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:

Peter

 

[micromass]

My problem is that surely F = \psi' (F') implies that \phi ( F (d) )) = \phi ( \psi' ( F' (d)))

This cannot be true. We know that $F^\prime\in \textrm{Hom}(D,L)$, and thus $F^\prime(d)\in L$. But the domain of $\psi^\prime$ is not $L$, but rather $\textrm{Hom}(D,L)$. So $\psi^\prime(F^\prime(d))$ makes no sense since $F^\prime(d)$ is not in the domain of $\psi^\prime$.

 

[Math Amateur]

Thanks ...

Yes, you are right ...

BUT, now I have two problems ...

Problem 1 ... to get \phi ( F (d) )) = \phi ( \psi' ( F' (d))) I simply replaced the F with \psi' (F') since we have that F = \psi' (F') ... but presumably I did something 'illegal' ... can you explain my error?

Problem 2 ... How then does it follow that \phi ( F (d) )) = \phi ( \psi ( F' (d))) for any d \in D (given that F = \psi' (F')? Can you help?

Peter

 

[micromass]

Thanks ...

Yes, you are right ...

BUT, now I have two problems ...

Problem 1 ... to get \phi ( F (d) )) = \phi ( \psi' ( F' (d))) I simply replaced the F with \psi' (F') since we have that F = \psi' (F') ... but presumably I did something 'illegal' ... can you explain my error?

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))

 

[Math Amateur]

Thanks ... appreciate your help ... given me the confidence to go forward from there in my attempt to understand projective modules

Thanks again,

Peter