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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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Motivating the definition of right derived functors in the context of derived categories.

Let $A$ and $B$ be abelian categories and let $F : A \to B$ be an additive functor. Let $K^+(F) : K^+(A) \to K^+(B)$ be the induced functor on the corresponding homotopy categories of left bounded complexes. Let $\pi_A$ and $\pi_B$ be the…
Partha Solapurkar
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How is cocone defined?

I am studying homological algebra and I am completely lost in sign conventions, I need someone to checjk the definitions I am using. Unfortunately the lecturer of my course is not very precise on this matter and we don't follow the notations of any…
arnett
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Weibel 1.5.8: Well-definedness of long exact sequence containing homology of $f$

I'm currently reading Weibel's An Introduction to Homological Algebra and I am a little stuck in section 1.5.8. In the sections before it is proven that the homology of each map $f : B_. \to C_.$ occurs in some long exact sequence, using mapping…
Tzimmo
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Split extensions and Ext functor

We consider the following exact sequences, first is a proyective resolution of $C$ and second is an extension $\xi$ of $A$ by $C$: $P_2\xrightarrow {d_2}{P_1}\xrightarrow{d_1}P_0\rightarrow C\rightarrow 0$ $0\rightarrow A\rightarrow B\rightarrow…
user74411
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Show that $\text{Ext}^n (X,Y) = 0$ for all $n \ge 2$ and give an example in which $X$ and $Y$ be two Abelian groups but $\text{Ext} (X,Y) \ne 0$

Problems: Let $X$ and $Y$ be two Abelian groups. Show that $\text{Ext}^n (X,Y) = 0$ for all $n \ge 2$. Give an example in which $X$ and $Y$ be two Abelian groups but $\text{Ext} (X,Y) \ne 0$. My attempt: Consider the exact sequence $$K \colon 0…
Minh
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Homotopy equivalent chain complexes

I am searching an example of (non-negative) chain complexes over a commutative ring $R$ which are not isomorphic but homotopy equivalent. Such an example could come from Algebraic Topology (for instance singular homology) but not necessarily. Thank…
bc87
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Homology of $H_1(X_m)$.

Let $X_m$ be a space obtained from $S^1$ by attaching $D^2$ through the map $f(z)=z^m$ around the boundary. I have computed the homology group of it by exact sequence $$\mathbb{Z} \cong H_1(S^1)\xrightarrow{\times m} H_1(X_m) \rightarrow…
Lev Bahn
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Projective dimension on a short exact sequence

Let $0 \to B \to P \to A \to 0$ be a short exact sequence of $R$-modules with $P$ is projective. By projective dimension of a module, I mean the length of the shortest projective resolution of it. A statement from Passman, A course in Ring Theory,…
Philip Johnson
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Homology of a simple chain complex

The question is to calculate the homology groups of the chain complex: $0 \to A \stackrel{n}{\to} A \to 0$, where $A$ is an Abelian group and $n \in \mathbb{N}$. I don't see a nice way to get anything out of this other than $H_1 = \mathrm{ker} \…
Juan S
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Short exact sequences whose long exact sequence in homology have the same boundary

I'm currently educating myself about homological algebra from Weibel's book (motivated by an interest in Floer and Khovanov knot homologies). After reading about long exact sequences in homology I became curious about what the boundary maps can tell…
Shai Deshe
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Naturality of connecting morphisms

It is well known that cohomology takes short exact sequences to long exact sequences. Moreover, the connecting morphisms are natural meaning that cohomology takes morphisms of short exact sequences to those of long exact sequences. Now,…
YYF
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Extensions are classified by $Ext^1(C, A)$: should one apply $Hom(\cdot, A)$ or $Hom(C, \cdot)$?

It is known that extensions $0 \to A \to B \to C \to 0$ are classified by $Ext^1(C, A)$. One can get such an element in two ways: applying $RHom(\cdot, A)$, one gets $$0 \to Hom(C, A) \to Hom(B, A) \to Hom(A, A) \to Ext^1(C, A)$$ and takes the…
evgeny
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Horseshoe lemma for Cartan-Eilenberg resolution

Let $$0 \to A^\bullet \to B^\bullet \to C^\bullet \to 0$$ be a short exact sequence of bounded complexes of objects in an abelian category with enough injectives. Does there exist Cartan-Eilenberg resolutions $A^\bullet \hookrightarrow…
HYL
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Ext for integer ring of torsion-free abelian groups

I am reading the book An Introduction of Homological Algebra by Rotman. I'm disturbed by Proposition 7.33 on page 427, which claim that If $F$ is a torsion-free abelian group and $T$ is a group of bounded order (i.e. $nT=\{0\}$ for some positive…
liwolf
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Homotopic homomorphisms of chain complexes induce the same homomorphisms of on the level of homology

I read this statement in Siegfried Bosch's "Algebraic geometry and commutative algebra", but it is not proved in the book. Basically it says that if $M_*$: $$\cdot\cdot\cdot \rightarrow M_{n+1} \xrightarrow{d_{n+1}} M_n \xrightarrow{d_{n}} M_{n-1}…
brick
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