualizing via \text{Hom}, a derived version of the "Künneth" \mathbb{Z} \cong \mathbb{Z} \otimes \mathbb{Z}. 4 Yes. First, recall that a suspension has no nontrivial cup products. By Poincaré duality, it follows that the integral cohomology of M is torsion in degrees 1 through n-1, and that the same is true of the integral homology. From universal coefficients we know that H_{n-1} is torsion-free, which here implies that it's trivial. By Poincaré duality ... 4 We can compute H_i(S^n\times X) for any space X and all i without using Künneth's theorem as follows: Step 1: There is a retraction r:S^n\times X\to \{x_0\}\times X=X given by pinching S^n. This means that r\circ i=id_X where i:X=\{x_0\}\times X\hookrightarrow S^n\times X is the inclusion. Then, by functoriality of H_i, the inclusion is ... 4 Instead of looking up the answer and back engineering, you can just use good old linear algebra. The image of a linear map \mathbb{Z}^m \mapsto \mathbb{Z^n} defined by an m \times n matrix is equal to the column space of that matrix, which in the case of \delta_2 means the column space of$$\begin{pmatrix} 1&-1&1\\ 1&-1&-1 ...

3

Since $M$ is orientable, without boundary and connected, we have $H_0(M;\mathbb{Z}) = H_3(M;\mathbb{Z}) = \mathbb{Z}$. Using Poincaré duality, you know that the torsion subgroup of $H_2(M;\mathbb{Z})$ is isomorphic to the torsion subgroup of $H_{3-2-1}(M;\mathbb{Z}) = H_0(M;\mathbb{Z}) = \mathbb{Z}$ and thus is zero. Using Poincaré duality again, you know ...

3

$\pi$ has to be torsions free: Let $M$ be a compact aspherical manifold and $C$ a finite cyclic subgroup of its fundamental group. As $M$ is aspherical, its universal covering $\tilde{M}$ is contractible, so the quotient $\tilde{M}/C$ provides a finite dimensional model for the classifying space $BC$ of $C$. In particular, $C$ has nonzero homology groups ...

3

There is a group of closely related necessary conditions on the homology and cohomology of $\pi$, in order for there to be a compact oriented $n$-dimensional manifold $K(\pi,1)$. Usually one fixes a choice of coefficient ring $R$, commutative and with unit element $1$. These conditions hold for any choice of $R$. (One can drop the "oriented" property but ...

3

Hint: There are no $\tau$-invariant simplices, because $\tau$ acts freely on $X'$. To show that a general $\tau$-invariant chain $\sum a_i\sigma_i$ can be written in the form $c+\tau(c)$, try grouping the $\sigma_i$ into their orbits under the action of $\tau$.

3

As Justin Young commented, there are many different ways of proving this (and which one is most comprehesible to you will depend on your background); here's one. Write $T_n(A)=H_n(K(\pi,1),A)$; then $T_n$ is a functor from the category of $\mathbb{Z}[\pi]$-modules to the category of abelian groups. Moreover, these functors together canonically have the ...

3

Tensor product is right exact: if $0 \to V' \to V \to V'' \to 0$ is an exact sequence of abelian groups, then $M \otimes V' \to M \otimes V \to M \otimes V'' \to 0$ is an exact sequence too (I'm looking at tensor product over $\mathbb{Z}$; if $f : X \to Y$ is a morphism, then the induced morphism $M \otimes X \to M \otimes Y$ is given on generators by $m ... 3 I assume you are parametrizing$S^1$as$\mathbb{R}/2\pi\mathbb{Z}$. To find such an$\eta$, consider the map$f:\partial\Delta^2\to S^1$whose zeroth edge is$\gamma$, first edge is$\sigma$, and second edge is$\gamma$(here I am using the standard orientation for boundary faces of a simplex, so if you orient them to all be going around counterclockwise, ... 3 If$Q$is a projective module and$f : Q \to S$is a homomorphism, then there exists a map$f' : Q \to P_S$with$f = \pi \circ f'$with$\pi : P_S \to S$the canonical epimorphism. The important thing to note is that if$f$is surjective, then$\text{im}(f') + \ker(\pi) = P_S$. As$\ker(\pi)$is superfluous (it is the radical of$P_S$) we conclude that$f'$... 3 As the closed manifold is orientable, we have the powerful tool of Poincaré Duality. This means we have an isomorphism by capping with the fundamental class $$H_{n-1}M \stackrel \sim \to H^1M.$$ Now it is easy to see that by definition (from here on we don't use any orientability by the way) or universal coefficients, that$H^1M$is just$Hom(H_1M,\mathbb ...

3

Consider the long exact sequence of the pair $(M,\partial M)$ and notice that Lefschetz duality gives an isomorphism $H_k(M)\cong H^{n-k}(M,\partial M)\cong H_{n-k}(M,\partial M)$, using that $M$ is orientable and the coefficients are over a field. Consider the truncated exact sequence $H_m(\partial M)\to H_m(M)\to H_m(M,\partial M) \to H_{m-1}(\partial ... 3 You need to consider the homotopy type of a map between a wedge of spheres and the previous stage in the CW decomposition, and there's no way to guarantee that this is another wedge of spheres of the same dimension. For example, suppose you want to understand complexes built by starting from a wedge of$k$-spheres and attaching an$(n+1)$-cell. Then you ... 3 Let$f \in \operatorname{Hom}(\Bbb{Z}_p , \Bbb{Z})$. Then by the first isomorphism theorem,$\Bbb{Z}_p / \ker f$is isomorphic to a subgroup of$\Bbb{Z}$. Recall that nontrivial subgroups of$\Bbb{Z}$are infinite cyclic. But$\Bbb{Z}_p$has no infinite quotients, so necessarily$\Bbb{Z}_p / \ker f$is trivial! This implies that$\ker f = \Bbb{Z}_p$, i. e. ... 3 Observe that a non-orientable manifold$M$is$R$-orientable iff$R$contains a unit of order$2$, which is basically same as$2=0$in$R$. So if$M$is not orientable , then$M$is not$R$orientable for$Z_m$where$m\geq 3$. Now if$M$(closed connected n-manifold) is not$R$orientable then as theorem$3.26$in Hatcher (pg 236) suggested that there ... 3 1) Try and write down what you think it should be and see what goes wrong. What you actually get is a comultiplication$H_*(X) \to H_*(X) \otimes H_*(X)$, at least with field coefficients. I think things go wrong without field coefficients but I don't remember clearly why - it's been a while since I fiddled with this. E: There are much more intelligent ... 3 It's a formal sum. So$2v$means precisely the coefficient$2$multiplied by the formal symbol$v$. You add formal sums in much the same way as you add polynomials, termwise. The fact that this is useful might be surprising, but formal sums and formal power series come up quite often as a tool for organizing or structuring information. 3 Formal sums are just that: they are form, not content. You shouldn't think of$2[v]$as anything more than the number$2$written next to the symbol$[v]$. Now, as you realized, there's a special way you can think about$2[e]$for the case that$[e]\$ is an edge (which points from one vertex to itself); loosely, as a path that traces this edge twice. This is ...

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