# Tag Info

7

Here is a counterexample: let $$C_i = \begin{cases} \bigoplus_{n \in \mathbb{N}} \mathbb{Z} & i = 0 \\ 0 & i \neq 0 \end{cases}$$ with the zero differential. Then $H_0(C)=\bigoplus_{n \in \mathbb{N}} \mathbb{Z}$ has countably infinite rank; $H^0(C) = \prod_{n \in \mathbb{N}} \mathbb{Z}$ has a rank $2^{\aleph_0}$ by a theorem of Nöbeling. It's ...

4

I will answer your question in what follows, but first it might be useful to say something about what derived functors are and what the point is. The idea of derived functors is that if you have a functor which preserves (say) left exactness when applied to a short exact sequence, but not right exactness (e.g. $Hom(X, \text{--})$ for a module $X$ ), then ...

3

Related: curved $A_\infty$-algebras $\mathcal A:=(A_\bullet, d)$ are endowed with a pseudo differential $d_1:A_n\rightarrow A_{n+1}$ s.t. $$d_1\circ d_1(a)\pm d_2(a,d_0(1))\pm d_2(d_0(1),a)=0$$ for all $a\in \mathcal A$, denoting by $d_0(1)\in A_2$ the curvature of the algebra ( $1$ is the unit) and with $d_2$ the binary product in $\mathcal A$. Clearly, ...

3

Of course the constant sheaf $\mathbb{Q}$ is not injective in general. Recall that for an injective sheaf $\mathscr{F}$, we have $H^1 ( X, \mathscr{F}) = 0$. However, for sufficiently nice spaces $X$ (say, paracompact), sheaf cohomology with coefficients in a constant sheaf $A$ is isomorphic to singular cohomology with coefficients in $A$; so in particular, ...

3

Recall that if $\Delta^n = [v_0, v_1, \ldots, v_{n-1}, v_n]$ then for a map $\sigma\colon\Delta^n\to X$ we define the boundary operator $\partial\colon C_n(X)\to C_{n-1}(X)$ by $$\partial(\sigma) = \sum_{j=0}^n (-1)^{j} \sigma \mid [v_0,v_1, \ldots, \hat{v}_j,\ldots, v_{n-1},v_n]$$ and if, for an arbitrary chain $\alpha$, we have ...

3

Explicit injective resolutions (hence also Cartan–Eilenberg resolutions) are hard to work with concretely. (They're usually not finitely generated, for example.) For the purposes of this calculation it is easier to use Čech cohomology. The standard affine open cover $\mathfrak{U} = \{ U, U' \}$ of $\mathbb{P}^1$ is a good open cover in the following sense: ...

3

Because $i_*$ is injective, the images of the boundary maps $\partial$ are zero (by exactness of the sequence given in axiom 4). The exactness of the sequence also means that $j_*$ is surjective. Hence the long exact sequence given by axiom 4 breaks up into short exact sequences $$0\overset{\partial}{\to} ... 3 This is a non-answer, but I don't think there is an answer. You want to compute the action directly from an H-module resolution, which as far as I know is not possible. (There is no reason it should be.) Instead, you must find a larger resolution. A resolution by \mathbb{Z}[G]-modules is a priori a resolution by \mathbb{Z}[H]-modules, and we know that ... 2 Let N be an R-module. We have to show that N \to N \otimes_R S is injective. As S is faithfully flat as an R-module, it suffices to prove this after tensoring with S. Hence it suffices to show that N \otimes_R S \to N \otimes_R S \otimes_R S, n \otimes s \mapsto n \otimes 1 \otimes s is injective. This is true because there is a section, ... 2 Are you familiar with the inflation restriction exact sequence? If N \unlhd G and G acts on module A, then$$0 → H^1(G/N, A^N) → H^1(G, A) → H^1(N, A)^{G/N} → H^2(G/N, A^N) →H^2(G, A)$$is exact. Taking N=H in your example, we have A^N=0, and |N| and |A| are coprime (I am assuming that Z_{p^k} denotes a cyclic group of order p^k), so ... 2 This was supposed to be a comment but it got too long. It is misleading to think of the semidirect product with \varphi = \text{id} as the categorical product. As I explain in this answer, the semidirect product is a (higher) colimit rather than a limit, and in particular its (higher) universal property involves maps out of it rather than maps into it. ... 2 I think it's only a matter of indexing. For example, according to the notation in Weibel's book we would have \cdots\to \mathscr R_1\to \mathscr R_0\to \mathscr G\to 0 in the projective resolution, rather than defining \mathscr R_0 = \mathscr G. That is, in a projective (or free) resolution, the final map \mathscr R_1\to\mathscr R_0 is typically ... 2 Any finitely generated module over a regular local ring satisfies the Tor-rigid property, a result due to Lichtenbaum ("On the vanishing of Tor in regular local rings"). The proof occurs within the first five pages in the paper. For a slightly less obvious but still perhaps too simple example, you could take the ring of power series k[[x]] over a regular ... 2 Finite direct sums are preserved since we deal with additive functors. Now use the fact that direct sums are directed colimits of finite direct sums. Explicitly, we have$$\bigoplus_{i \in I} M_i = \operatorname{colim}_{E \subseteq I \text{ finite}}\, \bigoplus_{i \in E} M_i.$$2 The f and g appearing in the proof are not the same as the f and g appearing in the axiom. We have g\circ f\sim\mbox{Id}_{X,A}\colon (X,A)\to (X,A) and so axiom 5 implies that (g\circ f)_*=(\mbox{Id}_{X,A})_*, the other axioms then give us that (g\circ f)_*=g_*\circ f_* by funtoriality, and (\mbox{Id}_{X,A})_*=\mbox{Id}_{H_k(X,A)} also by ... 2 Take any abelian category \mathscr A with insufficiently many projectives, and let \mathscr F be the zero functor. Flat modules are precisely those which are acyclic for Tor (in either variable). In fact, M is flat if and only if the functor \text{Tor}_1(-, M) vanishes, if and only if the functor \text{Tor}_i(-, M) vanishes for every i>0. As ... 2 I'm not sure what exactly you're asking for, but here's a blurb about derived functors that I hope is useful to you. Suppose we have a left-exact functor F on the category \cal{C} of modules over a given commutative ring. (The usual setting is abelian categories, but there's no harm in working with something concrete for a short description. Also, I'm ... 2 H^2(G, A) can also be written [BG, B^2 A], where for spaces X, Y, [X, Y] denotes the set of homotopy classes of maps X \to Y, BG denotes the Eilenberg-MacLane space K(G, 1), and B^2 A denotes the Eilenberg-MacLane space K(A, 2). If G is also abelian, then BG and B^2 A can both be modeled by simplicial abelian groups, and we can ... 2 Let's start by labeling the arrows:\require{AMScd}$$ \begin{CD} A @>f>> B @>g>> C\\ @VVaV @VVbV @VVcV\\ D @>>h> E @>>k> F \end{CD} $$The square (X+Y) is exact if and only if the "diagonal sequence" 0 \to A \xrightarrow{(gf, a)^t} C \oplus D \xrightarrow{(-c,kh)} F \to 0 is exact. The squares (X) and (Y) are ... 1 From the first definition we have that X=\stackrel{\circ}{X-U}\cup \stackrel{\circ}{A} because \overline{U}\subset \stackrel{\circ}{A} and \stackrel{\circ}{X-U}=X-\overline{U} then if i put B=X-U i obtain that H_n(B,B\cap A)=H_n(X-U,A-U) where A is the same and X-U=B in this case X=\stackrel{\circ}{A}\cup \stackrel{\circ}{B} so the two ... 1 See the Theorem 1 and 2 here, where there is equivalent. 1 For every i we have C_i=B_i/A. Taking direct limit, when B' denotes the limit of (B_i), we get$$C=B'/A, and $B'$ is a submodule of $B$. But we know that $C=B/A$, hence $B'=B$.

1

The trick to do (1) goes like this. If you have $x \in H_k(X)$, you apply $ir$ to get $ir(x) \in im i_*$. The cool thing is than you can then just check that $x - irx$ is in the kernel of $r$, and so $x = irx + (x - irx)$ is your decomposition. To show it's direct, you take $0 = iy + z$. Then applying $r$, we get $0 = riy = y$, and so the sum is direct. The ...

1

This is an application of a result of homological algebra which is: If $0\to N\stackrel{i}{\to} M\stackrel{j}\to P\to 0$ is an exact sequence and $r:M\to N$ such that $r\circ i=id_{N}$ then $\ker r\simeq P$ and $\text{Im}(i)\simeq N$, and $M=\text{Im}(i)\oplus \ker r \simeq N\oplus P$. Proof: Because $i$ is injective we have $\text{Im}(i)\simeq N$. For ...

1

It's helpful to first recall the construction of an induced map (cf. Brown, beginning of III/8 and II/6): Let $\varphi: H \to H'$ be a homomorphism of groups and let $F \to \mathbb{Z}\to 0$ resp. $F' \to \mathbb{Z}\to 0$ be a $H$- resp. $H'$-projective resolution. Via $\varphi$ any $H'$-module can be considered as a $H$-module. In this regard $F'$ is an ...

1

The theorem is applied (see the middle in $6.6$ ) to $X=\varphi^c\cap C$ and $X_i=\varphi^c\cap U_i$ which is a closed subset in $X$.($X_i=X\cap(\cup_{j\neq i}U_j)^c$ where the c's power is the complementary ).

1

The statement is true for vector spaces but not for arbitrary modules. Here are some steps, you can try filling in the gaps: Show that it is enough to prove that if $0 \rightarrow V_1 \rightarrow V_2 \rightarrow V_3 \rightarrow 0$ is exact then $0 \rightarrow V_1 \otimes W \rightarrow V_2\otimes W \rightarrow V_3\otimes W \rightarrow 0$ is exact. Next ...

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