# Tag Info

1

The introduction to this book (it is all about mod 2 (co)homology) gives a nice example of a result that was first proved with mod 2 homology. It provided the tools to generalize Poincaré duality to all closed manifolds, regardless of orientibility. The author also mentions that they are used to develop the theory of spin structures and in some cobordism ...

2

The answer is yes for the following reason: $\omega_n$ is the Euler class mod 2 Now use e.g. Theorem 4.7 here which says $e(\nu_N)([N])$ counts number of intersections, or you argue that the Thom class of $\nu_N$ in $M$ is the Poincaré dual of $N$ (follows from this exercise). Hence, by pulling back to the cohomology of $N$ the result follows. You ...

0

Hints: If $F$ is a field then it follows from the universal coefficient theorem (for homology and co-homology) that $H^*(X;F)= Hom_{F-mods}(H_*(X;F);F)$. This implies if $f_*$ is surjective then $f^*$ is injective.

1

The Solution is correct but more complicated than necessary. One can just show that $X$ and $\partial D$ are not homotopy equivalent by Computing their fundamental group or First homology. Namely, $\partial D$ is homeomorphic to $S^1$, so $\pi_1=H_1=Z$. But $X=T^2-D^2$ is homotopy equivalent to the wedge of 2 circles, so $\pi_1=Z*Z$ and $H_1=Z\oplus Z$.

1

The inclusion $* \times X \subset S^1 \times X$ is a cofibration (the singleton $*$ is a sub-CW-complex of $S^1$, hence $* \subset S^1$ is a cofibration, and by this question the product of a cofibration with an identity map is a cofibration), thus: $$H^*(S^1 \times X, * \times X) \cong \tilde{H}^*\bigl( (S^1 \times X) / (* \times X) \bigr)$$ (this is a ...

0

[Note that you should write something like $\mathbb{Z}\{x_0,x_1\}$ rather than $\mathbb{Z}[x_0,x_1]$, since the latter usually means the polynomial ring over $\mathbb{Z}$ in $x_0$ and $x_1$.] The definition of the reduced homology is not the quotient of $\mathbb{Z}\{x_0,x_1\}$ by the subgroup generated by $x_0-x_1$; rather, it is the subgroup of ...

0

This answer is due to Ammar Husain. It is well-known that $H_i(M;\mathbb Z^l) \cong H_i(BM; \mathbb Z^l)$, where $BM$ is the (fat or usual) geometric realization of the nerve $BM_\bullet$ of the category $M$. Via the set map \begin{eqnarray} BM_n &\rightarrow& (BM^{\rm op})_n \\ (m_n, \ldots, m_1) &\mapsto& (m_1, \ldots, m_n) \end{eqnarray} ...

1

You are considering $\mathbb{C}$ as a trivial $L$ module. So for any $X \in L$ and any 1-cocyle $\eta$ the value of $X\eta(Y)$ is 0. This is why you ignore all cross terms.

1

You know, since $\alpha$ is a cocycle, that $\alpha \circ d^A_n = 0$. However, you do not know that $\hat\alpha \circ d^X_n = 0$! So $\hat\alpha$ is not even a cocycle. What you know is that, if $i_* : C_*(A) \to C_*(X)$ is the inclusion, then $$\hat\alpha \circ d^X_n \circ i_{n+1} = \hat\alpha \circ i_n \circ d^A_n = \alpha \circ d^A_n = 0,$$ and so by ...

3

Take $C_*=0\to \mathbb{Z}\stackrel{2}\to\mathbb{Z}\to 0$ (with the $\mathbb{Z}$s in degree $0$ and $1$ and every other term in the complex $0$) and let $D_*=0\to\mathbb{Z}/2\mathbb{Z}\to 0$ (with the $\mathbb{Z}/2\mathbb{Z}$ in degree $0$ and every other term in the complex $0$). Then there is a unique nonzero chain map $f:C_*\to D_*$, and this induces an ...

1

Let $K$ be a number field and $G = \text{Gal}(\overline{K}/K)$ its absolute Galois group. The open subgroups of the profinite group $G$ are precisely the "lifts" of the identity $1\in\text{Gal}(L/K)$ (more precisely, kernels of projection maps $G\to \text{Gal}(L/K)$) for finite Galois extensions $L$, and there are countably many $K$-field extensions of any ...

3

Let $\alpha$ be an $n$-cell of $X$ and $\beta$ be an $n$-cell of Y. Then $f(\alpha)=\sum_{\beta\in J_n'} y_{\alpha \beta} \beta$. Here $J_n'$ denotes the set of $n$-cells of $Y$. We wish to determine the values of $y_{\alpha \beta}\in \mathbb Z$. Let $\varphi_\alpha:S^{n-1}\to X^{n-1}$ be the attaching map of $\alpha$. Let $\overline{f}:X^n/X^{n-1}\to ... 0 This really isn't even true when$M$is a point (this is really the only case I am familiar with). The symplectic blow-up is the natural generalization of the blowup in the complex setting (i.e. the birational isomorphism$\Bbb CP^2 \# \overline {\Bbb CP} ^2 \to \Bbb CP^2$). This is relatively clearly not a fiber bundle, and it would be completely impossible ... 1 Ok it seems I've found what I was looking for. It is a consequence of the axioms of a multiplicative cohomology theory. The question I've written may seem a little confused, but I think there is a splendid explanation on Tammo tom Dieck's Algebraic Topology at page 413, Proposition 17.3.1 clearly,$\sigma$is the suspension isomorphism he defined ... 1 Cohomology appears naturaly. Lots of problems are solved locally and then the local solutions glued together to construct global solutions, and cohomology is the precise description of how to do that. 0 Inverses are unique up to unique isomorphism: that is, if$F : C \to D$is a functor and$G_1, G_2 : D \to C$are two inverses to it, there's a unique natural isomorphism$G_1 \cong G_2$compatible with the data of being an inverse. (This might require some subtle modifications to the naive notion of "the data of being an inverse": a definition that should ... 1 The linear isomorphism$f_A: \mathbb{R}^n \to \mathbb{R}^n$given by multiplication by$A$is a diffeomorphism essentially by definition of the standard smooth structure on$\mathbb{R}^n$. Then it is a general fact that a diffeomorphism, when restricted to an open submanifold, is a diffeomorphism onto its image. In this case, the image of$(f_A) ...

1

First, instead of infinite products you should use infinite sums. Then simply define $$\partial \sum_{\alpha} c_\alpha \sigma_\alpha := \sum_{\alpha} c_\alpha \partial \sigma_\alpha$$ for each locally finite infinite sum.

1

What you've said is correct, though of course it requires proof.

0

I assume that you are working over $\mathbb{C}$, otherwise it does not make sense to talk about DeRham cohomology. By hodge decomposition theory, $H^2(X,\mathbb{C} )=H^{0,2}\oplus H^{1,1}\oplus H^{2,0}$. By GAGA principal, $H^{0,2}$ and $H^{2,0}$ are isomorphic to cohomology of sheaf of algebraic differentials. $H^2(X,\mathcal{O}_X)=0$ by Grothendieck ...

1

As explained in the other answer, if $(X, A)$ is a CW-pair, i.e., $X$ is a CW-complex and $A$ is a subcomplex, then one has the long exact sequence of homology groups $$\cdots \longrightarrow \tilde{H}_{k-1}(X/A) \longrightarrow \tilde{H}_k(A) \stackrel{i_*}\longrightarrow \tilde{H}_k(X) \stackrel{q_*} \longrightarrow \tilde{H}_k(X/A) \longrightarrow ... 2 What you wrote is in general completely false. Consider A = [0,1] and B = \{0,1\}. Then H_1(A) = 0 = H_1(B), but A/B is a circle and H_1(A/B) = \mathbb{Z}. In general what is true is that if the inclusion B \subset A is a cofibration (for example if B is a sub-CW-complex of A), then the reduced homology group \tilde{H}_k(A/B) is isomorphic ... 5 Others have given the first reasons that arose historically, and that relate to a first course in algebraic topology. But a further point comes up from a more advanced perspective: sheaf cohomology has a natural definition using injective resolutions, and these exist in any category of sheaves. You can define sheaf homology by projective resolutions but ... 2 You can compute this from the described pathspace spectral sequence, using the fact that P is contractible so every nontrivial term in the spectral sequence except the (0,0) term must eventually disappear. Since H^*(K(\mathbb{Z},1);\mathbb{Z}) vanishes unless *=0 or *=1, the only possible nontrivial differentials on the spectral sequence are the ... 16 Quote from Elements of Algebraic Topology by J. R. Munkres: One answer is that cohomology appers naturally when one studies the problem of classifying, up to homotopy, maps of one space in other. Another is that cohomology is involved when one integrates differential forms on manifolds. Still another answer is [...] that the cohomology groups have an ... 30 Fundamental reason why cohomology is more powerful is that singular cohomology has a natural ring structure over them. If \varphi and \psi are R-valued m and n-cochains on a topological space X, then one can define an (m+n) cochain \varphi \smile \psi by$$(\varphi \smile \psi)([v_0, \cdots, v_{m+n}]) = \varphi([v_0, \cdots, v_m]) ...

2

Your mistake is in the equality $\tilde{H}^2(S^2 \times (D^2, S^1)) = \tilde{H}^2(S^2 \times S^2)$ you wrote. It might be a bit easier to see why this is false in one dimension lower. Consider the trivial bundle $S^1 \times D^1 \to S^1$ instead. In the proof you would consider $(S^1 \times D^1, S^1 \times S^0)$, and it might be tempting to write $H^1(S^1 ... 1 Let$M$be a closed connected manifold, and$Y\subset M$a dense open subset, missing at least one point. Suppose Y is connected (otherwise$H_0$distinguishes$Y$and$M$). Then$H^n(M,\mathbb{Z}/2)=\mathbb{Z}/2$but$H^n(Y,\mathbb{Z}/2)=0$as it is a non-compact n-dimensional manifold. If one adds orientable to the assumptions of$M$then we get this for ... 4 A topological space$X$is called$k$-connected if$\pi_i(X) = 0$for$0 \leq i \leq k$. The Hurewicz theorem states that if$X$is$(n-2)$-connected, then the Hurewicz homomorphism$h_* : \pi_n(X) \to H_n(X)$is surjective. In your example, while$\pi_2(\mathbb{RP}^5) = 0$,$\pi_1(\mathbb{RP}^5) = \mathbb{Z}/2\mathbb{Z} \neq 0$, so$\mathbb{RP}^5$is not ... 1 For the hurewiczs homomorphism, ALL the homotopy groups$\pi_i, i\leq n$have to be zero to have a surjective morphism$\pi_{n+1}(M)\rightarrow H_n(M)$. In your case,$\pi_1(RP^5)$is not zero. 2 This is true if you assume$H_n(X)$is finitely generated for all$n$(all coefficients in this post will be$\mathbb{Z}$). In particular, this holds if$X$has the homotopy type of a CW-complex with finitely many cells in each degree. To prove this, we invoke the classification of finitely generated abelian groups, which says that$H_n(X)$is a finite ... 3 These conditions are equivalent if$X$is a levelwise finite CW complex (finitely many cells of each dimension), since this condition ensures that each homology group is finitely generated. You can prove this using the universal coefficient theorem as described here; it ensures that the torsion subgroup of$H_k(X, \mathbb{Z})$is isomorphic to the torsion ... 1 Let$i : S^1 \to X$be a closed curve$N$. We say$N$is nullhomologous in$X$if$i_*[S^1] \in H_1(X)$is zero, where$[S^1]$is the the fundamental class of$S^1$. Likewise, we say$S^1\times N$is nullhomologous in$S^1\times X$if$(\operatorname{id}_{S^1}, i)_*[S^1\times S^1] \in H_2(S^1\times X)$is zero. Under the Künneth isomorphism$H_2(S^1\times ...

1

First, you mixed up exponents and indices quite liberally in this question... There's a difference. The map $f$ is a cocycle in $\operatorname{Hom}(X,Y)$ of degree $p+q$. So first, by definition of the degree in the chain complex $\operatorname{Hom}(X,Y)$, it will map a $p$-cochain of $X$ to a $q$-cochain of $Y$.* So since $z_2 \in Z^p(X)$, $f(z_2) \in ... 0 To show that is an excision it suffices to prove that it induces isomorphisms in the homology. By a lemma in "Elements of algebraic topology" in Hatcher's book, a map of pairs$ f:(X,A) \rightarrow (Y,B)$induces isomorphisms at the homology level if$f:X \rightarrow Y$and$f|A \rightarrow B$are both homotopic equivalences.$\mathbb{S}^{n} ...

1

The answer is given in the comments. In general you can't say if a certain map is injective or surjective without further information. Knowing that certain groups are zero will show that certain maps are injective (the one after the map that has as a domain the zero group) or surjective the map before the one mapping to zero). It is also good to know that a ...

3

Here's a Mayer-Vietoris argument. Write $X$ as a union $\cup_{i=0}^\infty T_i$ where each $T_i$ is a torus with two boundary components (except $T_0$, which has only one boundary component). Let $A= \sqcup_{i=0}^\infty T_{2i}$ and $B= \sqcup_{i=0}^\infty T_{2i+1}$. Then $A \cap B$ is a countably infinite disjoint union of circles $\sqcup_{i=1}^\infty ... 4 Here is an answer without Mayer–Vietoris – maybe it's possible to use the technique you describe, but the following approach seems easier to me. Homology commutes with (some) directed colimits. You can e.g. find a precise statement in Hatcher's book Algebraic Topology, Proposition 3.33. Your space$X$can naturally be seen as a directed colimit ... 2 Actually, the answer is almost yes on the level of forms themselves. You've provided an inner product on$\Omega^k(M)$, and once you pass to its Hilbert space completion (the so-called space of$L^2$forms) it is true, by the Riesz representation theorem, that any continuous linear functional$\Omega^k_{L^2}(M) \to \Bbb R$is uniquely represented by$\alpha ...

5

The Hodge duality gives the non degeneracy of the pairing. The Hodge star define a duality on de Rham forms, that is, if $\alpha$ is a non-zero $k$-form then $\star \alpha$ is a non-zero $(n-k)$-form. The defining property of the $\star$ operator is indeed given by $\beta \wedge \star \alpha = \langle \alpha, \beta \rangle \operatorname{vol}$, where ...

5

Not in general. Suppose that all of the $C_n = \oplus_{i = 0}^{\infty} \mathbb{k}$, with differentials $0$, and let $G = \mathbb{k}$. Then $Hom(C_n, \mathbb{k}) = \Pi_{i = 0}^{\infty} \mathbb{k}$. Taking the $\mathbb{k}$ dual again gives something containing as a subspace $\Pi_{i = 0}^{\infty} \mathbb{k}$, with zero as the differentials. The cohomology ...

4

If $G = \mathbb R$ (or any field), and $C$ is the simplical chain complex associated to a countably infinite discrete space (e.g., the integers), then $H_0 = Z_0$ will be an infinite direct sum of copies of the reals, but $Z^0$ will be the dual space of that, and the double dual will not be isomorphic to the original (because of the inifinite dimensions), ...

1

First, note that $\mathrm{Hom}_{\mathbb{Z}}(\mathbb{Z}G,A)^S=\{\phi\in\mathrm{Hom}_{\mathbb{Z}}(\mathbb{Z}G,A)\mid s.\phi=\phi\}$. Now, the action of $S$ on $\mathrm{Hom}_{\mathbb{Z}}(\mathbb{Z}G,A)$ is $$(s.\phi)(g)=\phi(gs)$$ so $\phi\in\mathrm{Hom}_{\mathbb{Z}}(\mathbb{Z}G,A)^S$ if, and only if, it is constant on the cosets of $S$ in $G$ (i.e. ...

1

Using the cohomological description of Br(k) as $H^2$($k^s$/k, $k^s$ $^*$) (where $k^s$ denotes a separable closure), one can give a first approach to your problem for a Galois extension K/k with group G, via the Hochschild-Serre spectral sequence: because of Hilbert 90, we get a 5 term exact sequence which shows that the kernel Br(K/k) is isomorphic to ...

3

One has the following quotient map $q : (Y \times I, Y \times \partial I) \to (Y \times S^1, Y \times s_0)$ obtained from the equivalence relation $\sim$ on $Y \times I$ defined on $\partial I \times Y$ by $(y, 0) \sim (y, 1)$. One can prove that $q_*$ on cohomology is an isomorphism as follows. $$\require{AMScd} \begin{CD} H^k(Y \times I, Y \times ... 4 Since X is contractible, X/A is naturally homotopy equivalent to the suspension \Sigma A. (Explicitly, contractibility of X allows you to extend the inclusion A\to X to a cone CA on A, and then you get an induced map \Sigma A=CA/A\to X/A. From the long exact sequences on cohomology it is immediate that this map induces isomorphisms on ... 3 Yes, this is true over \mathbb{Z} by the universal coefficient theorem over \mathbb{Z}. For the question with "homology" and "cohomology" switched see this math.SE question. 1 This is true for any finitely generated module M over a commutative noetherian ring A: M is then itself a noetherian A-module. If f\colon M\rightarrow M is a surjective A-linear map, then the chain of A-submodules of M$$ \ker(f)\subseteq \ker(f^2)\subseteq\dotsb\subseteq \ker(f^n)\subseteq\dotsb  becomes stationary as $M$ is noetherian. ...

1

Intuitively, it just says that the image of $f_n$ goes $n$ times around the circle which is clear. Let $T$ be a triangulation of the circle that consists of $3|n|$ edges, and let $S$ be a triangulation of the circle that consists of $3$ edges, assuming the usual orientation. Note that $f_n$ is homotopic to a simplicial map $g_n: T\to S$ that sends each of ...

1

The definition $P_i = \mathbb{Z}\left[G^{i+1}\right]$ should be made for all $i \geq -1$, not only for $i \geq 0$. Thus, $P_{-1} = \mathbb{Z}\left[G^{0}\right]$. But $G^0$ is the trivial group consisting of the one element $\left(\right)$ (the empty list, a.k.a. the $0$-tuple). If you plug in $i = 0$ into the definition of $d_i$, you obtain (1) ...

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