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- Jun 22, 2012

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I am at present trying to understand the use of Theorem 6 (D&F, page 354 - see attachment) in Theorem 8 (D&F page 362, see attachment).

The proof of Theorem 8 in D&F Chapter 10 (see attachment) reads as follows:

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Proof: Suppose \(\displaystyle \phi : \ N \to L \) is an R-module homomorphism to the S-module L.

By the universal property of free modules (

**Theorem 6 in Section 3**) there is a Z-module homomorphism from the free Z-module F on the set S x N to L that sends each generator \(\displaystyle (s,n) \) to \(\displaystyle s \phi (n) \). ... ...

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I am trying to see how the use of Theorem 6 applies and operates in the proof of Theorem 8 - so I am essentially trying to "tie up" or "tie together" the symbols and structures of Therem 6 with those of Theorem 8.

Now we know that in Theorem 8 " ... there is a Z-module homomorphism from the free Z-module F on the set S x N to L ..."

Thus for the map \(\displaystyle A \to F(A) \) in Theorem 6

we have the map

\(\displaystyle S \times N \to F(S \times N) \) in Theorem 8, but this map is not showing in the diagram for Theorem 8 on page 362 (see attachment). ?

To link up to \(\displaystyle S \oplus_R N = F(S \times N)/H \) we would need to extend our map above from \(\displaystyle F(S \times N) \) to \(\displaystyle F(S \times N)/H \) and then apply \(\displaystyle \Phi \) to map to L, thus:

\(\displaystyle S \times N \to F(S \times N) \to F(S \times N)/H = S \oplus_R \to L \)

But how does this fit with the figure for Theorem 8 on page 362 which shows the maps

\(\displaystyle N \to S \oplus_R N \to L \) and \(\displaystyle N \to L \)

So basically I am asking for guidance and help on how the use of Theorem 6 "works" in Theorem 8. In other words, how exactly does it apply? What elements does it apply to and how are its preconditions satisfied?

Hoping someone can help.

Peter