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From the outset, I would use the fact that both connected components of $\mathrm{GL}_n(\mathbf R)$ are path-wise connected, assuming that $n>0$. Hence, it suffices to prove the statement for the identity matrix and the diagonal matrix $\mathrm{diag}(-1,1,\ldots,1)$. Then you apply your argument to these two maps. For the former you get the identity on ...

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If you'd like a super-summerized introduction you may want to check these course notes on singular homology : http://math.arizona.edu/~rwilliams/SingularHomology.pdf Again there is no proof or anything and it only covers the basic but I found it helpful just to, you know, see the big picture. Hope that helps.

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No, having a CW complex structure is absolutely not enough to be a manifold. Being a CW complex is very easy, being a manifold is hard. Take the wedge sum of two circles $S^1 \vee S^1$ for example: it's a CW complex, but not a manifold. If a space $M$ is a compact manifold without boundary, then you can read its dimension using singular homology: it is the ...

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$\newcommand{\Z}{\mathbb{Z}}$ The question is equivalent to when $X$ can be written as a product of eilenberg mclane spaces. First why the question is important: The motivation for introducing Postnikov towers is to show that every space can be written as a fibered product of Eilenberg MacLane spaces. It is desirable to know if the $k$ invariants, the ...

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All of the homology groups of $\mathbb{S}^n$ are trivial, except of top and bottom one. The induced map $H_0(\mathbb{S}^n) \to H_0(\mathbb{R}\mathbb{P}^n)$ will always be isomorphism (this is very easy to calulate). The top homology group $H_n(\mathbb{R}\mathbb{P}^n)$ is either trivial or $\mathbb{Z}$, depending on whether $\mathbb{R}\mathbb{P}^n$ is ...

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From a short exact sequence $$0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0,$$ which is easier than the case you describe, knowing $A$ and $B$ in general is not enough to compute $C$. This is the extension problem. Some examples $$0\rightarrow \mathbb{Z}\rightarrow \mathbb{Z}\rightarrow 0\rightarrow 0,$$ or $$0\rightarrow ... 1 Let \pi \colon M \to S^1 denote the usual projection and let C denote a non-zero homologous simple closed curve with no self intersections, i.e. an injective map c from S^1 to M, which induces something non-zero on homology. We can isotop c s.th. \pi \circ c is a covering of S^1 (This is not that hard, you just make the map locally ... 4 If f:M\to M is a diffeo, then there is a linear map f^p:H^pM\to H^pM and then we can consider the spaces$$H^p(M)_f=H^p(M)/(1-f^p)H^p(M)$$and$$H^p(M)^f=\ker(1-f^p).$$One can then show that there are exact sequences$$0\to H^{p-1}(M)_f\to H^p(S_fM)\to H^p(M)^f\to 0$$with SM the mapping torus of f. One speedy way of getting this, by the way, is to ... 4 Consider the space K given by the disjoint union of two circles. If \phi is the map interchanging the two components (preserving the orientations), then we see that K_{\phi} is a torus. However, K\times I is the disjoint union of two tori, hence the Betti numbers are obviously different:$$(b_0(K_\phi),b_1(K_\phi),\ldots)=(1,2,1,0,\ldots)$$... 2 I'm confused. H_1(\mathbb{R}P^3)=\mathbb{Z}/{2\mathbb{Z}}, but is orientable (on edit, this answers your question 2). EDIT: Maybe it refers to the fact that H^{m-1}(M) can only be torsion if the manifold M is non-orientable. It is Corollary 3.28 in Hatcher. 1 a) Cohomology should be regarded as a graded-commutative algebra, therefore is a finer invariant. This allows for some quite interesting definitions, e.g. the Hopf invariant. b) If M is a manifold of suitable dimension, then one has b_{n-k}=b_k for the Betti numbers (proved via Poincare duality, which involves cohomology). It is sometimes easier to ... 6 To elaborate on Dmitri Pavlov's comment, your questions have been answered in a recent paper by Yehonatan Sella. Sella give an example (Example 0.3) of a locally contractible space X for which \mathcal{S}^k(X) \xrightarrow{\mathrm{shf}_X} \tilde{\mathcal{S}}^k(X) is not surjective. Sella's space has 5 points, and the argument that \mathrm{shf}_X is ... 0 By the Kunneth theorem,$$H_i(X\times S^n) \cong \bigoplus_{j+k=i} H_j(X)\otimes H_k(S^n)$$This is because the Tor term in the short exact sequence vanishes because H_k(S^n) is always a free abelian group. Since H_k(S^n) = 0 when k \neq 0, n, it follows that$$H_i(X\times S^n) \cong (H_i(X)\otimes H_0(S^n)) \oplus (H_{i-n}(X) \otimes H_n(S^n)) \cong ...

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There is a homotopy $H_t: C_r \to \Bbb R^2 - 0$ given by $H_t(z) = f(tz)$. At time $1$ this is just $f(z)$ and at time $0$ it is the constant map $f(0)$. Note that $f(tz) \neq 0$ by our assumption on $f(z)$. Therefore $f$ induces the same homomorphism on homology as the constant map does, and the latter clearly induces the zero homomorphism. Edit: When I ...

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Write $S^n$ as the union of two open sets $U= S^n-N$, $V=S^n-S$where $N,S$ are the north and south pole. Note that the intersection $U\cap V$ is connected if $n\geq 2$. $U,V$ are contracile (diffeomorphic to $R^,$, and on each of these sets the form $\alpha$ is exact and andmits primitive $f_U, f_V$. On the intersection $d(f_U-f_V)=0$ and by connexity ...

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You can imitate the argument that shows $\pi_1(S^n)=0$ if $n>1$ by removing a point! Try it!

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I think the function you are looking for is $f(x)=\int_{\gamma_x} \alpha$ where $\gamma_x$ is a path from some chosen base point $x_0$ to $x$. But then for this to be well defined you have to say that all paths between $x$ and $x_0$ are homotopic, i.e. that your sphere is simply connected, so the whole argument gets a bit circular... P.S. Traying to write ...

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As OP correctly remarked in the comments, my answer was wrong! And his proof was right (I guess). Well done, there's just a trivial mistake you've made: $H_2(Y)=0$ would imply that the map $H_2(Y)\to H_1(A\cap B)$ is the zero map. On the other hand, the image of this map is the kernel of the map $\phi$, which you correctly described; the kernel of this map ...

1

Recall that the $H_0$ of a path-connected space $X$ is isomorphic to $\mathbb{Z}$ via the augmentation map, which counts the sum of the coefficients of a given (class of a) $0$-chain, i.e., a formal sum of points of $X$. Therefore the generator $1 \in \mathbb{Z}$ corresponds to the class of any point $x \in X$; alternatively, if you pick $x,y \in X$, then ...

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You used it when you wrote this equation: $$\frac{\ker \phi \oplus H}{\operatorname{im} \partial_1} = \frac{\ker \phi}{\operatorname{im} \partial_1} \oplus H$$ You're implicitly using the fact that $\operatorname{im} \partial_1 \subset \ker \phi \iff \phi \circ \partial_1 = 0$. Otherwise taking the quotient $\ker \phi / \operatorname{im} \partial_1$ doesn't ...

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Let $M$ be an orientable (connected, closed) $n$-manifold. Then classically $H_n(M;F) \cong F$. But moreover, $H_{n-1}(M;\mathbb{Z})$ is torsion-free. This is Corollary 3.28 in Hatcher's book. The proof basically goes like this: if $H_{n-1}(M;\mathbb{Z})$ weren't torsion free, by the UCT then $H_n(M; \mathbb{F}_p)$ would be bigger than just $\mathbb{F}_p = ... 3 I'm going to use singular cohomology. Recall that if$C_i(M)$denotes the collection of singular$i$-chains, then for an abelian group$G$,$C^i(M; G) = \operatorname{Hom}(C_i(M), G)$is the collection of singular$i$-cochains with values in$G$. Note, if$H$is a subgroup of$G$, then$C^i(M; H) \subseteq C^i(M; G)$. There is a coboundary operator ... 3 Note that the restriction of$p$to$S\setminus\delta S$is a homeomorphism onto$X$. So$X\cap Y$, being a subset of$X$, is homeomorphic to a subset of the open unit square. Which subset is that? Well,$p^{-1}(Y)$is just a small neighborhood of$\delta S$, so$Y\cap X$will look like a small open ring along the edge of the square, but not actually ... 0 The important thing is that all$H_1(-;V)$, and$H^1(-;V)$are functorial in the group. That is, if you have a group homormophism$G\to H$, then it induces$H_1(G;V)\to H_1(H;V)$respecting composition, inversions and identities. In particular it sends isomorphisms (of groups) to isomorphisms (of (co)homology). In particular, when given a$X\to Y$which ... 1 Why not directly prove it is isomorphic to the projective line? For example, changing variables to$(x_0+ix_1, x_0-ix_1,ix_2)$, this curve$M$is same as the one given by$x_0x_1-x_2^2=0$and you have a map$\mathbb{P}^1\to M$, by$(u,v)\mapsto (u^2, v^2, uv)$, which you can easily see to be an isomorphism. 1 From Universal Coefficient Theorem we have the following natural s.e.s. for homology we have $$0 \to H_1(X;\mathbb{Z})\otimes \mathbb{Z}_2 \to H_1(X;\mathbb{Z}_2) \to \text{Tor}(H_0(X;\mathbb{Z});\mathbb{Z}_2)\to 0$$ since$H_0(X;\mathbb{Z})$is$\mathbb{Z}$the s.e.s becomes $$0 \to H_1(X;\mathbb{Z})\otimes \mathbb{Z}_2 \cong H_1(X;\mathbb{Z}_2) \to 0$$ ... 1 This is the fundamental lemma in homological algebra (see here theorem$2.22$page$36$and the following corollary.) If you take the identity the thesis of the theorem says in particular that you obtain a chain homotopy 1 It is enough to show that$H_n(g):H_n(\Bbb RP^n)\to H_n(\Bbb RP^n)$is multiplication by an odd integer because$2\deg(f)=\deg(p\circ f)=\deg(g\circ p)=2\deg(g)$. By the UCT, this is equivalent to showing that $$H_n(g;\Bbb Z/2):H_n(\Bbb RP^n;\Bbb Z/2)\to H_n(\Bbb RP^n; \Bbb Z/2)$$ is an isomorphism, which in turn is equivalent to the analogous statement for ... 3 First, let us note that a map induces isomorphisms on cohomology with coefficients in all rings iff it induces isomorphisms on cohomology with coefficients in all abelian groups. Indeed, if$G$is any abelian group, consider the ring$R=\mathbb{Z}\oplus G$, with multiplication defined by$(m,g)\cdot(n,h)=(mn,mh+ng)$. Then there is an isomorphism ... 1 Write the two torus as$S^1 \times S^1$. Use that$H_0(S^1) = H_1 (S^1) = \mathbb Z$and$H_i (S^1) = 0$for$i > 1$. Also use the fact that$\mathbb Z \otimes_{\mathbb Z} \mathbb Z = \mathbb Z$. The answers you would get would be$H_0(T^2) = \mathbb Z$.$H_1(T^2) = \mathbb Z \oplus \mathbb Z$.$H_2(T^2) = \mathbb Z$.$H_i(T^2) = 0$for$i >2$. ... 1 For the two-dimensional torus, we have$T^2 = S^1\times S^1$. In general, we have $$T^k = (S^1)^k = \underbrace{S^1\times\dots\times S^1}_{k\ \text{times}}.$$ The Künneth formula generalises in the way one might expect: $$H_n(X_1\times\dots\times X_k) = \bigoplus_{r_1 + \dots + r_k = n}H_{r_1}(X_1)\otimes\dots\otimes H_{r_k}(X_k).$$ 2 The standard answer is to observe$S^1$is an Eilenberg-MacLane spaces, specifically that$S^1$is a model for$K(\Bbb Z, 1)$. (This relies on showing that all of the higher homotopy groups of$S^1$vanish.) In general, there is a natural isomorphism$H^n(X;G) \cong [X, K(G, n)]$. This is referred to as the representability of cohomology by ... 2$H_n(M,M-x;R)= H_n(M,M-x)\otimes R$. For each$r\in R$determines a covering space$M_r$of$M$consisting of points$\pm \mu_x\otimes r \in H_n(M,M-x;R)$where$\mu_x\in H_n(M,M-x)$is a generator. If$r=-r$i.e order$2$, then$M_r$is just a copy of$M$, otherwise it is isomorphic with the oriented$2-$sheeted cover$\bar{M}$. So observe that a ... 4 In terms of the logical process of inducing maps, you can think in the following way: A continuous map$f : X\rightarrow Y$induces chain maps$f_\# : C_k(X) \rightarrow C_k(Y)$(for each$k$) which induces maps on homology$f_* : H_k(X) \rightarrow H_k(Y)$. It's important to note that$f_*$exists only because$f_\#$sends homologous cycles in$X$to ... 2 Since$\alpha = (1,0)$and$\beta = (0,1)$generate$\Bbb{Z}^2$, let's see where your operator takes them. Let$T = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$. Then$T\alpha = (1,3)$and$T\beta = (2,4)$. Then for arbitrary,$a,b \in \Bbb{Z}$,$T(a \alpha + b \beta) = (a+2b, 3a+4b)$. As$a,b$independently range through all of the integers, you ... 0 By construction the edge homomorphism coincide with the map induced by the inclusion$\ast \to X$. See Davis and Kirk's book for example. 0 You're completely correct with everything you said. To show what you asked for, use the map$\mathbb R^n -0 \to S^{n-1}$which collapses the rays$\mathbb R_+x$. Now there are two ways; one elementary and one more algebraic (still being elementary if you're not purely a differential topologist). 1) Notice that the inclusion$S^{n-1} \to \mathbb R^n-0 \to ...

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Cargo's answer doesn't seem to give any geometric intuition for the statements he/she claims; let me try and remedy that. Let's just talk about $\mathbb{CP}^n$. There are ways of calculating $H^{\bullet}(\mathbb{CP}^n;\mathbb{Z}) = \mathbb{Z}[x]/x^{n+1}$, where $x$ has degree 2 (cellular homology + universal coefficients gives you the module structure; ...

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Your De Rham cohomology algebra is $H^*(\mathbb P^3_\mathbb C)=\mathbb R[w]/(w^4)$, where $w=[\omega]$ is the class of some closed $2$-form $\omega$. We then have $0\neq w^2=[\omega \wedge \omega]$, so that $\omega\wedge \omega$ is not exact and a fortiori $\neq 0$.

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I think the key here is de Rham's theorem, (part of) which can be pertinently paraphrased: A closed $k$-form $\omega$ is exact if and only if $\int_c \omega = 0$ for every closed $k$-chain $c$. The fact that $$\int_{c_1 \times c_2} \omega \wedge \eta = \int_{c_1} \omega \int_{c_2} \eta$$ is then all you need.

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$\require{AMScd}$I'd rather think about the negation of your first condition. So call a space bad if there exists a decomposition $A \cup B$ into connected open sets with disconnected intersection. Call a space semi-bad if there exists a decomposition into open sets $A \cup B$ such that the induced map on relative homology $\tilde H_0(A \cap B) \to \tilde ... 2 It is hard to know what will give you an intuition without knowing what you want and what background you already have. If you are asking roughly what kind of geometric or topological information Cech cohomology reveals about a space, then answer is the same as for all cohomology theories: it reveals the global connectivity of spaces with relatively simple ... 1 Your logic does follow, but you should include the coefficients - when (co)homology groups are written without coefficients, it is (often) implicit that$\mathbb{Z}coefficients are meant. You've shown \begin{align*} H^n(M; R) &\cong H_0(M; R)\\ H^{n-1}(M; R) &\cong H_1(M; R). \end{align*} However, more can be said, but that may be beyond the ... 1 One knows thatr=\min\{i:H_i({\bf x,y};M)\ne0\}$. Notice that$I=({\bf x,y})$. We also know that$I$contains a maximal$M$-sequence, namely${\bf y}$. If$IM\ne M$then one can apply the Corollary 17.12 and get$s=\min\{i:H_i({\bf x,y};M)\ne0\}$, so$s=r$. The only thing to show now is$IM\ne M$. Can you do this? 0 I think one can give definition of cubical maps of cubical complexes in a similar way to simplicial maps between simplicial sets. Cubical maps between cubical sets is a map of vertices which is compatible with face and degeneracy maps. 2 By simple computations, one checks$AB=BA$,$AC=CA$,$CB=BCA$is a presentation. So in the abelianization$A=1$,$BC=CB$the abelianized group is$\bf Z^2$. Note that$dy, dz$are invariant by$G\$ and form a base for the de Rham co-homology

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