Hot answers tagged

6

Yes, it is representable, consider $G_R$ the normal subgroup of $G_0$ generated by $R$, $F(G)=Hom(G_0/G_R,G)$


5

Let $G_1$ be the normal closure of $R$, the smallest normal subgroup of $G_0$ containing $R$. Since the kernel of a homomorphism is a normal subgroup, any element of $F(G)$ contains $G_1$ in its kernel. Conversely, any homomorphism from $G_0$ to $G$ whose kernel contains $G_1$ is obviously an element of $F(G)$. Hence,$$F(G) = \{\varphi \in \text{Hom}(G_0,G) ...


5

There are several elementary points that can be made : On the other hand, why consider only homology ? I don't see why one would be more natural than the other (actually for me cohomology seems more natural because I'm used to subjects where cohomology natuarally appears). There are cohomology theories that are clearly useful and natural and are related to ...


3

A major issue is the multiplicative structure that is around in cohomology. This allows you to distinguish spaces, which have the same homology. As an example, the $X:=\mathbb CP^2$ and $Y:=S^2\vee S^4$. Then both $X$ and $Y$ are CW-complexes with one cell in dimensions $0$, $2$, and $4$. Hence in both cases the homology with integral coefficients is ...


3

S = QQ[x,y,z,t] M = module S/ideal "xy2,x2z" for i from 0 to 4 list Ext^i (M,S) In general, if you have installed M2 property, viewHelp would direct you to a help page, and help would give you a short help message in M2.


2

First it is easy to see that $\operatorname{Hom}(M,\mathbb{C}[x](-1))\cong 0$ since $M$ has only elements in degree $0$ and must be sent to $C[x]_{-1}=0$. There is an easy projective resolution of $M$ given by: $$0 \to \mathbb{C}[x](-1) \xrightarrow{d_1} \mathbb{C}[x] \xrightarrow{d_0} M \to 0$$ $d_1(p(x))=xp(x)$ and $d_0$ is the quotient map. This ...


2

From this answer you can learn that if $R/\mathfrak m$ is $R$-flat then $\mathfrak m=\mathfrak m^2$. Since $\mathfrak m$ is finitely generated it must be generated by an idempotent, that is, $\mathfrak m=(0)$ so $R$ is a field.


2

No such groups exist. Suppose $\operatorname{Ext}(A,\mathbb{Z})=\operatorname{Hom}(A,\mathbb{Z})=0$. First, note that the functor $\operatorname{Ext}(-,\mathbb{Z})$ turns injections into surjections (this follows from the long exact sequence and the fact that $\operatorname{Ext}^2$ always vanishes for abelian groups). So we must have ...


2

Let $X$ be either $\mathbb{Q}$ or $\mathbb{Z}$. Hence $H_i(X) = 0$ for all $i > 0$, and thus by the long exact sequence in homology (recalling that $H_i(\mathbb{R}) = 0$ for $i > 0$ too) the following sequence is exact: $$0 \to H_1(\mathbb{R},X) \to H_0(X) \to H_0(\mathbb{R}) \to H_0(\mathbb{R},X) \to 0$$ Now, $H_0(\mathbb{R}) = \mathbb{Z}$, and ...


2

Note: in this answer I assume the singular homology theory. Sometimes $A$ is a subcomplex of $X$, but not in general. For example, we could consider the pair $(\mathbb R,\mathbb Q)$. In order for $H_n(X,A)=\tilde H_n(X/A)$ to be true, we need a neighborhood of $A$ in $X$ which deformation retracts to $A$. This is not true in many cases. For a simple ...


2

Suppose $f:C\to D$ is a quasi-isomorphism. If $f$ is either onto or injective, then the remaining (co)kernel must be acyclic by a LES argument. Thus for you to find a counterexample, you need to find a quasi-isomorphism that is neither injective nor a surjection to begin with. Consider the complex $D$ where $D_n= \Bbb Z_4$ and $\partial$ is multiplication ...


2

This is because, in a minimal free resoluion, we begin by choosing in $A$ a minimal set of . If the number of generators is $n$, we define a linear map from $L_0=R^n$ by mapping the canonical basis of $L_0$ onto the generators. Now, by Nakayama's lemma, this minimal set of generators yields a basis of the vector space $A\otimes_Rk$, hence the quotient map ...


1

It doesn't need "perfect". All the modules in this answer will be finitely generated. Suppose $\operatorname{gl}(A)=n<\infty$. Then there are modules $X$ and $Y$ with $\operatorname{Ext}^n(X,Y)\neq0$. Take a short exact sequence $0\to\Omega Y\to P\to Y\to0$ with $P$ projective. The long exact sequence of $\operatorname{Ext}(X,-)$ contains ...


1

$Hom(K,Hom(M,E))\cong Hom(K\otimes M,E)$ $F$ is flat iff $F\otimes -$ is exact. Let $E$ be injective cogenerator. Then $0 \longrightarrow X \longrightarrow Y \longrightarrow Z \longrightarrow 0$ is exact iff $0 \longrightarrow Hom(Z, E) \longrightarrow Hom(Y, E) \longrightarrow Hom(X, E) \longrightarrow 0$ is exact. Using 1 and 2, you can see ...


1

So we can actually build up the answer inductively here: let's label the simplices of $\Delta_2$ as $v_0, v_1, v_2$, and the corresponding vertices of the top and bottom faces of $\Delta_2 \times I$ as $t_0, t_1, t_2, b_0, b_1, b_2$. So let's define $P$ on $C_0(\Delta_2)$. Here $\partial P + P\partial = \partial P$ (as $\partial$ vanishes on 0-simplices) ...


1

In the notation of the question, if $X$ is a finite dimensional $A$-module then by induction on the length of $X$ it's easy to prove that $$\dim_K\operatorname{Hom}_A(P_2,X)=[X:L_2].\dim_K\operatorname{End}_A(L_2),$$ where $[X:L_2]$ denotes the multiplicity of $L_2$ as a composition factor of $X$, and dually ...


1

Hint: The statement of the lemma is using the wrong $n$; you'll want to use it for $n-1$ rather than $n$ (so the $P$s go up to $P_{n-1}$). If $M$ has projective dimension $n$, then it has some such resolution in which $K$ is projective. Now what does this tell you about the $K'$ you get by truncating any other projective resolution?


1

$\ker(B)/\operatorname{im}(A)$ is just the kernel of the map $$\Phi:\operatorname{cok}(A)=R^n/\operatorname{im}(A)\to R^o$$ induced by $B$, from the cokernel of $A$ to $R^o$. But $\operatorname{cok}(A)$ is isomorphic to $$R/a_1R\oplus\dots\oplus R/a_r R \oplus R^{n-r},$$ and $\ker(\Phi)$ must contain the torsion submodule $R/a_1 R\oplus\dots\oplus R/a_r ...


1

Your proof of invariance of domain doesn't work, because you have given no justification that $\tilde{f}$ is proper. In fact, it probably won't be proper. Indeed, if (as you suggest) you replace $\epsilon$ by $\epsilon+\xi$ for some $\xi>0$, then $\tilde{f}$ will definitely not be proper, since $\psi^{-1}(\overline{B(\psi(y), \epsilon)})$ is a compact ...


1

It is enough to construct a chain complex $K$ such that $H_n(K)=A$ for a given $n$ and $A$, and $H_j(K)=0$ for $j\neq n$ (then you pass to a direct sum of such $K$'s). This can be done by defining a left free resolution $\ldots E^2\stackrel{\partial_2}{\to} E^1\stackrel{\partial_1}{\to} E_0 \stackrel{\epsilon}{\to} A\to 0$, letting $E$ to be the chain ...



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