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2

$S^2 \to T^2$ be a continuous map. As $S^2$ is simply connected, you can lift this map to the universal cover of $T^2$. But any map $S^2 \to \Bbb R^2$ is nullhomotopic by straightline homotopy. Now pushdown this homotopy to get a nullhomotopy of $S^2 \to T^2$. So every map from sphere to torus induces the zero map on the 2nd homology group. Yes, consider ...

1

Off the top of my head, the only examples I know where this holds are abelian varieties; in this case $X$ is topologically a torus, and we in fact have $H^k(X, \mathbb{Z}) \cong \bigwedge^k H^1(X, \mathbb{Z})$. As Roland says in the comments, $\mathbb{CP}^n$ for $n \ge 1$ is a counterexample.

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This is only possible for $n = 1$. When $n \ge 2$, $H^{2n}(\mathbb{CP}^n)$ is generated by $\alpha^n$ where $\alpha \in H^2(\mathbb{CP}^2)$, but since $H^2(S^{2n}) = 0$, $\alpha$ must map to zero in $H^{\bullet}(S^{2n})$, and hence so must $\alpha^n$.

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E: Sorry, I misread the original post. Let $M$ be a closed $n$-manifold. Then a degree $1$ map $S^{n} \to M$ is a homotopy equivalence. Proof: First, $M$ is simply connected; for otherwise we could factor this map through $\tilde M$. If $\tilde M$ is noncompact then $H_n(\tilde M) = 0$, so the map is degree zero; if $\tilde M$ is compact then the cover is ...

3

For $\textbf{part 2}$: The linear algebra is a bit messy, but you should get that $\ker \partial_1 = \langle a,b,c,d-i,d-h,f-g,e-f,d-e \rangle$ and $\text{Im } \partial_2 = \langle 2a-d+f,2b-f+h,2c+d-h, -a+c+2d-e-i, -d+2e-f,a-b-e+2f-g \rangle$ But, what's important is that $\partial_2(S_1 + S_2 + \ldots + S_6) = (a+e-d) + (a+f-e) + (b+g-f) + (b+h-g) + ... 0 I'd like to add a more general approach to this problem. Given a subcomplex$L$of a simplicial complex$K$, the homomorphism$C_n(L)\to C_n(K)$which sends any simplex in$L$(as a generator of the simplicial chain group) to the same simplex in$K$gives a chain map which fits into a commutative diagram below the relative chain group$C_n(K,L)$is ... 1 Define a map$f:C_n(K)\to\mathbb{Z}$by sending$\sigma$to$1$and every other$n$-simplex to$0$. Since$\sigma$is not a face of any$(n+1)$-simplex,$f$vanishes on any boundary, and thus induces a map$C_n(K)/B_n(K)\to\mathbb{Z}$. Restricting this to the subgroup$H_n(K)\subseteq C_n(K)/B_n(K)$, we get a map$\psi:H_n(K)\to \mathbb{Z}$. You might ... 1 Your confusion stems from the fact that you are using the wrong definition of$Z^n$and$B^n$. They are not defined to be$\ker(\partial^n)$and$\operatorname{im}(\partial^{n-1})$. Rather, they are defined to be the duals of$Z_n$and$B_n$; that is,$Z^n=\operatorname{Hom}(Z_n,G)$and$B^n=\operatorname{Hom}(B_n,G)$. The map$i^n$is then the map that ... 1 By exactness of the long exact sequence,$\ker(\delta_k)$is equal to the image of the map$H_k(C_*)\to H_k(C_*/F_0C_*)$, which is in turn the quotient of$H_k(C_*)$by the image of$H_k(F_0C_*)\to H_k(C_*)$. This is exactly your description of$G_1$. 1 If I understand your first question correctly, you ask whether you can find a$\varphi \colon \Gamma(T^{*}M) \rightarrow \Gamma(T^{*}M)$such that$\bigwedge^{k}(\varphi) \colon \bigwedge^k\Gamma(T^{*}M) \rightarrow \bigwedge^k\Gamma(T^{*}M)$coincides with the Laplacian acting on$k$-differentiable forms under the identification of ... 0 A lot of confusion arises in this problem due to an index shift performed during the calculation. I'm going to change the notation used slightly, but I'll hopefully define everything below (let me know if I miss something). First, I'll write the i-th face map$d^i:\Delta^{n-1}\rightarrow\Delta^n$: ... 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) = ...

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Every odd-dimensional manifold has vanishing Euler characteristic, so $$0 = \chi(M) = b_0 - b_1 + b_2 - b_3.$$We have $b_0 = 1$, and $b_3 = 0$ since $M$ is nonorientable. Hence, $b_1 > 0$ and thus $H_1(M, \mathbb{Z})$ is infinite.

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[This argument is stolen from the end of this answer (which handles the case $q=1$).] Let $1\leq i\leq n-1$ and $\alpha\in H_i(M,\mathbb{Z})$. By Poincare duality, there exists $\beta\in H^{n-i}(M,\mathbb{Z})$ such that $\alpha=z\cap\beta$. Since $H^{n-i}(S^n,\mathbb{Z})=0$, $f^*(\beta)=0$. Thus $0=f_*(i_n\cap ... 2 The transfer homomorphism of the orientation cover, composed with the pushforward, is multiplication by$2$. The image is free abelian since the$(n-1)$-integral homology of the orientation cover is by my answer here, so the only torsion in$H_{n-1}(X, \mathbb{Z})$is$2$-torsion. Applying the universal coefficient theorem for$H_n$with$\mathbb{Z}/2$... 2 Yes, this is true. The natural map$H_n(X,\mathbb{Z})\otimes \mathbb{Z}_p\to H_n(X,\mathbb{Z}_p)$is an isomorphism for$X=S^n$and$X=M$. We thus have the following commutative diagram, where the vertical maps are isomorphisms: $$\require{AMScd} \begin{CD} H_n(S^n,\mathbb{Z})\otimes\mathbb{Z}_p @>{f_*\otimes 1}>> H_n(M,\mathbb{Z})\otimes ... 2 You should be able to find a proof in most books covering homology theory. A specific reference is Theorem 4.59 of Hatcher's Algebraic Topology (p. 399-402). As Mike Miller commented, the idea of the proof is to show that any (co)homology theory can be computed by a "cellular chain complex" associated to that theory, and that these cellular chain complexes ... 1 A closed orientable n-manifold is called homology sphere if it has homology (or equivalently cohomology) of a sphere. Note that for a closed connected orientable 3-manifold we have always H_0M\cong \mathbb Z \cong H_3M and also H_2M \cong H^1M \cong Hom(H_1M,\mathbb Z). You see that H_1M=0 forces it to be a homology sphere. 5 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 ... 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 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 ... 5 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 ... 2 This is just the commutativity of the restriction maps with the cup product on the level of cochains combined with the fact that \tilde{H}_\bullet(X) = H_\bullet(X, *). 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 ... 2 From the commutative diagram here, H_\bullet(X, A \cup B) = H_\bullet(X, X) = 0, so if we can show the left arrow is surjective, we are done. However, on each factor, it is just a restriction map, so this is clearly the case. 0 If \mathbb{C} \setminus U has only finitely many components then you can proceed roughly as follows: Choose \varepsilon > 0 less than the distance of K to any other component of \mathbb{C} \setminus U. Then consider the set of all squares$$ Q_{k, l} = \{ x+iy \mid (k-1)\varepsilon \le x \le k\varepsilon, (l-1)\varepsilon \le y \le l\varepsilon ... 2 The problem is that you are performing an operation which is "not allowed": The cup product$\alpha\smile\alpha$is a cohomology class in$H^2(X;\mathbb Z)$. Now such a class is determined only up to coboundaries: If$φ$and$\psi$represent the same class, then$φ-\psi$is a coboundary, and that means it vanishes on cycles. So if$C$is a cycle in$C_2(X)$, ... 1 It is natural to assume that any reasonable definition of a$\Delta$-map should be based on the idea that we assign to every simplex in$X$a simplex in$Y$such that this assignment is compatible with the face operator. To see what that means, I'll introduce some combinatorics: Let us think of a$\Delta$-complex as a sequence$X_0,X_1,\dots$of sets, and ... 1 The claim that it is sufficient to show$\tilde H_{\le n}(Y)=0$if$Y^n$is a point is directly linked to the isomorphism$H_k(X,X^n)\cong \tilde H_k(X/X^n)$. We want to show that$H_k(X,X^n)=0$for$k\le n$. By the isomorphism, it suffices to show that$\tilde H_k(X/X^n)=0$for$k\le n$. Now$X/X^n$is homeomorphic to a CW complex whose$n\text{-skeleton}$... 0 I've done it myself but the full proof is a little bit long so here is the pdf file: https://www.dropbox.com/s/d368qulnyzn82v4/Poisson.pdf?dl=0. The idea is that there exists a bracket on differential forms that is trivial on de Rham cohomology for any Poisson structure and that coincides with Schouten-Nijenhuis bracket if Poisson structure is ... 2 It's completely immediate from the definition of the relative cup-product (cf. the answer to your other question). The relative cup-product is defined as the map induced in cohomology by the unique map that makes the following diagram commute (it's the upper horizontal one, the second horizontal map is a quasi-isomorphism so you can invert it in cohomology): ... 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: $$(\varphi \smile \psi)(\sigma) = \varphi(\sigma_{\mid[v_0, \dots, v_k]}) \cdot ... 3 As the closed manifold is orientable, we have the powerful tool of Poincaré Duality. This means we have an isomorphism by capping with the fundamental class$$H_{n-1}M \stackrel \sim \to H^1M.$$Now it is easy to see that by definition (from here on we don't use any orientability by the way) or universal coefficients, that H^1M is just Hom(H_1M,\mathbb ... 3 Observe that a non-orientable manifold M is R-orientable iff R contains a unit of order 2, which is basically same as 2=0 in R. So if M is not orientable , then M is not R orientable for Z_m where m\geq 3. Now if M (closed connected n-manifold) is not R orientable then as theorem 3.26 in Hatcher (pg 236) suggested that there ... 2 No, this is not correct. For instance, in H^2(BO(n);\mathbb{Z}/2), we have not just w_2, but also w_1^2, which is linearly independent from w_2 (over \mathbb{Z}/2). More generally, a basis for H^i(BO(n);\mathbb{Z}/2) over \mathbb{Z}/2 is given by the set of all monomials w_1^{m_1}w_2^{m_2}\dots w_n^{m_n}, where m_1,\dots, m_n are ... 1 Your part 1) is correct, however there is a justification that you failed to mention: all six vertices of the hexagon are identified to a single vertex in the quotient space. This needs to be said or else your Van Kampen application would be incorrect (I do see that you indicated this pretty clearly in your picture). For part 2), the six vectors you get ... 1 To do this by Fitting, P_S= \operatorname{im} (qp)^r \oplus \ker (qp)^r for some r, and one of these is zero as P_S is indecomposable. Since \lambda (qp)^r=\lambda we can't have (qp)^r=0 so \operatorname{im}(qp)^r=P_S and q is surjective. Now as Matthias says$$ 0 \to \ker q \to P_n \to P_S \to 0$$is exact and$P_S$projective, so$P_S | P_n$. ... 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$f'$... 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 ...

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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 ...

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

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

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" $\mathbb{Z} \cong \mathbb{Z} \otimes \mathbb{Z}$.

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I would suggest that you carefully read Example 3.7. in Hatcher's Algebraic Topology. And by carefully I mean really carefully and multiple times. It explains the case for a closed surface of genus $\geq 1$, and for $=1$ you get $S^1 \times S^1$. It covers the topic really well. If you want more intuition you can also read his introduction to cohomology (ch. ...

1

As soon as you have a cellular structure, it gets even easier. Pick a cellular structure on $X$, then $\bar X$ inherits a cellular structure. You get the chain complexes as $C_k(X)=\mathbb Z^{i_k}$ and $C_k(\bar X) = \mathbb Z[\pi]^{i_k}$. Considering that we have free $R$-modules for $R=\mathbb Z,\mathbb Z\pi$, the result gets pretty obvious. You can also ...

1

Let $P$ be the set of all functions $\pi\to A$. Make $P$ an abelian group by pointwise addition, and a $\mathbb{Z}[\pi]$-module by $(g\varphi)(h)=\varphi(hg)$ for $\varphi\in P$ and $g,h\in\pi$. Note that there is a canonical monomorphism $i:A\to P$ given by $i(a)(g)=ga$. So it suffices to show that $H^q(K(\pi,1),P)=0$ for all $q>0$. To show this, ...

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I do it using singular cochains. Functions from $\tilde\sigma:\Delta^n\rightarrow\tilde X$ to $A$ that are invariant under $\mathbf Z[\pi]$ are precisely those functions that are invariant under deck transformations, which is how $\pi$ acts on $\tilde X$. Hence there is a natural map collapsing orbits under the deck action that puts such functions in ...

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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

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

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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