# Tag Info

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Given $f \in {\rm Hom}(B,X)$, the element $\phi$ mapping onto $f$ has to be a $G$-homomorphism from $B$ to ${\rm Hom}({\mathbb Z}G,X)$. So, for $g \in G$, $b \in B$, we must have $\phi(gb) = g\phi(b)$. So, $f(gb) = \phi(gb)(1) = (g\phi(b))(1) = g(\phi(b)(g^{-1})$, and hence $\phi(b)(g^{-1}) = g^{-1}f(gb)$. So the inverse map you are looking for is $f ... 2 This is a partial answer: The push-pull (or projection) formula says that$f_*(\mathcal E \otimes f^* \mathcal F) = f_*\mathcal E \otimes \mathcal F,$if$\mathcal F$is locally free of finite rank on the target, and$\mathcal E$is coherent on the source. (See the exercises in Ch. II.5 of Hartshorne.) In your particular case, it gives that$f_*(\mathcal ...

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By the Künneth formula, we have $$H^*(S^1 \times S^1; \Bbb Z) \cong \Lambda_\Bbb Z[\alpha, \beta],$$ the exterior algebra on two variables, with $|\alpha| = |\beta| = 1$. This can also be shown by direct computation using simplical homology. By the formula for the cohomology ring of a wedge sum, we have $$\tilde H^*(S^1 \vee S^1 \vee S^2; \Bbb Z) \cong ... 0 Hint: Use the two classes in the 1st-cohomology coming from each S^1 factor. 7 Certainly. To find h^0: what is the dimension of the vector space of homogeneous polynomials of degree c in b+1 variables? Using for instance the Stars & Bars method, one can see that the answer is$${b+c \choose c}$$Example: {3+2 \choose 2}=10. As for higher cohomology, see Hartshorne Chapter III, which shows that h^a(\mathbf P^b, ... 1 Well, your question is exactly what motivates people to study homotopy theory. I would advise you to look at Model Category Theory. To answer your question, I think it always works when f or g is a fibration (and if your ring of coefficents is a field). Let me also give you a counterexample. Let X and Y be points, Z be the closed interval, and ... 3 Note that O_X(1) isn't well-defined unless you specify an embedding in projective space. But yes, the dual of a very ample line bundle (or even just a nontrivial effective line bundle) cannot have nonzero global sections: if L and L^* both had nonconstant global sections, we could multiply them to get a nonconstant section of O_X. 2 Here's one way to see it. This is probably overkill with the machinery. Finite morphisms are proper, and the higher direct images of a coherent sheaf under a proper morphism are still coherent. Thus R^1f_*\mathcal{O}_X is coherent. But now by the Theorem on Formal Functions, the completion of the stalks are just cohomology of the fibers: ... 2 The cochain complex of sheaves$$0 \to \mathbb{R} \to \Omega^0 \to \Omega^1 \to \cdots$$is exact: this follows from the Poincaré lemma. (Any closed differential (n+1)-form on a sufficiently small open neighbourhood must be the exterior derivative of some differential n-form.) Thus, the cochain complex$$\Omega^0 \to \Omega^1 \to \Omega^2 \to \cdots$$... 0 Isn't it clear considering the fundamental group? Every isolated point you delete from \mathbb{R}^2 gives you a generator of the fundamental group of your space, which will be the free group with \mathbb{Z}^2 generators. Then taking the abelianizated of it, and you find \mathbb{Z}^{\mathbb{Z}^2}, take the tensor product with \mathbb{R}, and you got ... 2 The complex you define is a special case of the complex$$C^*(X;G) of maps of groups $C_n(X;\Bbb Z)\to G$ defined generally for any topological space $X$ and abelian group $G$ (your example being of course $G=\Bbb R$). Its elements are called "$G$-valued cochains on $X$", and the associated cohomology groups are called cohomology with coefficients in $G$.

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If $M$ is a $\Bbb Z$-module, then $\text{Hom}(M,\Bbb R)\cong\text{Hom}(M,\Bbb Z)\otimes \mathbb R$. Integrating any $k$-form over a (smooth) chain gives a practical such "real" cochain.

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Since $X$ is projective, a global function is a constant, so it's killed by $d$, which gives the injectivity of the first map. For the surjectivity of the second, note that a projective curve can always be covered by two open affines (since a point is ample by Riemann-Roch). Write down the Cech complex with respect to a double cover (i.e. resolve ...

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No; for instance, there are acyclic spaces (i.e. with vanishing cohomology in dimension $>0$) which are not contractible (i.e. not homotopic to a $0$-dimensional manifold). For instance, the Poincaré sphere with a point deleted from it is acyclic, but it is not contractible because it has nonvanishing $\pi_1$. (Without deleting a point it has only ...

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