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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 ...


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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 = ...


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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 ...


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$S^0 \subset S^2$ 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. ...


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 ...


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.


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$\require{AMScd}$ 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 ...


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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.


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First, the theorem is about strict commutativity and associativity. The spaces that represent cohomology theories in a fixed degree (namely infinite loop spaces) are almost never strictly commutative and associative (precisely because they are almost never products of Eilenberg-MacLane spaces); for example, the infinite loop space $\mathbb{Z} \times BU$ ...


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You can say a little bit more, the following proposition is also proved in Hatcher's Algebraic Topology Proposition. Let $f:S^n \to S^n$ be a continuous map, if $f$ is not surjective then deg$(f) = 0$ Proof. Choose a point $x_0$ not in the image of $f$, then you can factor $f$ with the inclusion of $S^n\setminus \{x_0\}$ which is homeomorphic to ...


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 ...


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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 ...


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 ...


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Since $H_n(X;Z)$ finitely generated free abelian for each $n≥0$ So the Tor part will vanish in the short exact sequence for homology.Another observation: Since homology free this implies $H_n(X) = H^n(X)$.Hence the result follows.


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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).


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


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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 ...


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), ...


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 ...


0

So you want to find the element $d_i$ which is dual to $x^i$. Now your "orientation class" will be either $+$ or $-x^m$. Then, since you know the product structure, you see that $x^i \smile x^{m-i} = x^m \in H^{2m}\mathbb CP^m$, so we get $\pm x^{m-i}$ as $d_i$ (the sign depending on orientation and the element $x$).


2

We may assume $X=U \cup V$. The Mayer-Vietoris sequence gives an exact sequence $$H^i(U,F) \oplus H^i(V,F) \to H^i(U \cap V,F) \to H^{i+1}(X,F) \to H^{i+1}(U,F) \oplus H^{i+1}(V,F)$$ which simplifies to $H^i(U \cap V,F) \xrightarrow{\,\cong\,} H^{i+1}(X,F)$. Since $H^{i+1}(X,F) \neq 0$ may happen when $i+1 \leq \dim(X)$, the answer is probably "no". (But I ...


0

I will assume that you know about connections on the tangent bundle. These connections induce connections on the tensor bundle $\mathcal{T}^{r,s}M$ of $(r,s)$ tensor fields. Given $∇ : Γ(TM) \to Γ(TM \otimes T^{*}M)$ there is a unique connection $d_∇ : Γ(\mathcal{T}^{r,s}M) \to Γ(\mathcal{T}^{r,s}M \otimes T^{*}M)$ satisfying $d_∇ = ∇$ on $TM$, $(d_∇X)f = ...


0

The loops A and B determine the two Z2 cohomology classes by transverse intersections. The self intersection of A is zero since you can move it parallel to itself slightly and get a disjoint loop. So its square is zero. Its intersection with B is 1 so their product is non-zero. The self intersection of B is 1 because of the half twist. So the square of ...


1

We describe a linear functional $\alpha : \Omega^p \mathbb R^n \to \Omega^{p-1} \mathbb R^n$. The space alternating $p$-linear functions on $\mathbb R^n$ has dimension $n \choose p$, and you can write the basis as $dx_{i_1} \wedge dx_{i_2} \wedge \cdots \wedge dx_{i_p}$ where $1 \leq i_1 < i_2 < \cdots < i_p \leq n$. If $I = (i_1, i_2, \cdots, ...


1

Let $X$ be a (nice) topological space, $i:Z\subset X$ a closed subset and $j:U\subset X$ the open complement. The sheaves $\mathbb{Q}_X$ and $j_!\mathbb{Q}_U\oplus i_*\mathbb{Q}_Z$ are not isomorphic in general, for instance if $Z$ is a point in $X=S^1$ (they are obviously isomorphic when $U$ and $Z$ are complementary clopens). However, both sheaves have ...



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