Questions tagged [exact-sequence]

A sequence of morphisms where the image of one is the kernel of the next. It is a useful thing to examine in the study of abstract algebra and homological algebra. Use this tag if your question is about the general theory of exact sequences, not just because an exact sequence appears in it. For sequences of numbers use the tag (sequences-and-series) instead.

An exact sequence in a category a sequence of of morphisms in that category

$$ \dotsb \xrightarrow{\;\;\varphi_{i-1}\;\;} X_i\xrightarrow{\;\;\varphi_{i}\;\;} X_{i+1}\xrightarrow{\;\;\varphi_{i+1}\;\;} \dotsb $$

such that the image of $\varphi_j$ is equal to the kernel of $\varphi_{j+1}$ for any $j$. Any (long) exact sequence can be decomposed in a reasonable way into short exact sequences, so these are more often the objects that we examine. A short exact sequence is a sequence

$$ 0 \to B \xrightarrow{\;\;\varphi\;\;} C \xrightarrow{\;\;\psi\;\;} A \to 0 $$

Such that $\mathrm{Im}(\varphi) = \mathrm{Ker}(\psi)$, $\varphi$ is an monomorphism, and $\psi$ is an epimorphism. The object $C$ is referred to as an extension of $A$ by $B$. Exact sequences are major objects of study in the broader areas of abstract algebra and homological algebra.

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Sequences of Abelian Groups Split

Show that an exact sequence of abelian groups $$0 \rightarrow \mathbb{Q} \rightarrow M \rightarrow N \rightarrow 0$$ always splits. If $f: \mathbb{Q} \rightarrow M$ and $g: M \rightarrow N$, what needs to be found is a map $h: N \rightarrow M$…
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Showing that a certain map in a commutative diagram with exact rows is injective

This exercise is from Dummit and Foote, Section 10.5. (Exercise 1, part d) The following diagram is commutative with exact rows. We know that $\alpha,\gamma$ are surjective, and $\beta$ is injective. I want to show that $\gamma$ is injective. I…
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Exact Sequences

Consider the following exact sequences: $$ 0 \to A \to B$$ and $$B \to C \to 0$$ The first is exact at $A$ iff the kernel of the map from $A$ to $B$ is $\{ 0 \}$. The second is exact at $C$ iff the image of the map from $B$ to $C$ is all of $C$.…
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exactness of $A \to B \to C \to 0$

Suppose we want to prove the exactness of the following sequence in an abelian category: $A \to B \to C \to 0$ and we already have the exactness at $C$. Suppose now the cokernel $B \to Y$ of $A \to B$ factors through $C$, how to show the exactness…
scsnm
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Proof about exact sequence

Here is the sequence I have the next sequence where the row is exact and $ h \circ f =0$. I need to prove that exists a homomorphism $k: C \to D$ uniquely determined such that $ k \circ g = h$
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Is this an exact sequence split?

I am reading about split exact sequence and come out with a question. Consider the following sequence. $0 \to A\to A\oplus B\oplus C\to B\to0$ It seems to me that it's exact. So, is it an split exact sequence? To be more explicit, just set $A\to…
Yuyi Zhang
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Why every short exact sequence of finitely generated $\mathbb{C}G$-modules splits?

Let $G$ be a finite group and $\mathbb{C}$ be the complex field. $L,M,N$ are finitely generated $\mathbb{C}G$-modules (Where $\mathbb{C}G$ denotes a group ring). Show if $$0\rightarrow L \rightarrow M \rightarrow N\rightarrow 0$$ is a short exact…
Maddy
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Mayer Vietoris - arbitrary homology

Section 2.2, exercise 38 in Hatcher's algebraic topology textbook is: I can show that the final sequence $$\cdots \to E_{n+1} \to B_n \to C_n \oplus D_n \to E_n \to B_{n-1} \to \cdots$$ has the property that composing two maps gives zero, but I'm…
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Understanding splitting lemma

Read this. If the sequence is short exact, then it already has arrows $0 \to \dots$ and $\dots \to 0$ which imply injectivity and surjectivity respectively, right? So in that case there is already always such a map $t$. So what am I getting…
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Directs sum in exact sequences

Let $0\to L\stackrel{\alpha}\to M\stackrel{\beta}\to N\to 0$ be a splitting s.e.s. where $\alpha$ has a retraction $r$. (a) Show that in this case $M=\alpha(L)\oplus\ker(r)$. I am a little bit confused because we have defined $\oplus$ as "direct…
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First function in a short exact sequence

Let $0\to L\to M\to N \to0$ be short exact sequence. How does the function $0\to L$ look like? And what does $0$ mean here? Is it the zero of $A$ (a commutative ring s.t. $L,M,N$ are $A$-modules)?
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Proof that sequence is not split using left inverse

For the following exact sequence, i am trying to proof that it is not split $0\to 2\mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0$ I want to show that for injective function $i : 2\mathbb{Z} \to \mathbb{Z}$ if there exists left inverse…
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Why induced exact sequence implies injectivity in question

N is a module and M & L are submodules of N I want to show that $\beta$ is injective but am having trouble understanding why the exact sequence above implies injectivity.
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Regarding the exact sequence

Consider the exact sequence $$ \cdots \rightarrow 0 \rightarrow \mathbb{Z} \overset{\times p}{\rightarrow} \color{blue} {\mathbb{Z}} \twoheadrightarrow \mathbb{Z} /p \mathbb{Z} \rightarrow 0 \rightarrow \cdots$$ Show that in the middle group…
MAS
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Exact sequence $A\hookrightarrow B\to C\to\cdots$ and kernel

Given an exact sequence $$ 0\to A\to B\overset\alpha\to C\to\dots $$ is $$ \ker\alpha\cong A $$ true? I feel that the answer is trivially "yes, by definition" but I couldn't find this expression anywhere so now I'm in doubt. Anyway, I'm interested…
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