# Exact Sequences - Isomorphisms Resulting from Exact Sequences

#### Peter

##### Well-known member
MHB Site Helper
I am reading Dummit and Foote Section 10.5 Exact Sequences - Projective, Injective and Flat Modules.

I need help with some of the conclusions to Example 2, D&F Section 10.5, pages 379-380 - see attached. However, note that the question is essentially about isomorphisms. However, I would like someone to confirm the correctness of the exact sequences context.

To establish the context of the example as I interpret it ...

On pages 378-379 D&F, as I see it, establish the following:

If we have

$$\displaystyle A \stackrel{\psi}{\longrightarrow} B \stackrel{\phi}{\longrightarrow} C$$

where $$\displaystyle \psi$$ is an injective homomorphism

and $$\displaystyle \phi$$ is a surjective homomorphism and $$\displaystyle \text{ ker } \phi = \psi (A)$$

then

(1) $$\displaystyle A \cong \psi (A)$$ because the homomorphism $$\displaystyle \psi \ : \ A \to \psi (A)$$ is both injective and surjective

and

(2) by the First Isomorphism Theorem $$\displaystyle C \cong B/ \text{ker } \phi \text{ or } C \cong B/ \psi (A)$$

so that C is isomorphic to the quotient of B by an isomorphic copy of A

or, in other words B is an extension of C by A.

Can someone please confirm that my interpretation of the situation regarding extensions of C by A as outlined by D&F is correct. I would definitely feel more confident if someone was to confirm the above as OK!

Now Example 2 is a special case of Example 1 - see bottom of page 379 and top of page 380 on the attachment.

In Example 2 D&F give two short exact sequences (see attachment).

The first exact sequence, I believe, establishes the following isomorphism:

$$\displaystyle \mathbb{Z} / n \mathbb{Z} \cong ( \mathbb{Z} \oplus ( \mathbb{Z} / n \mathbb{Z}) ) / \text{ ker } \phi$$

that is,

$$\displaystyle \mathbb{Z} / n \mathbb{Z} \cong ( \mathbb{Z} \oplus ( \mathbb{Z} / n \mathbb{Z}) ) / i ( \mathbb{Z} )$$ ... ... ... (3)

The second exact sequence, I believe, establishes the following isomorphism

$$\displaystyle \mathbb{Z} / n \mathbb{Z} \cong \mathbb{Z}) ) / \text{ ker } \phi$$

that is

$$\displaystyle \mathbb{Z} / n \mathbb{Z} \cong \mathbb{Z}) ) / n ( \mathbb{Z} )$$ ... ... ... (4)

Now it seems (intuitively) that given (3) and (4) above we should have

$$\displaystyle ( \mathbb{Z} \oplus ( \mathbb{Z} / n \mathbb{Z}) ) / i ( \mathbb{Z} ) \cong \mathbb{Z} / n ( \mathbb{Z} )$$ ... ... ... (5)

since they are both isomorphic to $$\displaystyle \mathbb{Z} / n \mathbb{Z}$$

BUT

D&F say that this is explicitly not the case ... indeed, they write:

"Note that the modules in the middle of the previous two exact sequences are not isomorphic even though the respective "A" and "C" terms are isomorphic. Then there are at least two "essentially different" or "inequivalent" ways of extending $$\displaystyle \mathbb{Z} / n \mathbb{Z}$$ by $$\displaystyle \mathbb{Z}$$.

Can someone please clarify this situation for me.

Peter

Last edited:

#### Peter

##### Well-known member
MHB Site Helper
I am reading Dummit and Foote Section 10.5 Exact Sequences - Projective, Injective and Flat Modules.

I need help with some of the conclusions to Example 2, D&F Section 10.5, pages 379-380 - see attached. However, note that the question is essentially about isomorphisms. However, I would like someone to confirm the correctness of the exact sequences context.

To establish the context of the example as I interpret it ...

On pages 378-379 D&F, as I see it, establish the following:

If we have

$$\displaystyle A \stackrel{\psi}{\longrightarrow} B \stackrel{\phi}{\longrightarrow} C$$

where $$\displaystyle \psi$$ is an injective homomorphism

and $$\displaystyle \phi$$ is a surjective homomorphism and $$\displaystyle \text{ ker } \phi = \psi (A)$$

then

(1) $$\displaystyle A \cong \psi (A)$$ because the homomorphism $$\displaystyle \psi \ : \ A \to \psi (A)$$ is both injective and surjective

and

(2) by the First Isomorphism Theorem $$\displaystyle C \cong B/ \text{ker } \phi \text{ or } C \cong B/ \psi (A)$$

so that C is isomorphic to the quotient of B by an isomorphic copy of A

or, in other words B is an extension of C by A.

Can someone please confirm that my interpretation of the situation regarding extensions of C by A as outlined by D&F is correct. I would definitely feel more confident if someone was to confirm the above as OK!

Now Example 2 is a special case of Example 1 - see bottom of page 379 and top of page 380 on the attachment.

In Example 2 D&F give two short exact sequences (see attachment).

The first exact sequence, I believe, establishes the following isomorphism:

$$\displaystyle \mathbb{Z} / n \mathbb{Z} \cong ( \mathbb{Z} \oplus ( \mathbb{Z} / n \mathbb{Z}) ) / \text{ ker } \phi$$

that is,

$$\displaystyle \mathbb{Z} / n \mathbb{Z} \cong ( \mathbb{Z} \oplus ( \mathbb{Z} / n \mathbb{Z}) ) / i ( \mathbb{Z} )$$ ... ... ... (3)

The second exact sequence, I believe, establishes the following isomorphism

$$\displaystyle \mathbb{Z} / n \mathbb{Z} \cong \mathbb{Z}) ) / \text{ ker } \phi$$

that is

$$\displaystyle \mathbb{Z} / n \mathbb{Z} \cong \mathbb{Z}) ) / n ( \mathbb{Z} )$$ ... ... ... (4)

Now it seems (intuitively) that given (3) and (4) above we should have

$$\displaystyle ( \mathbb{Z} \oplus ( \mathbb{Z} / n \mathbb{Z}) ) / i ( \mathbb{Z} ) \cong \mathbb{Z} / n ( \mathbb{Z} )$$ ... ... ... (5)

since they are both isomorphic to $$\displaystyle \mathbb{Z} / n \mathbb{Z}$$

BUT

D&F say that this is explicitly not the case ... indeed, they write:

"Note that the modules in the middle of the previous two exact sequences are not isomorphic even though the respective "A" and "C" terms are isomorphic. Then there are at least two "essentially different" or "inequivalent" ways of extending $$\displaystyle \mathbb{Z} / n \mathbb{Z}$$ by $$\displaystyle \mathbb{Z}$$.

Can someone please clarify this situation for me.

Peter
I am making a quick reply to clarify my post ... I still need an issue clarified ... but I also believe I did not read D&F clearly enough ...

I still think that my contention in (5) in my previous post, namely:

$$\displaystyle ( \mathbb{Z} \oplus ( \mathbb{Z} / n \mathbb{Z}) ) / i ( \mathbb{Z} ) \cong \mathbb{Z} / n ( \mathbb{Z} )$$ ... ... ... (5)

is correct. Do members agree?

However, D&F are correct in saying that the respective "A" and "C" terms in the two exact sequences are isomorphic - indeed they are the same (see attachment - bottom of page 379 and top of page 380) - respectively, they are $$\displaystyle \mathbb{Z}$$ and $$\displaystyle \mathbb{Z} / n \mathbb{Z}$$ in both cases. BUT also as D&F say, the modules in the middle of the two exact sequences (the "B" terms), namely $$\displaystyle \mathbb{Z} \oplus ( \mathbb{Z} / n \mathbb{Z})$$ and $$\displaystyle \mathbb{Z}$$ are different - not isomorphic. Yes, I can see this ... ... and that probably means that D&F's statement that "thus there are (at least) two "essentially different" or "inequivalent" ways of extending $$\displaystyle \mathbb{Z} / n \mathbb{Z}$$ by $$\displaystyle \mathbb{Z}$$" is correct ....

BUT ... ... it still seems to me that the two (different?) extensions of $$\displaystyle \mathbb{Z} / n \mathbb{Z}$$ by $$\displaystyle \mathbb{Z}$$ ... ... that is the actual extensions (as distinct from the ways of extending) namely $$\displaystyle ( \mathbb{Z} \oplus ( \mathbb{Z} / n \mathbb{Z}) ) / i ( \mathbb{Z} )$$ and $$\displaystyle \mathbb{Z} / n ( \mathbb{Z} )$$ are isomorphic ... ...

Is this correct?

Peter

#### Deveno

##### Well-known member
MHB Math Scholar
I think you are mis-understanding what is being said:

the extensions are, in the two cases:

$\Bbb Z \oplus \Bbb Z_n$ and $\Bbb Z$, and these two abelian groups are definitely NOT isomorphic.

In this scenario, "extension" actually means "reverse quotient" or "pre-image of a quotient".

It IS true that:

$(\Bbb Z \oplus \Bbb Z_n)/(\Bbb Z \oplus \{\}) \cong \Bbb Z_n$

this is what D&F mean by "the $C$'s are isomorphic".

What is usually more useful is a "split extension":

$0 \to A \to B \leftrightarrows C \to 0$

where the composition of the arrows between $B$ and $C$ (starting at $C$) yields the identity on $C$.

#### ThePerfectHacker

##### Well-known member
On pages 378-379 D&F, as I see it, establish the following:

If we have

$$\displaystyle A \stackrel{\psi}{\longrightarrow} B \stackrel{\phi}{\longrightarrow} C$$
Quite simply an exact sequence is when the kernel agrees with the image. So look at the middle term, $B$. You have a map going into $B$ by $\psi$ and a map going out of $B$ by $\varphi$. To be an exact sequence you need that $\ker \varphi = \text{im} \psi$.