Exact Sequences - Lifting Homomorphisms - D&F Ch 10 - Theorem 28

[Math Amateur]

I am reading Dummit and Foote, Chapter 10, Section 10.5, Exact Sequences - Projective, Injective and Flat Modules.

I need help with a minor step of D&F, Chapter 10, Theorem 28 on liftings of homomorphisms.

In the proof of the first part of the theorem (see image below) D&F make the following statement:

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Conversely, if F is in the image of \(\displaystyle \psi'\) then \(\displaystyle F = \psi' (F') \) for some \(\displaystyle F' \in Hom_R (D, L) \) and so \(\displaystyle \phi ( F (d) )) = \phi ( \psi ( F' (d))) \) for any \(\displaystyle d \in D\). ...

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My problem is that surely \(\displaystyle F = \psi' (F') \) implies that \(\displaystyle \phi ( F (d) )) = \phi ( \psi' ( F' (d))) \) and NOT \(\displaystyle \phi ( F (d) )) = \phi ( \psi ( F' (d))) \)?

Hoping someone can help>

Theorem 28 and the first part of the proof read as follows:

https://www.physicsforums.com/attachments/2474

Peter

NOTE: This has also been posted on the Physics Forums in the forum Linear and Abstract Algebra.

 

[Math Amateur]

I am reading Dummit and Foote, Chapter 10, Section 10.5, Exact Sequences - Projective, Injective and Flat Modules.

I need help with a minor step of D&F, Chapter 10, Theorem 28 on liftings of homomorphisms.

In the proof of the first part of the theorem (see image below) D&F make the following statement:

-------------------------------------------------------------------------------

Conversely, if F is in the image of \(\displaystyle \psi'\) then \(\displaystyle F = \psi' (F') \) for some \(\displaystyle F' \in Hom_R (D, L) \) and so \(\displaystyle \phi ( F (d) )) = \phi ( \psi ( F' (d))) \) for any \(\displaystyle d \in D\). ...

----------------------------------------------------------------------------

My problem is that surely \(\displaystyle F = \psi' (F') \) implies that \(\displaystyle \phi ( F (d) )) = \phi ( \psi' ( F' (d))) \) and NOT \(\displaystyle \phi ( F (d) )) = \phi ( \psi ( F' (d))) \)?

Hoping someone can help>

Theorem 28 and the first part of the proof read as follows:

https://www.physicsforums.com/attachments/2474

Peter

NOTE: This has also been posted on the Physics Forums in the forum Linear and Abstract Algebra.

Just a note to say I received some excellent help from micromass on the Linear and Abstract Algebra forum of the Physics forum.

The solution/answer due to micromass was as follows:

If you simply substituted, then you would have gotten

$$F(d) = \psi^\prime(F^\prime)(d)$$

Thus \(\displaystyle \psi^\prime(F^\prime)\) acts on \(\displaystyle d\). You would not get $$F(d) = \psi^\prime(F^\prime(d))$$

Now, by definition, we have \(\displaystyle \psi^\prime(F^\prime) = \psi\circ F^\prime\). Thus

$$F(d) = \psi^\prime(F^\prime)(d) = (\psi\circ F^\prime)(d) = \psi(F^\prime(d))$$

 

[Deveno]

It's easy to get confused about these things:

There are 3 "layers" here:

Layer 1: individual $R$-modules.

Layer 2: homomorphisms (arrows) between different modules.

Layer 3: arrows between sets of arrows (the "primed" maps).

As such, when you see something like $f(-)$ ("$f$" of something), you have to check the "something" is on the right layer.

It may, or may not, be helpful to see the sequence (10) as being something like a cone, with the sequence:

$0 \to L \to M \to N$ at its "base", and $D$ at the apex.

So, for example, the map $\psi'$ is actually a triangular commutative diagram:

$\begin{array}[l]{ccc}D\\ \downarrow\rlap{f}&\stackrel{\psi'(f)}{\searrow}\\L&\stackrel{\psi}{\longrightarrow}&M \end{array}$

(my apologies for the poor drawing).

 

[Math Amateur]

It's easy to get confused about these things:

There are 3 "layers" here:

Layer 1: individual $R$-modules.

Layer 2: homomorphisms (arrows) between different modules.

Layer 3: arrows between sets of arrows (the "primed" maps).

As such, when you see something like $f(-)$ ("$f$" of something), you have to check the "something" is on the right layer.

It may, or may not, be helpful to see the sequence (10) as being something like a cone, with the sequence:

$0 \to L \to M \to N$ at its "base", and $D$ at the apex.

So, for example, the map $\psi'$ is actually a triangular commutative diagram:

$\begin{array}[l]{ccc}D\\ \downarrow\rlap{f}&\stackrel{\psi'(f)}{\searrow}\\L&\stackrel{\psi}{\longrightarrow}&M \end{array}$

(my apologies for the poor drawing).

Thanks Deveno, appreciate your thoughts ...

... Helpful in my attempt to understand projective, injective and flat modules ...

Peter

 

[blue_raver22]

Dear Peter,

Thank you for bringing this to my attention. I agree with your observation that the statement in the proof should be \phi ( F (d) )) = \phi ( \psi' ( F' (d))) instead of \phi ( F (d) )) = \phi ( \psi ( F' (d))). This is a minor error in the proof and does not affect the overall validity of the theorem. However, it is important to correct this in order to maintain the accuracy of the proof.

To see why this correction is necessary, let us consider the definition of the image of a homomorphism. The image of a homomorphism \psi : A \rightarrow B is defined as the set of all elements of B that can be obtained by applying \psi to elements of A. In other words, if we have an element b \in B, then b is in the image of \psi if and only if there exists an element a \in A such that \psi(a) = b. In the proof of Theorem 28, we are trying to show that if F is in the image of \psi', then F = \psi' (F') for some F' \in Hom_R (D, L). This means that F is an element of Hom_R (D, L) that can be obtained by applying \psi' to elements of D.

Now, let us look at the statement in question: \phi ( F (d) )) = \phi ( \psi' ( F' (d))). As we established, F is an element of Hom_R (D, L) and \psi' is a homomorphism from D to L. This means that \psi' ( F' (d)) is an element of L, and since \phi is a homomorphism from L to M, \phi ( \psi' ( F' (d))) is an element of M. Therefore, \phi ( F (d) )) and \phi ( \psi' ( F' (d))) are both elements of M, and the statement is correct.

On the other hand, if we were to use \phi ( F (d) )) = \phi ( \psi ( F' (d))) instead, this would imply that F is in the image of \psi, which is not necessarily true. In fact, if we assume that F is in the image of \psi, then