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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.

5114 questions
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$f$ can be extended iff $\partial f = 0$

If $0\rightarrow{A'}\rightarrow{A}\rightarrow{A''}\rightarrow{0}$ is an exact sequence of modules, then there exists an exact secuence $0\rightarrow{}Hom(A'',B)\rightarrow{}Hom(A,B)\rightarrow{}Hom(A',B)\xrightarrow…
user73564
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$f$ and $g$ is homotopic $\iff$ $g-f$ is null-homotopic?

In the projective model structure on the category of chain complexes, is the homotopicness between $f:A\to B$ and $g:A\to B$ is equivalent to null-homotopicness of $g-f:A\to B$ ? I think it's intuitively true but can not find the proof.
Yuta
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Submodules of homology modules

I have been dealing with certain subgroups of group cohomology, and the following general question comes to my mind. Suppose $C$ is a chain complex of $R$-modules and $H_n(C)$ its $n$-th homology module. Now let M be an $R$-submodule of $H_n(C)$.…
paul les
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homology of acyclic complex and left exact functor

Let $F$ be a left exact functor and $A^\bullet$ be a chain complexes: $A^\bullet: \cdots \xrightarrow{f^{i-1}} A^i \xrightarrow{f^i} A^{i+1} \xrightarrow{f^{i+1}} A^{i+2} \cdots$. Moreover, assume that each $A^i$ is $F$-acyclic. Then, I guess…
aerile
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Uniqueness in the proof of the comparison theorem for injective resolutions

$\require{AMScd}$ I'm trying to prove the comparison theorem for injective resolutions: Suppose $\mathcal{A}$ is an abelian category, $A\in\mathcal{A}$ and $A\xrightarrow{\varepsilon} I$ is an injective resolution. Then given $f^\prime:A\to B$ and…
SeraPhim
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Weibel 1.2.7: Existence of exact sequences of complexes.

If $C$ is a complex, show that there are exact sequences of complexes: $$ 0 \longrightarrow Z(C) \longrightarrow C \stackrel{d}{\longrightarrow} B(C)[-1] \longrightarrow 0; $$ $$ 0 \longrightarrow H(C) \longrightarrow C / B(C)…
Ryze
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Does Hom commute with mapping cone?

I was wondering is there an isomorphism $$\mathrm{Hom}\big(\mathrm{cone}(A\xrightarrow{f}B),\,C\big)\,\cong\,\mathrm{cone}\big(\mathrm{Hom}(B,\,C)\xrightarrow{f^\ast}\mathrm{Hom}(A,\,C)\big)\,?$$ On the level of graded vector spaces, one has the…
Yining Zhang
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Contractible complexes and resolutions

This is probably a very simple question, and but I'm stuck on a loophole I can't get wrap my head around. A (co)chain complex of $R$ modules (here $R$ is $k$ algebra, where $k$ is a field) $C^{\bullet}$ is called contractible if there exists a map…
Federico
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Chain equivalence proof

Let be $R$ a principal ideal domain. Let be $h: C \to D $ a homomorphism between complex chains, so $h= \{f_n: C_n \to D_n | n \in \mathbb{Z}\}$ and $h$ induces an isomorphism $g: H_n (C) \to H_n (D)$. Where $H_n(C)= Ker(\partial) / Im(\partial)$…
erika21148
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Describe the map $\text{Tor}_1^R (k,k)\otimes \text{Tor}_1^R(k,k)\to\text{Tor}_2^R(k,k)$.

Let $R=k[x,y]$, describe the map $\text{Tor}_1^R (k,k)\otimes \text{Tor}_1^R(k,k)\to\text{Tor}_2^R(k,k)$. The Koszul complex gives a projective resolution of $k$: $$ 0\to R\stackrel{\binom{-y}{x}}{\longrightarrow} R\oplus R\stackrel{(x\ \…
Ryze
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Compute $\text{Tor}_i^R(k,\mathfrak{a}^n)$

Let $R=k[x,y]$, $\mathfrak{a}=(x,y)$ be an ideal of $R$, how to compute $\text{Tor}_i^R(k,\mathfrak{a}^n)$ for all $i\ge 0$ and $n\ge 1$? Here $\mathfrak{a}^n$ denotes the ideal obtained by taking product of $\mathfrak{a}$ with itself, $n$…
Ryze
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Weibel IHA Exercise 1.2.4

About the Exercise 1.2.4 Suppose that $A.\xrightarrow{f} B.\xrightarrow{g} C.$ I know that $\ker{f}=\{\ker{f_n}\}$, $\mathrm{im}{f}=\{\mathrm{im} f_n\}$ are chain complexes of the abelian category $\mathcal{A}$, where…
闫嘉琦
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Projective Resolution and Homology

Let $A$ be an abelian category with enough projectives. We construct a projective resolution of an object C as follows: since $A$ has enough projectives we have an epimorphism from a projective object $P^0$ onto C. Let $ΩΑ$ denote its kernel. Given…
mits314
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A computation of $Ext^1(M,N)$ (how to derive the commutative diagram for $Ext^1(M,N)$?).

I am trying to understand the proof of Proposition 5.6 in the paper on page 17. How to derive the commutative diagram for $Ext^1(M,N)$? Using the projective resolution : \begin{align} \cdots \to \oplus_{v \in V} P_v \overset{D}{\to} \oplus_{u \in U}…
LJR
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Is the edge operator derived from the short exact sequence of the following chain complexes commutative?

As shown in the figure, $A_i $, $B_i$ and $C_i $on the left are all chain complexes, which make up 3 × 3 short exact sequences. Is the diagram on the right necessarily commutative? ($\partial $ is the homologous edge operator derived from snake…
GCauchy
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