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3

Let me write $B^n A$ for $K(A, n)$. The Yoneda lemma implies that $[B^n \mathbb{Z}, B^{np} \mathbb{Z}]$ is precisely the set of natural transformations $H^n(-, \mathbb{Z}) \to H^{np}(-, \mathbb{Z})$. An obvious candidate for such a natural transformation is the $p^{th}$ cup power $$H^n(-, \mathbb{Z}) \ni \alpha \mapsto \alpha^p \in H^{np}(-, \mathbb{Z})$$ ...


4

If $X \to Y$ is a finite Galois cover with Galois group $G$ then it's known that the induced map $H^{\bullet}(Y, \mathbb{Q}) \to H^{\bullet}(X, \mathbb{Q})$ is injective and induces an isomorphism $$H^{\bullet}(Y, \mathbb{Q}) \cong H^{\bullet}(X, \mathbb{Q})^G.$$ (This has nothing to do with $X$ and $Y$ being compact or manifolds.) So here the answer is ...


3

Let $x \in S^n$ not be in $A$. Then $A \subset S^n \setminus \{x\} \cong \Bbb R^n$, and this is contractible. The the map $H_n(A) \to H_n(S^n)$ factors through $H_n(A) \to H_n(\Bbb R^n) \to H_n(S^n)$, and must therefore be zero. So the sequence $0 \to H_n(S^n) \to H_n(S^n,A)$ is exact, and $H_n(S^n,A)$ is nontrivial as desired.


-1

Because while there is a short exact sequence at level $n$, you have to remember that homology is also about the relationship between what happens at degree $n$ and what happens at degree $n-1$ and $n+1$. Essentially you have a diagram like this (something which is usually "hidden" in the notation $0 \to C_*(U) \to C_*(X) \to C_*(X, U) \to 0$): ...


0

You're right that for any pair $(X, A)$, there is a split short exact sequence of chain complexes $$0 \to C_\bullet(A) \to C_\bullet(X) \to C_\bullet(X, A) \to 0$$ But it is not true that having short exact sequences at the chain level implies that the snake maps $\partial$ at the homology levels are zero, i.e., you get a short exact sequence at the ...


3

Consider the pair $(\Bbb T, A)$ where $A \subset \Bbb T$ is one of the longitudal circles in $\Bbb T$. We thus have the long exact sequence $$\cdots \to H_{k+1}(\Bbb T) \to H_{k+1}(\Bbb T, A) \stackrel{\partial}{\to} H_k(A) \to H_k(\Bbb T) \to H_k(\Bbb T, A) \stackrel{\partial}{\to} H_{k-1}(A) \to \cdots$$ $(\Bbb T, A)$ has homotopy extension property, as ...


0

I will outline a proof that the cohomology groups are preserved under homotopy equivalence. Every function is $C^{\infty}$. Let $f : M \to N$. Then $f^*$ (pullback map) induces a map between $H^k_{de}(N) \to H^k_{de}(M)$. Prove a generalization of Poincare's lemma that states that if $f,g : M \to N$ are homotopic then $f(\omega) - g(\omega)$ is an exact ...


7

There are many more 3-manifolds that arise as the boundary of compact contractible 4-manifolds than just those, including some Brieskorn spheres like $\Sigma(2,3,5)$. The keyword you want is Mazur manifold. It is my impression that, and I would be surprised if it weren't true, there is not much known about what 3-manifolds arise as the boundary of a Mazur ...


2

A very good, more recent introduction to Khovanov homologies was written by Dolotin and Morozov. Introduction to Khovanov Homologies. I. Unreduced Jones superpolynomial. Introduction to Khovanov Homologies. II. Reduced Jones superpolynomials. Introduction to Khovanov Homologies. III. A new and simple tensor-algebra construction of Khovanov-Rozansky ...


0

Let's concentrate on your case: you are trying to find an explicit homology between $h$ and $g$. Draw the line $0\times I$ which connects the two loops' basepoints. If you cut along this line the space becomes a square; draw one diagonal of this square. (Don't actually make the cut! I'm not trying to alter your space. I just invoked it to describe the ...


0

Purely algebraically, you know that $H_0=\bigoplus\mathbb{Z}$ is free of finite rank $n$, and $\tilde{H}_0\subset H_0$ is a submodule. By the structure of modules over a principal domain, any submodule of a free module is again free, hence $\tilde{H}_0$ is free and it will be of rank $n-1$. In fact, you don't need to know the structure of modules over a ...


2

The category $\mathcal{O}$ is closed under quotienting, submodules and finite direct sums, but not under extensions. Hence $Y$ need not be in $\mathcal{O}$ in general (see here).


1

There are two reasonable ways to define Cech cohomology, and they give the same answer in end. The exact answer to your question depends on which route you take. Let $X$ be a topological space. Let $U_i$ be an open cover, with the index $i$ running through some set $I$. For $i_0$, $i_1$, ..., $i_p \in I$, write $U_{i_0 i_1 \cdots i_p}$ for $U_{i_0} \cap ...


0

I don't understand what you mean by relative homology preserving injectivity and surjectivity, so consider instead the following general point. No interesting homotopy-invariant functor on spaces can preserve injections or surjections. The reason is that every map is homotopy equivalent to an injection and also to a surjection! There are very nice explicit ...


0

Your example already answers your question. $\iota_*: H_*(S^{n-1})\to H_*(S^n)$ is not injective. Also, $\pi: S^2\to\mathbb{RP}^2$ is surjective but the induced map on homology is not.


1

You can always construct a non vanishing volume form, e.g. by choosing a Riemannian metric. Call this $\eta$. Since $H^n_{dR}(M)$ is a one-dimensional vector space, any nonzero class $[\omega]$ is a multiple of $[\eta]$, say, $[\omega] = c[\eta]$, $c\neq 0$. So $[\omega] = [c\eta]$ and you have a nowhere vanishing representative of $[\omega]$.


3

Well, if you know already that $M$ is orientable, then there is $\eta$ a nowhere zero $n$-form on $M$. Let $C = \int_M \eta \in \mathbb R$. Now for any $[\omega] \in H^n(M)$, consider $D:= \int_M \omega$. This is nonzero. Then $[\frac{D}{C} \eta] = [\omega]$ as $$\int_M \frac DC \eta = D = \int_M \omega . $$ Note that $\frac DC \eta$ is nowhere ...


4

The Yoneda lemma says that any natural transformation such as $a$ between two representable functors ($H^i(-, \Bbb Z/2\Bbb Z)$ and $H^{i+k}(-, \Bbb Z/2\Bbb Z)$ in this case) is determined uniquely by maps between the representing objects. In our case this is a map $K(\Bbb Z/2\Bbb Z, i) \to K(\Bbb Z/2\Bbb Z, i+k)$, which corresponds precisely to the element ...


1

Let $x_1x_2\ldots x_n$ denote the element in $H^{n}(\Bbb R P^\infty \times \dots \times \Bbb R P^\infty; \Bbb Z/ 2\Bbb Z)$ that I am assuming is the image of the generator of $H^n(K(\Bbb Z/2\Bbb Z, n); \Bbb Z/ 2\Bbb Z)$ Instead of trying to compute $\operatorname{Sq}^I(x_1x_2\ldots x_n)$ completely, if you just want to demonstrate a monomorphism consider ...


5

The Hodge star is an isomorphism and you've transported the differential along that isomorphism, so you just get de Rham cohomology with the indices backwards. A more fun thing to do is to study both differentials at the same time, which leads to Hodge theory.


0

We have inclusions $(X_i,A\cap X_i)\hookrightarrow(X,A)$, inducing a map $H_n(X_i,A\cap X_i)\to H_n(X,A)$, and these maps together give a map $\iota:\oplus H_n(X_i,A\cap X_i)\to H_n(X,A)$. On the other hand, given a relative $n$-cycle $\gamma+C_n(A)$, the chain $\gamma$ can be expressed as $\gamma=\sum_i \gamma_i$, where $\gamma_i$ is the sum of those ...


0

I assume $n\geq 3$ below. The morphism $f:X\to Y$ in your question is a smooth projective family of $n-1$-dimensional hypersurfaces in $\mathbb P^n$. (You say in your first paragraph that the fibers are smooth hypersurfaces.) The cohomology of a single hypersurface of dimension $n-1$ in $\mathbb P^n$ in degree $\neq n$ is quite simple to compute by the ...


2

Let $a,b\in A\,.$ If $[a]_X=[b]_X\,,$ then $[a]_X-[b]_X$ is a boundary, hence there is a path in $X$ connecting $a$ to $b\,,$ so that $a,b$ lie in the same path connected component of $X\,.$ Since each path connected component of $X$ contains at most one path connected component of $A\,,$ $a,b$ can also be connected by a path in $A\,,$ hence we also have ...


1

I think your question is a little ambiguous so I will answer both potential questions. 1 An inclusion of spaces does not induce an inclusion of groups. As pointed out in the comments you can include a space with non-trivial homology into a space with trivial homology (i.e. same homology as a point). That would imply that there has to be an induced map in ...


3

An element of $H_0(X)$ is basically just a point. Take any $a\in A\,,$ then $a\in X$ generates $H_0(X)$ since any other point $x\in X$ can be connected to $a$ since $X$ is path connected, hence $x-a$ is a boundary, ie. $x-a\equiv 0\,,$ hence $x\equiv a\,.$


2

By the second isomorphism theorem, the condition that $HK=G$, or probably in your case $H+K=G$ (if we're using additive notation) is sufficient.


2

Presumably, you mean $\sigma_{i_0,\ldots,i_p} = -\sigma_{i_0,\ldots,i_{q-1},i_{q+1},i_q,i_{q+2},\ldots,i_p}$. Let us consider the case $p=1$. Then we want to show that $\sigma_{i_0,i_1}=-\sigma_{i_1,i_0}$ for any $(i_0,i_1)$. But $\delta\sigma = 0$ means that for all $i$, we have $$\sigma_{i_1,i}|_{V_i}-\sigma_{i_0,i}|_{V_i}+\sigma_{i_0,i_1}|_{V_i} = 0,$$ ...


0

First, I think it should be stated somewhere that $K$ needs to be abelian. Otherwise $H^2(Q,K,\theta)$ doesn't make much sens. In this case $\theta$ is a group morphism from $Q$ to $Aut(K)$. Secondly a generalisation is known and when $K$ needs not to be abelian. In that case $\theta$ is a group morphism from $Q$ to $Out(K)$, and the group cohomology ...


2

As Stefan Hamcke has indicated in his comments, one can obtain $S^2/A$ step-by-step by first pinching the equator to a point to get $S^2 \vee S^2$ and then pinching the pole to the wedged point to have $S^2 \vee S^2 \vee S^1$. This is homotopy equivalent to our space. Hence, $\widetilde{H_n}(S^2/A) \cong H_n(S^2 \vee S^1 \vee S^1)$ which is $\Bbb Z^2$ when ...


2

One place to look at is Topospaces (The Topology Wiki), which was suggested by a now-deleted user in a now-deleted answer. The description says This is a pre-alpha stage topology wiki primarily managed by Vipul Naik, a Ph.D. in Mathematics at the University of Chicago. We have over 400 articles including some material in basic point-set topology. As ...


2

First off, yes, you're right: contractibility is a very strong hypothesis, and is overkill if all you want to know is that closed $k$-forms are exact. You have heard about de Rham cohomology. Let me quickly recall what it is all about. Given a manifold $M$, the space $\Omega^k(M)$ is the (vector) space of $k$-forms on $M$. You have the linear map $d_k : ...


1

You are right that an extra degree is needed in the hypotheses. Conceptually, you don't expect the zero-th groups of something contractible to vanish, because you should have the homology of a point. Note that $H_0(P_*)$ is only zero if $P_1 \to P_0$ is onto. If there existed a $D_{-1}: 0 \to P_0$ satisfying $D_{-1}d_0 + d_1D_0 = \operatorname{Id}$, then ...


2

The Mayer-Vietoris sequence goes $\cdots \rightarrow H^{n - 1}(U) \oplus H^{n - 1}(V) \rightarrow H^{n-1}(U \cap V) \rightarrow H^{n}(M) \rightarrow H^{n}(U) \oplus H^{n}(V) \rightarrow 0$. Note that since $V \cong \mathbb{R}^{n}$ we have $H^{n}(V) = 0$. We also already know that $H^{n-1}(U \cap V)$ and $H^{n}(M)$ are both $\mathbb{R}$, so in particular they ...


0

You have the right idea, but you're line of thought is not entirely correct. The map $$S^{n-1} \hookrightarrow U \setminus \{ x \} \hookrightarrow \Bbb{R}^{n} \setminus \{ x \}$$ is indeed a homotopy equivalence, so the composition $$H^{n-1}_{dR}(S^{n-1}) \rightarrow H^{n-1}_{dR}(U \setminus \{ x \}) \rightarrow H^{n-1}_{dR}(\Bbb{R}^{n} \setminus \{ x \})$$ ...


0

This actually can be done by some basic advanced calculus technique. In particular the Stokes Theorem. First by a translation and a scaling, assume that $x$ is the origin and the unit sphere $S$ is in $U$. Let $$\eta = \sum_{i=1}^n (-1)^{i-1} x^i dx^1 \wedge \cdots \wedge \hat{dx^i} \wedge \cdots dx^n,\ \ \ \ \omega = \frac{1}{r^n} \eta,$$ where $r = ...


3

In general it really depends on how well you understand $f$ and its action on the cohomology of $X$. Here are two nice special cases where you only need to understand the action of $f$ on a single cohomology group and then everything else is determined by the cup product. Tori Let $\Gamma$ be a lattice in $\mathbb{R}^n$. (You can take $\Gamma = ...


0

Here is the simplest way I know to see this. Consider the case where $E = L_1\oplus\dots\oplus L_n$. Then with the direct sum connection on $E$, the curvature $2$-form $F^\nabla$ will be the matrix with the curvature $2$-forms $F^{\nabla_1},\dots,F^{\nabla_n}$ of the respective line bundles on the diagonal. So $\text{tr}\,F^{\nabla} = \sum\limits_{j=1}^n ...


0

Let $F$ be a locally free sheaf on an $n$-dimensional complex manifold $X$. The obvious is to consider the complex $$ O_X(F) \to A_X^{0,0,r}(F) \to A_X^{0,1,r-1}(F)\to \dots \to A_X^{0,n,r-n}(F) \to 0 $$ where $A_X^{p,q,s}(F)$ denotes the space of $(p,q)$ forms on $X$ with values in $F$ of class $C^s$, and all differentials are $\overline \partial$. ...


2

The first relation: take any vector space with subspaces $C, Z\supseteq B$. The natural surjection $Z/B \to (Z+C)/(B+C)$ has kernel $(Z\cap C +B)/B$ which is isomorphic to $Z\cap C /B\cap C$ so the isomorphism you want follows from the first isomorphism theorem. $\partial^0$ is the map formed as the direct sum of the maps $C_{dp}/C_{d,p-1}\to ...


0

@ Sadok Kallel: The book Topology and Groupoids has Chapter 11 on "Orbit spaces, orbit groupoids". The main result is to give circumstances under which the morphism of fundamental groupoids $\pi_1(X) \to \pi_1(X/G)$ presents the latter as an orbit groupoid. As a consequence, 11.5.4 calculates the fundamental group of a symmetric square.



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