New answers tagged

0

You are correct. The first displayed formula on page $95$ is $$ \delta^{p-1} : \operatorname{Hom}(\mathsf{C}_{p-1}, G) \to \operatorname{Hom}(\mathsf{C}_p, G)$$ and on the first line of that page (before the displayed formula), $\mathsf{C}^p$ is defined as $\operatorname{Hom}(\mathsf{C}_p, G)$. With this definition, the above can be rewritten as $$\delta^...


7

Following Mike's nice hint, note that $\pi_{1}(\mathbb{R}\mathbb{P}^{3}) = \mathbb{Z}/2\mathbb{Z}$, and $\pi_{1}(S^{2} \times S^{1}) \cong \pi_{1}(S^{2}) \times \pi_{1}(S^{1}) \cong \mathbb{Z}$, so the induced map $\pi_{1}(\mathbb{R}\mathbb{P}^{3}) \to \pi_{1}(S^{2} \times S^{1})$ must be the zero map. Recalling that $p \colon S^{2} \times \mathbb{R} \to S^{...


5

The cohomology ring of $\Bbb{CP}^n$ is $\Bbb Z[x]/(x^{n+1})$, with $|x|=2$. Any map $f: \Bbb{CP}^n \to S^2$ induces zero on cohomology when $n>1$, because if $z \in H^2(S^2)$ is a generator, then $0 = f^*(z^2) = f^*(z)^2$, so $f^*(z)$ must be zero. Now use the fact that the universal coefficient theorem is natural to see that the induced map on $H_2$ is ...


4

Let $X=[0,1]$ and $A=\{0,1,1/2,1/3,\cdots\}$. Then the quotient $X/A$ is homeomorphic to the Hawaiian earring which has uncountable $H_1$ (try to prove this). On the other hand, $H_1(X,A)$ is isomorphic to $\bigoplus_{i=1}^\infty \Bbb Z$, which is countable.


2

Let $H\subseteq G$ be a subgroup. There are two standard maps on the group cohomology associated to this: $res_H^G:H^*(G, M)\to H^*(H, M|_H)$ and $tr^G_H:H^*(H, M|_H)\to H^*(G, M)$. We have the relation $tr\circ res(x)=[G:H]x$. Now we wish to show that in the case $H=S$, a Sylow-group of $G$, we have that $res_S^G$ is an injection. Letting $x\in Ker(res_S^G)$...


2

They mean to have the locally path-connected assumption, presumably, but didn't feel the need to state it (often "space" can be taken to mean "non-terrible space"). As a counterexample in general, take the Warsaw circle $X$. This has trivial homotopy groups and homology groups (exercise), but collapsing the 'bad part' gives us a non-null map $X \to S^1 =: Y$...


1

I've recently struggled with the same sort of things, so I'll try to explain how I understand it. This is all supposing you understand the technical tools such as the isomorphism $H_n(X,A) \cong H_n(X/A)$, naturality, degree theory, and the likes. First, let me say that there are a variety of levels you can make this argument so "the right way to see it" ...


3

A quick proof using Stiefel–Whitney classes: a manifold $M$ is orientable iff the first SW class $w_1(M) \in H^1(M;\mathbb{Z}/2\mathbb{Z})$ is zero. But by the universal coefficient theorem, $$H^1(M;\mathbb{Z}/2\mathbb{Z}) = \operatorname{Hom}_\mathbb{Z}(H_1(M;\mathbb{Z}), \mathbb{Z}/2\mathbb{Z}) = 0.$$ Of course under the hood I don't think there's ...


9

For any connected manifold $M$, there is a homomorphism $\pi_1(M)\to\mathbb{Z}/2$ which sends a loop to $0$ if going around the loop preserves orientation and sends the loop to $1$ if going around the loop reverses orientation. This homomorphism is trivial iff $M$ is orientable. Since $\mathbb{Z}/2$ is abelian, this homomorphism factors through the ...


0

I think that I can give an answer, at least for the very first part of my question. Assume that $GL_{n}(\mathbb{R})$ is the general linear over $\mathbb{R}$. It works almost in the same fashion for $\mathbb{C}$ as well. Then, someone can prove that the "natural map" $\rho:V_{n}{\mathbb{R}^{\infty}} \rightarrow G_{n}{\mathbb{R}^{\infty}}$ from the $n$-frames ...


4

It is not true that $(-(j_U)_* \omega , (j_V)_*\omega)=0$ iff $\omega=0$. Note that $(j_U)_*\omega$ and $(j_V)_*\omega$ here are cohomology classes in $H^1_c(U)$ and $H^1_c(V)$, not just $1$-forms. So we have to consider the possibility that they might be coboundaries. A $1$-form on $U$ is a coboundary iff its integral is $0$, and similarly for $V$. So $...


2

Rudyak's article The problem of realization of homology classes from Poincare up to the present seems to be a good starting point. The original paper is Thom's Quelques propriétés globales des variétés différentiables.


2

The trouble is that an object of the functor category $[\cal B,\cal C]$ need not be projective even if every object in the image is. I think what you would get would be a weak homotopy (homotopy at each object of $\cal B$). For more on this subject, we my book titled Acyclic Models, which deals with every version of the theorem I am aware of.


5

Consider an embedding of a torus $N$ inside $M=(\mathbb R^3 \text{with a line removed})$, such that the torus is the boundary of a regular neighborhood of a curve that goes around the line once. Since $M$ is homotopy equivalent to a circle, the induced map on $H_2(N)\to H_2(M)$ is trivial. Yet on the level of $H_1$ we have a nontrivial map $\mathbb Z\oplus \...


1

Functoriality in the first variable only says that for $k = 0$ (morphisms have degree $0$). If $f$ is a morphism of degree $k$ then it induces a map $$[X[k], Y] \to [W, Y]$$ and there are some signs involved in moving this shift $[k]$ around.


1

So Roland put you onto the right track, but just so you have something to check against later (and for those who may come after): Let $\sigma: \Delta^2 \to X$ be defined by first projecting $\Delta^2$ onto the edge $[v_0, v_2]$ (orthogonally, so that $v_1$ ends up mapping to the midpoint of $[v_0, v_2]$), and then mapping via $f\cdot g: [v_0, v_2] \to X$. ...


1

Yes, the candidate is $BG$ the classifying space of $G$. You can endow $G$ with the discrete topology, you have the universal bundle $EG\rightarrow BG$ which is a covering and $EG$ is contractible $BG$ is a $K(G,1)$, apply Serre 1.5 p.91 Serre, Jean Pierre. Cohomologie des groupes discrets. http://www.maths.ed.ac.uk/~aar/papers/serrecoh2.pdf


1

When you apply the Kunneth theorem, the Tor is being taken over $\mathbb{F}_p$, not over $\mathbb{Z}$, and so it vanishes. Over a field the Kunneth theorem just says that the homology of a product is a graded tensor product of homologies. The assumption that $p > 3$ does not help you in any way.


3

A. De Rham cohomology. i) some general topology (basic notions: what topological spaces are, compactness, connectedness) ii) some smooth manifold theory (basic notions: what manifolds are, tangent spaces) iii) some linear algebra (basic notions: vector spaces, exact sequences, quotient spaces) B. Sheaf cohomology. i) some sheaf theory (basic notions: what ...


0

You nearly have solved your problem. Observe the connecting homomorphisms $\partial_*$ of the long exact sequence you used $$\require{AMScd} \begin{CD} \cdots @>{\partial_*}>> H_i(X,pt) @>{i_{1*}}>> H_i(X \vee Y,pt) @>{}>> H_i(X \vee Y, X) @>{\partial_*}>> \cdots\\ @.@.@VV\pi_{2*}V @A{\cong}Ak_*A \\ @.@.H_i(Y,pt)@= H_i(Y,...


2

In the case of the Floer homology of the cotangent bundle the answer is yes. You should have a look at this: The Viterbo transfer as a map of spectra by Thomas Kragh.


2

Following Mike Miller's suggestion, consider the cylinder $X =S^1 \times \mathbb{R}$ (as a Riemann surface, you may view it as either $\mathbb{C} \setminus \{0\}$ or $\mathbb{C}/\mathbb{Z}$). As this deformation retracts onto the base circle and homology is a homotopy invariant, we know that $H_2(X;\mathbb{C}) \cong H_2(S^1;\mathbb{C}) =0$. As $\mathbb{C}$ ...


3

All the constructions that you used to define the isomorphism are natural/functorial: Given a map $X \to Y$, you have a natural map that respect inclusions, which gives a starting point for all the applications of naturality to come: $$(\Sigma X, C_+ X, C_- X, X \times \{0\}) \to (\Sigma Y, C_+ Y, C_- Y, Y \times \{0\});$$ The long exact sequence of a pair ...



Top 50 recent answers are included