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To finish the proof for the theorem, we are tasked with proving that condition (ii) implies condition (i), i.e., that **if every monomorphism \( J \to M \) splits**, then **\( J \) is injective** as an \( R \)-module.

### Key Definitions
- **Injective module**: An \( R \)-module \( J \) is injective if, for every injective (monomorphic) map \( f: J \to M \) and for every map \( g: J \to N \), there exists an extension \( g' \) of \( g \) over \( M \), such that the following diagram commutes:
  \[
  \begin{matrix}
  J & \xrightarrow{g} & N \\
  \downarrow{f} & \nearrow{g'} & \\
  M & &
  \end{matrix}
  \]
- **Splitting of a monomorphism**: A monomorphism \( J \to M \) is said to split if there exists a map \( M \to J \) (a retraction) such that the composition \( J \to M \to J \) is the identity map on \( J \). In other words, \( J \) is a direct summand of \( M \).

### Proof Outline for (ii) \( \Rightarrow \) (i)

1. **Assume that (ii) holds**: That is, every monomorphism \( f: J \to M \) splits. This means that for every such \( f \), there exists a map \( g: M \to J \) such that \( g \circ f = \text{id}_J \), where \( \text{id}_J \) is the identity on \( J \).

2. **Show that \( J \) is injective**: To prove that \( J \) is injective, we need to show that any monomorphism \( f: J \to M \) has the extension property.

3. **Extension of maps**: Since \( f: J \to M \) splits, there exists a retraction \( g: M \to J \), i.e., a map such that \( g \circ f = \text{id}_J \). Now, given any map \( h: J \to N \), we want to extend \( h \) over \( M \).

4. **Define the extension**: Define the map \( h' = h \circ g \), where \( g: M \to J \). This composition gives us the required extension since:
   \[
   h' \circ f = h \circ g \circ f = h \circ \text{id}_J = h.
   \]
   Thus, \( h' \) is an extension of \( h \) over \( M \), which shows that \( J \) has the injective property.

5. **Conclusion**: Since \( J \) satisfies the extension property for every monomorphism \( J \to M \), we conclude that \( J \) is injective.

This completes the proof that (ii) implies (i).

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Do you want more details on any step? Here are 5 related questions to deepen your understanding:

1. How does the concept of splitting relate to direct sum decompositions of modules?
2. Can you give an example of an injective module in the category of abelian groups?
3. What are some key applications of injective modules in homological algebra?
4. How does the proof change if the module \( J \) is not assumed to be injective?
5. Can you explain the dual concept of projective modules and their relationship to surjections?

**Tip**: When dealing with splitting maps, always think of direct summands—they often provide insight into module decompositions.

To finish off the proof of the theorem below from class, prove that (ii) implies (i). Theorem T.F.A.E. for an R-module J: (i) J is injective (ii) every monomorphism J → M splits

Explore the proof of why the splitting of every monomorphism J → M implies that J is injective for an R-module. This advanced abstract algebra theorem covers key concepts like injective modules and the splitting lemma, suitable for advanced undergraduate and graduate students.

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