Tag Info

Hot answers tagged

4

Restricting to the bounded derived category, this is equivalent to asking the following Question: Given $X\in\textbf{D}^b({\mathscr A})$, is $X\cong\bigoplus_n \Sigma^{-n}\text{H}^n(X)$ in $\textbf{D}^b({\mathscr A})$? Answer: This is true for all $X$ precisely if ${\mathscr A}$ is hereditary, i.e. $\text{Ext}_{\mathscr A}^2(-,-)\equiv 0$. Before ...


4

The complex $(A\oplus B,D)$ is the mapping cone of the chain map $$d_1: (A[-1],d_0)\to(B,d_0),$$ and the fact you want follows from the resulting long exact sequence in cohomology: $$\dots H^{i+1}(A)\to H^i(B)\to H^i(A\oplus B)\to H^i(A)\to H^{i-1}(B)\to\dots$$


3

No, there are examples with non-integer Herbrand quotient. Here I give two of them: 1- $G=\mathbb{Z}/n\mathbb{Z}$ and $M$ an $n$-divisible abelian group with finite nontrivial $n$-torsion elements. Then the Herbrand quotient of the trivial action is $\frac{1}{|\{x\in M:\,nx=0\}|}$. 2- Consider the only nontrivial action of $G=\mathbb{Z}/2\mathbb{Z}$ on ...


2

iii) Note that we still assume that $M$ is free with basis $B$. Let $B_2=\overline{(B\setminus B_1)}$. Then $\overline M$ is free with basis $B_2$. Indeed given any map $B_2\to A$, we can lift it to a map $B\setminus B_1\to A$, extend this by zero to a map $B\to A$, hence get a homomorphism $M\to A$ that vanishes on $B_1$ and hence also on $N_1$, hence this ...


2

Let $f\in\text{Hom}(M,N)$ be in the kernel of $\overline{u}$. Therefore, for all $m'\in M'$, $(f\circ u)(m')=0$. Let $m_1,m_2\in M$ be such that $m_1-m_2\in M'$. Then $f(m_1-m_2)=0$, so $f(m_1)=f(m_2)$. Therefore, $f$ is constant on quotients $M/M'\simeq M''$. Therefore, we can define a lift $\widetilde{f}\in\text{Hom}(M'',N)$ as follows. Let $m''\in ...


2

If the functor you want to compute is the derived functor of $-\otimes K$, a projective resolution of $M$ is $\dots\to0\to0\to0\to0\to M\to M\to 0$. Then… If you want to compute the derived functor of $M\otimes-$, then recall that projective modules are flat. So take a projective resolution $\dots\to K_{2}\to K_{1}\to K_{0}\to K\to 0$. Tensoring it with $M$ ...


2

The internal hom is defined via maps into it. You know nothing about the classification of maps into a coproduct, but you know how to classify the maps into a product, by the very definition of a product. You can also see how the product comes up in the following calculation: Let $A,B,C$ be graded $R$-modules (the case of chain complexes is similar, but ...


2

A much simpler diagram suffices (or the symmetrical subdiagram with $Z$ instead of $T$): $$\begin{matrix} &&&0\\ &&&\downarrow\\ 0\to&H&\to &W_1&\to&T\\ &\downarrow&\searrow&\downarrow\rlap{\scriptstyle\subseteq}&&\|\\ 0\to &W_2&\stackrel\subseteq\to &V&\to &T \end{matrix} $$ ...


2

If you think $H,W_1,W_2$ as subspaces of $V$, then you have $H = W_1 \cap W_2$. Clearly $H \subseteq W_1 \cap W_2$ (simply look at the diagram). Viceversa, if $w \in W_1 \cap W_2$, then you have $$w \in \ker (V \longrightarrow T) \cap \ker (V \longrightarrow Z)$$ so $$w \in \ker (W_1 \longrightarrow T) \cap \ker (W_2 \longrightarrow Z) = \operatorname{Im} ...


2

Schanuel’s Lemma. In the book "An Introduction to Homological Algebra" by Rotman, You can find this: Remark. There exists also a dual for Schanuel’s Lemma. For this, see Exercise 3.14 of the book


2

(1) $$ K\rightarrow K^2,\ z\mapsto (zb,-za ) $$ and $$ K^2\rightarrow \langle a,b \rangle,\ (x,y)\mapsto xa+yb $$ (2) Since $K,\ K^2$ are free so they are projective.


1

Hre's one way of looking at (a) of your posted question: $\beta$ injective means that you can replace $B'$ by $B$ (we've identified $B$ with $\beta(B)$), as long as you now interpret $\psi'$ as $\psi'|_{{\rm image}(\alpha)}$. Now there's no point in separating the two occurrences of $B$ nor keeping the map $\beta$; once we identify the two occurrences of ...


1

The comments (which should have been posted as answers) contain more general statements. If $\mathcal{C}$ is a groupoid, then you can explicitly write down an equivalence $\mathcal{C} \simeq Q(\mathcal{C})$. There is always a canonical functor $\mathcal{C} \to Q(\mathcal{C})$ which is bijective on objects. So it remains to check that it is fully faithful ...


1

If $k$ is a field as you say it is, it is not difficult to see using the standard tensor-hom adjunction that $Ext^*_R(k,k)\cong (Tor_*^R(k,k))^*$ as $k$-graded modules, where $R$ is a $k$-algebra. I've learned this from these Tor-Ext notes by May. Here's the relevant passage, which contains other comments which might be useful to you. It's at the very end ...


1

For $n=2$ let $M$ be the module with graded parts $$M_{i,j}=\begin{cases} K&\mbox{ if }(i,j)=(1,1),(1,0)\mbox{ or }(0,1)\\ 0&\mbox{ otherwise.} \end{cases}$$ and with $x_1$ and $x_2$ acting as isomorphisms $M_{0,1}\to M_{1,1}$ and $M_{1,0}\to M_{1,1}$ respectively. Then your conjecture is not true for $M$. If $n>2$ (possibly also $n=2$) then ...


1

First note that if either $g$ or $k$ is an isomorphism, then the proposition is just the right exactness of tensor functor. Now rewrite $g\otimes k$ as $(g\otimes N'')(M\otimes k)$, where we rewrite $1_M$ as $M$ for abbreviation, or note that $-\otimes N''$ is a functor, therefore applies to a morphism $g$. By right exactness of tensor functor, we have ...


1

This is not exactly an answer to your question, except that it is relevant to the question of homotopical invariants preserving at least some colimits. For example, the fundamental group of based spaces does not preserve all colimits: the usual Seifert-van Kampen Theorem determines the fundamental group of a union $X=U \cup V$ of based spaces if $U,V$ are ...


1

As darij grinberg mentions, the exterior algebra is the free graded-commutative algebra on a vector space. It's not the free algebra or the free graded algebra. This doesn't rule out the possibility of "unexpected" right adjoints. Your dimension-counting argument doesn't work, because in order for $\operatorname{dim}(A/B) = \operatorname{dim} A - ...


1

Let $K$ be the kernel of $L_{r-1} \to L_{r-2}$. (I must have been sleeping.)


1

It's allmost trivial if you know the following fact of homology, wich you can found in any standard text (Hatcher for example): Theorem. Homology groups are homotopy invariant, that is, if $X$ is homotopic to $Y$ then $H_n(X)\cong H_n(Y)$ So now you only need to see that every convex set is contractible. Lemma. Let $X$ be a convex set, then $X$ is ...


1

Okay, so the idea is that the map $M_0 \to N_0$ will be onto on $H_0$ and also hit the image which is quotiened out. (assuming the chain complex is bounded at zero) Since chain maps commute with the boundary operators and $M_1 \to N_1$ is onto, we know that $im(M_0 \to N_0)= im(N_1 \to N_0)$. We have an isomorphism on homology so we know $M_0/im(M_1 \to ...



Only top voted, non community-wiki answers of a minimum length are eligible