MHB 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 $$ \psi'$$ then $$ F = \psi' (F') $$ for some $$ F' \in Hom_R (D, L) $$ and so $$ \phi ( F (d) )) = \phi ( \psi ( F' (d))) $$ for any $$d \in D$$. ...

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

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 $$ \psi'$$ then $$ F = \psi' (F') $$ for some $$ F' \in Hom_R (D, L) $$ and so $$ \phi ( F (d) )) = \phi ( \psi ( F' (d))) $$ for any $$d \in D$$. ...

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

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

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

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