# Tag Info

4

First, recall that $\mathbb{R}^{\times} \cong \mathbb{R} \times \mathbb{Z}_2$, and hence $$H^2(G, \mathbb{R}^{\times}) \cong H^2(G, \mathbb{R}) \times H^2(G, \mathbb{Z}_2).$$ Second, since $\mathbb{R}$ is a $\mathbb{Q}$-vector space and $G$ is finite, $H^2(G, \mathbb{R}) = 0$. So the problem reduces to the computation of $H^2(G, \mathbb{Z}_2)$. At this ...

4

The short answer is yes. For example, $\text{Spec }H^{\bullet}(\mathbb{CP}^{\infty})$ (here the cohomology is concentrated in even degree so there is no issue with applying $\text{Spec}$ in the classical sense) can be identified with the formal affine line. The group structure on $\mathbb{CP}^{\infty}$ given by tensoring line bundles induces a group scheme ...

4

$$\require{AMScd}\begin{CD} 0@>>>R@>1>>R\\ @VVV@VV\begin{pmatrix}1\\0\end{pmatrix}V @VV1V\\ R@>\begin{pmatrix}1\\1\end{pmatrix}>>R^2@>\begin{pmatrix}1&-1\end{pmatrix}>>R\\ @VV1V @VV\begin{pmatrix}1&0\end{pmatrix}V @VVV\\ R@>1>>R@>>>0\\ \end{CD}$$

3

It is enough to show that the homomorphism $\operatorname{Hom}_{R_S}(R_S,I_S) \rightarrow \operatorname{Hom}_{R_S}(J_S,I_S)$ is surjective for every ideal $J$ of $R$. We have an exact sequence $0 \rightarrow J \rightarrow R \rightarrow R/J \rightarrow 0$. Since $I$ is injective, the functor $\operatorname{Hom}_R(-,I)$ is exact. Hence we have an exact ...

3

As is written in Wikipedia, the equality $b_1 = m - n + k$ makes sense and is true for topological graphs, not arbitrary spaces. A full triangle and an annulus are not graphs, so the equality isn't true (the fact that you get the right result for the annulus is more of a coincidence than anything). For an arbitrary CW complex, the Euler characteristic ...

3

In an abelian category (or in a different direction a pointed category with coequalizers so that I can talk about $\mathsf{Grp}$), you can define something called the cokernel: the cokernel of a morphism $f : X \to Y$ is the coequalizer of $f$ and the zero morphism $0 : X \to Y$. It's is generally denoted $Y / X$ when $X \to Y$ is a monomorphism, ie. when ...

3

Your reasoning is correct and gives the first equality in the chain $$\text{projdim}_{{\mathscr O}_{X,x}}(k(x))=\text{dim}({\mathscr O}_{X,x})=\text{codim}(\{x\},X)\leq\text{dim}(X)$$ in which the second equality can be checked directly for $X$ affine, and in which the third inequality is clear but strict in general: Take $R$ a discrete valuation ring with ...

2

You don't need the homology exact sequence, you merely need functoriality of homology: $H(r\circ j)=H(r)\circ H(j)$ and $H(id)=id_H$. The map $H(A)\rightarrow H(X)$ is not always injective, but the map on the chain level $S(A)\rightarrow S(X)$ is always injective (and is often viewed as an inclusion). Let me guess why you think that the map $H(A)\rightarrow ... 2 Take the projective resolution $$0\to0\to\mathbb{K}[x]\to\mathbb{K}[x]\to\mathbb{K}\to0$$ of$\mathbb{K}$, which you can regard as a length$2$extension of$\mathbb{K}$by$0$, and take the direct sum with $$0\to\mathbb{K}\to\mathbb{K}\to0\to0\to0$$ to get $$0\to\mathbb{K}\to\mathbb{K}\oplus\mathbb{K}[x]\to\mathbb{K}[x]\to\mathbb{K}\to0,$$ which is a length ... 2 Well, this is a classical result of homological algebra, the cohomology LES. You get your sequence by applying the Snake lemma to the commutative diagram with exact rows having as columns exact sequences of the form$0\rightarrow H^n(C)\rightarrow C^n/B^n(C)\rightarrow Z^n(C)[1]\rightarrow H^n(C)[1]\rightarrow 0$(note that$H^n(C)$is the kernel of the ... 2 Yes, more generally, you have the following (which applies to your situation when setting${\mathcal D} := {\mathcal C}^{\text{op}}$). Let${\mathcal C},{\mathcal D}$be finite length abelian categories and let${\mathscr F}\dashv{\mathscr G}$be an adjunction between the exact functors${\mathscr F}:{\mathcal C}\rightleftarrows {\mathcal D}: {\mathscr ...

2

Let $(R,\mathfrak m,k)$ be a noetherian local ring and $M$ an $R$-module of finite length. Then $(0)$ is irreducible in $M$ if and only if $\dim_k\operatorname{Soc}(M)=1$. "$\Rightarrow$" If $\dim_k\operatorname{Soc}(M)\ge2$, then pick two one-dimensional subspaces of $\operatorname{Soc}(M)$ which intersect trivially. "$\Leftarrow$" First note that ...

2

You are right, it finally involves cokernels (=coimages of the next arrows). Your first line is correct: taking cokernel of both sides of $\def\im{\rm im\,} \im f=\ker g$, we'll get $\def\coker{\rm coker\,} \def\coim{\rm coim\,} \coker f=\coim g$. For the converse, take kernel of both sides. Then, for the objects, your second line should read $${\rm ... 2 In my answer to your previous question I have constructed a free resolution of k as an R-module$$\cdots\longrightarrow R^2\stackrel{v}\longrightarrow R^2\stackrel{u}\longrightarrow R^2\stackrel{v}\longrightarrow R^2\stackrel{u}\longrightarrow R^2\stackrel{f}\longrightarrow R\longrightarrow 0$$where f(r_1,r_2)=r_1x+r_2y, u(a_1,a_2)=(a_1y,a_2x), and ... 2 Yes, there is a surjective homomorphism of A-modules L\to N and denote its kernel by K. Then, by the fundamental isomorphism theorem we have N\simeq L/K. The diagram chasing isn't that hard: let x\in M\otimes N which is sent to 0; since \beta is surjective there is y\in M'\otimes L such that x=\beta(y). Now let f=(M\otimes L\to M\otimes ... 1 Consider the diagram of the proof:$$\require{AMScd} \begin{CD} {} @. M'\otimes K @>f>> M\otimes K @>g>> M''\otimes K \\ @. @VVhV @VViV @VV{\alpha}V \\ 0 @>>> M'\otimes L @>j>> M\otimes L @>k>> M''\otimes L \\ @. @VV{\beta}V @VVlV \\ {} @. M'\otimes N @>m>> M\otimes N \\ @. @VVV @VVV \\ {} @. 0 @. 0 ...

1

$$0=Tor^1(M^{\prime\prime},N)\to M^\prime \otimes N\to M\otimes N\to M^{\prime\prime} \otimes N\to 0$$ $0=Tor^1(M^{\prime\prime},N)$, since $M^{\prime\prime}$ is flat.

1

According to Corollary 4.2 of Leif Melkersson's paper Cohomological Properties of Modules with Secondary Representations, any Artinian module over a commutative Noetherian ring is pure injective. In particular, any commutative Artinian ring is pure injective over itself, since Artinian rings are Noetherian by the Hopkins-Levitzki theorem. An example which ...

1

In this answer I assume that what you mean by "locally finite space" means "homology groups are finitely generated" (sometimes people call a graded module "locally finite" when it is are finitely generated in each degree), if that's not the case please clarify. Since you talk about the universal coefficient theorem I assume the base ring is a PID. I'm ...

1

If $X$ is a topological space, it is standard to put $C_i(X)=0$ whenever $i<0$.

1

For general semisimple ${\mathfrak g}$, if ${\mathbb Z}\Phi$ is the root lattice, then any ${\mathfrak g}$-module $X$ in ${\mathcal O}$ decomposes as $X = \bigoplus\limits_{C\in {\mathfrak h}^{\ast}/{\mathbb Z}\Phi} X_C$, where $X_C := \bigoplus\limits_{\lambda\in C} X_\lambda$. As vector spaces, this is ok since it is only nested way of writing the weight ...

1

I am assuming we are dealing with a functor $F: R\mathrm{Mod} \to S\mathrm{Mod}$ where $R$ and $S$ are commutative rings, although the result may hold in more general settings that I am not sufficiently familiar with. A split exact sequence $0 \to A \xrightarrow{i} B \xrightarrow{p} C \to 0$ can be characterized by 4 functions and 5 equations: \begin{align} ...

1

You can always break up a long exact sequence into small exact sequences. Suppose your long exact sequence is: $$\dots \to A_{n+1} \xrightarrow{f_{n+1}} A_n \xrightarrow{f_n} A_{n-1} \xrightarrow{f_{n-1}} A_{n-2} \to \dots,$$ then you can split it as: $$0 \to \operatorname{coker} f_{n+1} \xrightarrow{f_n} A_{n-1} \xrightarrow{f_{n-1}} \operatorname{im} ... 1 Suppose \ker i_*\neq 0. let a\in\ker (i_*)\setminus \{0\}. We have 0=\ker\mbox{Id}_{H_n(A)}=\ker(r_*\circ i_*). But note that$$(r_*\circ i_*)(a)=r_*(0)=0 so $a\in\ker\mbox{Id}_{H_n(A)}$. This is a contradiction by the non-triviality of $a$ so $\ker i_*=0$ hence $i_*$ is an injection. You can similarly show that $r_*$ is a surjection and you should ...

1

If you know a few formal properties of homology, you might argue as follows: Realize ${\mathbb S}^n$ as ${\mathbb D}^n / \partial{\mathbb D}^n$ with basepoint $\overline{\partial{\mathbb D}^n}$, you have a pinch map ${\mathbb S}^n \to {\mathbb S}^n\vee{\mathbb S^n}$ collapsing $\{x_n = \frac{1}{2}\}$ to a point, and which you can prove to induce the ...

1

Disclaimer: not an answer. Since I have no idea what you really are talking about, I may just play around with what I guess you are talking about using representation theory (which can be regarded as homological algebra in some sense.) It will not be easy to explain the exact mathematics, but just watch the combinatorics evolve: Your first definition ...

1

This is consequence of the following fact : Let $(A,\mathfrak m,k)$ be a noetherian local ring. If $\operatorname{inj dim}_A k$ is finite, then $A$ is regular, where $\operatorname{inj dim}_A k$ is the injective dimension. As far as I remember, one variation on this assertion is called "global dimension theorem", referring to a (JP) Serre's result on the ...

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