Exact Sequences and short exact sequences - basic question

In summary: The first three terms of the sequence 0 \longrightarrow \alpha (X) \longrightarrow Y actually mean:1. The group $0$ is mapped to the group $\alpha(X)$ by the map $\alpha(X)$.2. The group $\alpha(X)$ is mapped to the group $Y$ by the map $\alpha(X)$.3. The group $Y$ is mapped to the group $Y/ \ker \beta$ by the map $\beta$.It is not talking about the image of $0$ in $\alpha(X)$, but rather the image of the group $0$ under the map $\alpha(X)$.I hope this helps clarify the statement for you. In
  • #1
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In Dummit and Foote Section 10.5 Exact Sequences (see attachment) we read the following on page 379:

"Note that any exact sequence can be written as a succession of short exact sequences since to say

[TEX] X \longrightarrow Y \longrightarrow Z [/TEX]

[where the homomorphisms involved are as follows; [TEX] \alpha \ : \ X \longrightarrow Y [/TEX] and [TEX] \beta \ : \ Y \longrightarrow Z [/TEX]

is exact at Y is the same as saying that the sequence

[TEX] 0 \longrightarrow \alpha (X) \longrightarrow Y/ {ker \beta} \longrightarrow 0 [/TEX]

is a short exact sequence.

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I am trying to get an understanding of this statement.

Can someone please demonstrate formally that this is true.

To enable me to get an understanding of this it would help enormously if someone could devise an example of this.

Peter

[This has also been posted on MHF]
 
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  • #2
Peter said:
In Dummit and Foote Section 10.5 Exact Sequences (see attachment) we read the following on page 379:

"Note that any exact sequence can be written as a succession of short exact sequences since to say

[TEX] X \longrightarrow Y \longrightarrow Z [/TEX]

[where the homomorphisms involved are as follows; [TEX] \alpha \ : \ X \longrightarrow Y [/TEX] and [TEX] \beta \ : \ Y \longrightarrow Z [/TEX]

is exact at Y is the same as saying that the sequence

[TEX] 0 \longrightarrow \alpha (X) \longrightarrow Y/ {ker \beta} \longrightarrow 0 [/TEX]

is a short exact sequence.

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I am trying to get an understanding of this statement.

Can someone please demonstrate formally that this is true.
You have misquoted Dummit and Foote. What they actually write is: "to say [TEX] X \stackrel{\alpha}{\longrightarrow} Y \stackrel{\beta}{\longrightarrow} Z [/TEX] is exact at $Y$ is the same as saying that the sequence

[TEX] 0 \longrightarrow \alpha (X) \longrightarrow Y \longrightarrow Y/\ker (\beta) \longrightarrow 0 [/TEX]

is a short exact sequence."

To see why this is true, remember that the definition of [TEX] X \stackrel{\alpha}{\longrightarrow} Y \stackrel{\beta}{\longrightarrow} Z [/TEX] being exact at $Y$ is that $\alpha(X) = \ker(\beta)$.

In the short exact sequence [TEX] 0 \longrightarrow \alpha (X) \longrightarrow Y \longrightarrow Y/\ker (\beta) \longrightarrow 0 [/TEX], there are three places where exactness must be checked. For the first three terms of that sequence, exactness for [TEX] 0 \longrightarrow \alpha (X) \longrightarrow Y [/TEX] means that (the image of) $0$ in $\alpha(X)$ is the kernel of the inclusion map $\alpha (X) \subset Y$ (which is obviously true). For the last three terms of the sequence, exactness for [TEX] Y \longrightarrow Y/\ker (\beta) \longrightarrow 0 [/TEX] means that the image of $Y$ in the quotient $Y/ \ker (\beta)$ is the kernel of the map taking everything in $Y/ \ker (\beta)$ to $0$ (which is again obviously true, since both those things are the whole of $Y/ \ker (\beta)$). Finally, exactness for the middle three terms of the short exact sequence says that the image of $\alpha$ is equal to the kernel of the quotient map $\beta$. That is the same as saying that [TEX] X \stackrel{\alpha}{\longrightarrow} Y \stackrel{\beta}{\longrightarrow} Z [/TEX] is exact.

Putting those things together, you see that exactness for the short exact sequence is equivalent to exactness for [TEX] X \stackrel{\alpha}{\longrightarrow} Y \stackrel{\beta}{\longrightarrow} Z [/TEX].
 
  • #3
Opalg said:
You have misquoted Dummit and Foote. What they actually write is: "to say [TEX] X \stackrel{\alpha}{\longrightarrow} Y \stackrel{\beta}{\longrightarrow} Z [/TEX] is exact at $Y$ is the same as saying that the sequence

[TEX] 0 \longrightarrow \alpha (X) \longrightarrow Y \longrightarrow Y/\ker (\beta) \longrightarrow 0 [/TEX]

is a short exact sequence."

To see why this is true, remember that the definition of [TEX] X \stackrel{\alpha}{\longrightarrow} Y \stackrel{\beta}{\longrightarrow} Z [/TEX] being exact at $Y$ is that $\alpha(X) = \ker(\beta)$.

In the short exact sequence [TEX] 0 \longrightarrow \alpha (X) \longrightarrow Y \longrightarrow Y/\ker (\beta) \longrightarrow 0 [/TEX], there are three places where exactness must be checked. For the first three terms of that sequence, exactness for [TEX] 0 \longrightarrow \alpha (X) \longrightarrow Y [/TEX] means that (the image of) $0$ in $\alpha(X)$ is the kernel of the inclusion map $\alpha (X) \subset Y$ (which is obviously true). For the last three terms of the sequence, exactness for [TEX] Y \longrightarrow Y/\ker (\beta) \longrightarrow 0 [/TEX] means that the image of $Y$ in the quotient $Y/ \ker (\beta)$ is the kernel of the map taking everything in $Y/ \ker (\beta)$ to $0$ (which is again obviously true, since both those things are the whole of $Y/ \ker (\beta)$). Finally, exactness for the middle three terms of the short exact sequence says that the image of $\alpha$ is equal to the kernel of the quotient map $\beta$. That is the same as saying that [TEX] X \stackrel{\alpha}{\longrightarrow} Y \stackrel{\beta}{\longrightarrow} Z [/TEX] is exact.

Putting those things together, you see that exactness for the short exact sequence is equivalent to exactness for [TEX] X \stackrel{\alpha}{\longrightarrow} Y \stackrel{\beta}{\longrightarrow} Z [/TEX].

Thanks to Opalg for this help in the past ...

I am revising exact sequences and must admit to still being uneasy about the above example taken from this remark in Dummit and Foote:View attachment 5817Basis questions are as follows:Question 1

D&F write that saying [TEX] X \stackrel{\alpha}{\longrightarrow} Y \stackrel{\beta}{\longrightarrow} Z [/TEX] is exact at \(\displaystyle Y\)

is the same as saying that

[TEX] 0 \longrightarrow \alpha (X) \longrightarrow Y \longrightarrow Y/\ker (\beta) \longrightarrow 0 [/TEX]

is a short exact sequence ...... BUT ... they do not specify the particular homomorphisms involved ... how are we to fully understand the nature of a short exact sequence when the homomorphisms involved are not specified ...?

Question 2

in the above post to me, Opalg writes:

" ... ... For the first three terms of that sequence, exactness for [TEX] 0 \longrightarrow \alpha (X) \longrightarrow Y [/TEX] means that (the image of) $0$ in $\alpha(X)$ is the kernel of the inclusion map $\alpha (X) \subset Y$ (which is obviously true). ... ... "
But the above seems to assume that the homomorphism from \(\displaystyle \alpha (X)\) to \(\displaystyle Y\) is the inclusion map ... ... BUT ... ... how do we know it is the inclusion map ... ... ? Do we have to deduce it is thus, in order that the sequence is exact ... ?

Question 3

In the above post to me, Opalg writes:

" ... ... Finally, exactness for the middle three terms of the short exact sequence says that the image of $\alpha$ is equal to the kernel of the quotient map $\beta$. That is the same as saying that [TEX] X \stackrel{\alpha}{\longrightarrow} Y \stackrel{\beta}{\longrightarrow} Z [/TEX] is exact. ... ... "
My problem is as follows:

... ... how does exactness for the middle three terms of

[TEX] 0 \longrightarrow \alpha (X) \longrightarrow Y \longrightarrow Y/\ker (\beta) \longrightarrow 0 [/TEX]

say that, or mean that ...

... the image of $\alpha$ is equal to the kernel of the quotient map $\beta$?

Hope someone can help ... ...

Thanks again to Opalg for the previous help ...

Peter

Peter
 
  • #4
To describe a short exact sequence $ 0 \longrightarrow A \stackrel{\theta}{\longrightarrow} B \stackrel{\phi}{\longrightarrow} C \longrightarrow 0 $ it is usually necessary to specify the homomorphisms $\theta$ and $\phi$ as well as the spaces $A,B,C$. (But notice that even here it is not necessary to specify the homomorphisms $ 0 \longrightarrow A$ and $C \longrightarrow 0 $, since these are uniquely determined.)

However, in the case where $A$ is contained in $B$ people often write $ 0 \longrightarrow A \longrightarrow B \longrightarrow B/A \longrightarrow 0 $ without specifying any of the homomorphisms, on the understanding that these are the natural inclusion and quotient maps.

I assume that is why Dummit and Foote omit specifying the homomorphisms when they write $0 \longrightarrow \alpha (X) \longrightarrow Y \longrightarrow Y/\ker (\beta) \longrightarrow 0$.
 
  • #5
Opalg said:
To describe a short exact sequence $ 0 \longrightarrow A \stackrel{\theta}{\longrightarrow} B \stackrel{\phi}{\longrightarrow} C \longrightarrow 0 $ it is usually necessary to specify the homomorphisms $\theta$ and $\phi$ as well as the spaces $A,B,C$. (But notice that even here it is not necessary to specify the homomorphisms $ 0 \longrightarrow A$ and $C \longrightarrow 0 $, since these are uniquely determined.)

However, in the case where $A$ is contained in $B$ people often write $ 0 \longrightarrow A \longrightarrow B \longrightarrow B/A \longrightarrow 0 $ without specifying any of the homomorphisms, on the understanding that these are the natural inclusion and quotient maps.

I assume that is why Dummit and Foote omit specifying the homomorphisms when they write $0 \longrightarrow \alpha (X) \longrightarrow Y \longrightarrow Y/\ker (\beta) \longrightarrow 0$.
Thanks Opalg ... I appreciate your help ...

Peter
 
  • #6
Remark 1
Given R-maps $\alpha : X\longrightarrow Y$ and $\beta : Y\longrightarrow Z$ then

The sequence $X\longrightarrow _\alpha Y\longrightarrow _\beta Z$ is exact in Y if and only if $0\longrightarrow \alpha X \longrightarrow _i Y\longrightarrow _\pi Y/ \ker \beta \longrightarrow 0$ is a short exact sequence. (i is the natural inclusion, $\pi$ is the natural projection.) This is very easy to prove.

Remark 2
I think D&F are really too concise here. I think what they meant here, is better stated by Rotman in exercise 2.6 on p.65 (Rotman - An Introduction to Homological Algebra 2nd edition 2009).
 
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FAQ: Exact Sequences and short exact sequences - basic question

1. What is an exact sequence?

An exact sequence in mathematics is a sequence of objects and morphisms between them that captures the notion of a chain of inclusions or a chain of surjections. In other words, it is a sequence where the image of one object is exactly equal to the kernel of the next object in the sequence.

2. What is the importance of exact sequences?

Exact sequences are important in various fields of mathematics, including algebra, topology, and homological algebra. They provide a powerful tool for studying the relationships between objects and their properties, and for proving important theorems and results.

3. What is a short exact sequence?

A short exact sequence is a special type of exact sequence in which there are only three objects and two morphisms, and the image of the first morphism is equal to the kernel of the second morphism. This type of sequence is particularly useful in algebraic structures such as groups, rings, and modules.

4. How do you determine if a sequence is exact?

To determine if a sequence is exact, you need to check if the image of one object is equal to the kernel of the next object in the sequence. This can be done by applying the corresponding homomorphisms to elements in the objects and checking if the resulting elements are equal.

5. Can you give an example of a short exact sequence?

One example of a short exact sequence is 0 → ℤ → ℤ² → ℤ → 0, where the first morphism is the inclusion of the integers into the 2-dimensional integer lattice, and the second morphism is the projection onto the first coordinate. This sequence is exact because the image of the first morphism (the integers) is exactly equal to the kernel of the second morphism (the elements with 0 as the second coordinate).

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