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6

Let me stick to the case that everything has coefficients in $\mathbb{Z}$. Then universal coefficients says that the sequence $$0 \to \text{Ext}^1(H_{n-1}(X), \mathbb{Z}) \to H^n(X) \to \text{Hom}(H_n(X), \mathbb{Z}) \to 0$$ is exact, and so the natural morphism $H^n(X) \to \text{Hom}(H_n(X), \mathbb{Z})$ is an isomorphism iff $$\text{Ext}^1(H_{n-1}(X), ...


3

Almost by definition, $f : M \to N$ is orientation-preserving iff it sends the fundamental class $[M] \in H_n(M)$ to a positive multiple of the fundamental class $[N] \in H_n(N)$. Since homotopic maps induces the same morphism in homology, it follows that $f$ is orientation preserving iff $g$ is.


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 = ...


2

John Baez in "Circuit Theory" states: "In 1923, Hermann Weyl published a paper in Spanish which described electrical circuits in terms of the homology and cohomology of graphs (W). In this approach, Kirchhoff’s voltage and current laws simply say that voltage is a 1-coboundary and current is a 1-cocycle. Furthermore, the electrical resistances labelling ...


2

HINTS: This is going to be a common, everyday surface, so you need to do some algebraic manipulations and more multivariable analysis. You should first prove that the surface is compact. Then you will need to recognize that this surface is a branched double cover of the $2$-sphere and use basic facts about Euler characteristic.


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 ...


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 ...


2

$S^0 \subset S^2$ be two antipodal points marked on $S^2$. One can use the relative homology long exact sequence for $(S^2, S^0)$ and note that it contains the short exact sequence $$0 \to {H_i}(S^0) \to {H_i}(S^2) \to {H_i}(S^n, S^0) \to 0$$ Thus, $H_i(S^2, S^0) \cong H_i(S^2)$, hence $H_i(S^2, S^0) \cong \Bbb Z$ if $i = 2$ and trivial otherwise. ...


2

Yes this is true. I quickly recall how you show that the derived functor can be computed by choice of acyclic resolutions, because the explicit description of this isomorphism is important. Suppose you have any effecable $\delta$-functor ${\mathscr T}^{\ast}$ in non-negative degrees (e.g. the right derived functors of a left exact functor defined by ...


1

The usual embedding of $S^{n-1}$ in $S^n$ is as the equatorial sphere; repeating that gives an embedding $S^0 \to S^2$. You can take any two distinct points in $S^2$ for the embedding here, though. The long exact sequence in homology is given by: $$\cdots \tilde H_*(S^0) \to \tilde H_*(S^2) \to \tilde H_*(S^2, S^0) \to \tilde H_{*-1}(S^0) \to \cdots$$ Since ...


1

Here is another approach that doesn't use the long exact sequence for homology. From the realization of $S^2$ as a CW complex, we see that $S^0$ is identified with two antipodal points in $S^2$. Since $H_i(S^2, S^0) \cong \tilde H_i(S^2 / S^0)$ and $S^2 / S^0$ is homotopy equivalent to $S^2 \vee S^1$, we see that $H_i(S^2, S^0) \cong \mathbb Z$ for $i = ...


1

This question has been asked and answered on MathOverflow. I have replicated David Speyer's accepted answer below. Here is some nonsense that I find useful: On a complex manifold, $$\frac{\mbox{locally constant functions}}{\mbox{smooth functions}} \approx \frac{\mbox{locally constant functions}}{\mbox{holomorphic functions}} \cdot ...


1

The universal coefficient theorem says not only that the two are isomorphic as groups but that a natural map between them, namely the reduction $\bmod p$ map $H^{\bullet}(X, \mathbb{Z}) \otimes \mathbb{Z}_p \to H^{\bullet}(X, \mathbb{Z}_p)$, is an isomorphism. This map always respects cup products (with no hypotheses).


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 ...


1

With field coefficients you have that cohomology is the dual of homology. By linear algebra, we get an isomorphism. Proof for my statement is the fact that $Ext_{R}(H_i(M;R),R)=0$ for a field $R$ (because dualizing is exact for vector spaces), which you apply to the universal coefficient theorem. About the reasoning in the first part, you were completely ...


1

$\require{AMScd}$ Note that we have a commutative diagram $$\begin{CD} @. 0 @. 0 @. 0 \\ @. @VVV @VVV @VVV \\ 0 @>>> B_0(X) @>>> B_0(X) @>>> 0 @>>> 0\\ @. @VVV ...


1

$\newcommand\Z{\mathbb{Z}}$ Let $P = \mathbb{CP}^2 \times \mathbb{CP}^2$ be the cartesian product and $S = \mathbb{CP}^2 \vee \mathbb{CP}^2$ be the wedge sum. Then the smash product is $\mathbb{CP}^2 \wedge \mathbb{CP}^2 = P / S$. Since $S \subset P$ is a sub-CW-complex, $\tilde{H}^n(P/S) \cong H^n(P,S)$. It is well known that $H^*(\mathbb{CP}^2) = ...


1

It's allmost trivial if you know the following fact of homology, wich you can found in any standard text (Hatcher for example): Theorem. Homology groups are homotopy invariant, that is, if $X$ is homotopic to $Y$ then $H_n(X)\cong H_n(Y)$ So now you only need to see that every convex set is contractible. Lemma. Let $X$ be a convex set, then $X$ is ...


1

Since you are using simplicial homology, you first need to equip $\Bbb R$ with a $\Delta$-complex structure. You can do so by choosing the integers $\tau_n:\Delta^0\to\{n\}$ as the $0$-simplices, and the closed intervals $\sigma_n:\Delta^1\to[n,n+1]$ as the $1$-simplices. Then $\partial_1:C_1(X)\to C_0(X)$ sends $\sigma_n$ to $\tau_{n+1}-\tau_n$. Can you see ...


1

Start with any $X$ which has torsion in first homology.


1

You have the following two results in the book Simplicial homotopy theory of Goerss and Jardine: Corollary III.2.7. Suppose that A is a simplicial abelian group. Then there are isomorphisms $$\pi_n(A,0) \cong H_n(NA) \cong H_n(A),$$ where $H_n(A)$ is the $n$th homology group of the Moore complex associated to $A$. These isomorphisms are natural ...


1

You're almost right: if you can show that $H_k(X,X^n;\mathbb Z)$ is trivial for $k\leq n$ you will be done, by applying the universal coefficient theorem. We want to use that $X/X^n$ has only a point as $n-1$ skeleton, it is in particular connected. Because then we know by excision that $H_k(X,X^n) \cong \tilde H_k (X/X^n) =0$, since we know that cellular ...



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