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Questions tagged [homological-algebra]

Homological algebra studies homology and cohomology groups in a general algebraic setting, that of chains of vector spaces or modules with composable maps which compose to zero. These groups furnish useful invariants of the original chains.

A chain complex is a sequence of abelian groups, vector spaces, or modules, with linear maps connecting them which compose to zero.

Homological algebra is the study of chain complexes and their homology groups.

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Cohomology of a complex included in a big commutative diagram

Suppose you have a complex $0 \to C^1 \to C^2 \to ... \to C^n \to 0$ which can be included in a commutative diagram of the following type (see image) where all rows and columns (except the complex itself) are exact and all rows sufficiently high and…
kvardekkvar
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Zero map between spectral sequences

Let $E^{pq}_2\implies H^{p+q}$ and $\hat{E}^{pq}_2\implies \hat{H}^{p+q}$ be first quadrant spectral sequences of abelian groups. Suppose we are given a morphism $h:H^*\to \hat{H}^*$ compatible with a morphism $f:E\to \hat{E}$ of the spectral…
Marco Flores
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Zeroth left derived functor of right exact functor

I have some difficulty understanding the following proof (source): Claim: If $T: \mathcal{A} \rightarrow \mathcal{B}$ is a right exact functor of two Abelian categories, then $L_0 T$ and $T$ are canonically naturally equivalent. Proof: Take a…
gen
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Free module implies trivial Tor

Let $A$ be a commutative ring. If $M$ or $N$ aree free $A$-module then $Tor_{n}^{A}(M,N)=0$. Since $Tor_{n}^{A}(M,N)=Tor_{n}^{A}(N,M)$ it suffices to deal with the case say when $N$ is flat right? Take a projective resolution of $N$: $...…
user6495
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Internal $Hom^{\bullet}$ of filtered complexes

Gelfand and Manin give the definition of an internal $Hom^{\bullet}(A^{\bullet}, B^{\bullet})$ complex for two cochain complexes $A^{\bullet}, B^{\bullet}$: $$Hom^n (A^{\bullet}, B^{\bullet})=\prod_{i} Hom (A^i, B^{i+n}).$$ Then in exercise IV.2.2.b…
Bananeen
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Sign convention for total complex

Let $C^{\bullet,\bullet}$ be a double complex with differentials $d,e$ both of degree $+1$.. I use the convention that the squares commute. I am mainly thinking of $C^{i,j}=A^i\otimes B^j$ or $C^{i,j}=hom(A^{-i},B^j)$ for cochain complexes…
Rene Recktenwald
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$\operatorname{Ext}^\bullet_R(R/rR,M)$ and $0 \to A[r] \to B[r] \to C[r] \to A/rA \to B/rB \to C/rC \to 0$

$\newcommand{Ext}{\operatorname{Ext}}\newcommand{Hom}{\operatorname{Hom}_R}$Let $R$ be a commutative ring with unity and $r \in R$. Applying snake lemma to the following diagram: $$\begin{array}{c} 0 & \longrightarrow & A~~ & \longrightarrow & B~~ &…
Kenny Lau
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Does the canonical map of the universal coefficient theorem induce an isomorphism?

Suppose $C$ is a free chain complex over a PID $R$, and $H(C)$ is finitely generated. From the universal coefficient theorem, there is a short exact sequence: $$ 0 \rightarrow \text{Ext}^1_R(H(C), R) \rightarrow H(C^*) \overset{h}{\rightarrow}…
Taketo Sano
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Why are there not $4$ kinds of derived functors/

Let $\mathcal{A}$ be an abelian category with enough injectives and projectives. Let $T: \mathcal{A} \to$ Ab be an additive functor into the category of abelian groups. You can assume that $T$ is left or right exact, if you wish. In Joseph Rotman's…
Future Algebraist
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DGA formality via $A_\infty$

Given an $A_\infty$-quasi-isomorphism $A\rightarrow B$ of dgas A and B how does one get a zig-zag of dga-quasi-isomorphisms $A\rightarrow \cdot \leftarrow \ldots \leftarrow \cdot \rightarrow B$? This is one implication in an equivalence on page 7 in…
Pavel
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Proving that Induction Maps Projective modules to Projective modules.

I'm trying to prove that that induction maps projective modules to projective modules but am getting a little stuck. I've read Module induced from projective is projective but I don't quite understand the details as he only talks about free modules…
Rhoswyn
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What is the reason direct product of injective modules are injective, while direct sum not necessarily?

In reviewing my algebra class material, I "discovered" a strange phenomenon, that the direct product of injective modules is injective, while if $R$ is not noetherian then the direct sum is not necessarily injective. This made me felt very…
Bombyx mori
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Exact sequence of total complexes

I'm working though Weibel right now, and I'm getting stuck on Exercise 1.3.6: If $0 \to A \to B \to C \to 0$ is an exact sequence of double complexes, show that $0 \to Tot(A) \to Tot(B) \to Tot(C) \to 0$ is exact. Obviously each $0 \to A_{p,q} \to…
Henry Swanson
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Question about definition of injective module

An $A$-module $N$ is injective if and only if $\text{Ext}_{A}^1(M,N)=0$ for any $A$-module $M$. I asked a similar question and got a hint but I am still stuck, this is how far I have got: If $N$ is injective we simply consider the injective…
user117449
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How does the long exact homology sequence coming from the mapping cone look like?

3.3.1. Let $f \colon P_\bullet \to Q_\bullet$ be a chain map of complexes. We define the mapping cone in the following way. Let $M_n = P_{n-1} \oplus Q_n$, and define $d_n^M \colon M_n \to M_{n-1}$ by $$ d_n^M(x,y) = ( -d_{n-1}^P(x),…
conrad
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