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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 $h^a(\mathbf P^b, ... 3 Note that$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 $$\Omega^0 \to \Omega^1 \to \Omega^2 \to \cdots$$ ... 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 ...
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 $f ... 1 This is a great question! I know no references for the following facts and I am pretty sure one can make more general statements, but these are the ones that I have seen used in practice. 1) Yes, at least when$R$is a ring. Indeed,$H_{*}(X, R)$can be defined as homology of$C_{*}(X, R)$, the chain complex which in degree$n$is the free$R$-module ... 1 Well, your question is exactly what motivates people to study homotopy theory. I would advise you to look at Model Category Theory. To answer your question, I think it always works when$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 ... 1 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 ... 1 Assume that$A \neq 0$. Since$pA=0$we can consider it as a vector space over$F_p$, and by taking the$G$-submodule generated by some nonzero element, we can assume that it is finite dimensional. In particular an action of$G$is just a homomorphism$\varphi:G \to GL_m(F_p)$. Suppose first that$G$is cyclic generated by$g$. Since$g^{p^m}=e\$ where ...