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Exact Sequences and short exact sequences - basic question

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
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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Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
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].
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
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 5817


Basis 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
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
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 $0 \longrightarrow \alpha (X) \longrightarrow Y \longrightarrow Y/\ker (\beta) \longrightarrow 0$.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
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
 

steenis

Well-known member
MHB Math Helper
Jul 30, 2016
250
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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