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