# Properties of an A-module

I must show that the following properties for an $$A$$-module $$P$$ are equivalent:

1) The functor $$Hom(P,-)$$ is exact.

2) There is an $$A$$-module $$Q$$ such that $$P \oplus Q$$ is free.

3) Every short exact sequence of $$A$$-modules of the form $$0 \rightarrow N \rightarrow M \rightarrow P \rightarrow 0$$ splits.

4) For every epimorphism $$p: M \rightarrow Q$$ of $$A$$-modules and every homomorphism $$f:P \rightarrow Q$$, there is a homomorphism $$g: P \rightarrow M$$ such that $$f = p \circ g$$.

I could go this far:

$$(2) \Rightarrow (4)$$

Let $$Q$$ be a module such that $$P \oplus Q$$ is a free module and let $$B = {b_i}_{i \in I}$$ be a basis of $$P \oplus Q$$. Since $$g$$ is an epimorphism for every $$i \in I$$, we can find $$m_i \in M$$ such that $$g(m_i) = f(b_i)$$. Define $$\overline{h}: P \oplus Q \rightarrow M$$ by $$\overline{h}(\sum_i r_ib_i) := \sum_i r_im_i$$. Since $$B$$ is a basis of $$P \oplus Q$$, the map is a well-defined $$A$$-module homomorphism and $$g \circ \overline{h} = f$$, as desired.

$$(4) \Rightarrow (3)$$

Since $$g$$ is an epimorphism, there is $$h: P \rightarrow M$$ such that $$g\circ h = id_P$$. Therefore, by Theorem 3.6.5, the sequence splits.

$$(3) \Rightarrow (2)$$

Consider the canonical epimorphism of $$A$$-modules $$f: \bigoplus_{p \in P} R \rightarrow P$$. This gives a short exact sequence $$0 \rightarrow Ker(f) \rightarrow \bigoplus_{p\ in P} R \rightarrow P \rightarrow 0$$. By assumption on P this sequence splits, so we obtain $$P \oplus Ker(f) \simeq \bigoplus_{p \in P} R$$ and, thus, P is projective.

\begin {itemize}

\item {$$(2) \Rightarrow (4)$$}

Let $$Q$$ be a module such that $$P \oplus Q$$ is a free module and let $$B = {b_i}_{i \in I}$$ be a basis of $$P \oplus Q$$. Since $$g$$ is an epimorphism for every $$i \in I$$, we can find $$m_i \in M$$ such that $$g(m_i) = f(b_i)$$. Define $$\overline{h}: P \oplus Q \rightarrow M$$ by $$\overline{h}(\sum_i r_ib_i) := \sum_i r_im_i$$. Since $$B$$ is a basis of $$P \oplus Q$$, the map is a well-defined $$A$$-module homomorphism and $$g \circ \overline{h} = f$$, as desired.

\item {$$(4) \Rightarrow (3)$$}

Since $$g$$ is an epimorphism, there is $$h: P \rightarrow M$$ such that $$g\circ h = id_P$$. Therefore, by Theorem 3.6.5, the sequence splits.

\item {$$(3) \Rightarrow (2)$$}

Consider the canonical epimorphism of $$A$$-modules $$f: \bigoplus_{p \in P} A \rightarrow P$$. This gives a short exact sequence $$0 \rightarrow Ker(f) \rightarrow \bigoplus_{p\ in P} A \rightarrow P \rightarrow 0$$. By assumption on P this sequence splits, so we obtain $$P \oplus Ker(f) \simeq \bigoplus_{p \in P} A$$ and, thus, P is projective.

\item {$$(2) \Rightarrow (1)$$} Given a short exact sequence $$0 \rightarrow Ker(f) \rightarrow \bigoplus_{p\in P} A \rightarrow P \rightarrow 0$$, we now that the functor $$\mathcal{F} = Hom(P,-)$$ is LEFT-exact and, also, that $$\mathcal{F}(\bigoplus A) = \bigoplus \mathcal{F}(A)$$ and, so, we have the canonical epimorphism $$\mathcal{F}(f): \bigoplus \mathcal{F}(A) \rightarrow \mathcal{F} (P)$$ and, so, the sequence $$0 \rightarrow \mathcal{F}(Ker(f)) \rightarrow \bigoplus \mathcal{F}(A) \rightarrow \mathcal{F}(P) \rightarrow 0$$ is also right exact and, thus, $$\mathcal{F}$$ is an exact functor.

\item {$$(1) \Rightarrow (4)$$}

It is clear from the previous step. If $$\mathcal{F}$$ is exact, it is also right exact and preserves epimorphisms. Thus, we have that for every epimorphism $$\mathcal{F}(p): \mathcal{F}(M) \rightarrow \mathcal{F}(Q)$$ and every homomorphism $$\mathcal{F}(f): \mathcal{F}(P) \rightarrow \mathcal{F}(Q)$$, there is a homomorphism $$\mathcal{F}(g): \mathcal{F}(P) \rightarrow \mathcal{F}(M)$$ such that $$\mathcal{F}(f) = \mathcal{F}(p) \circ \mathcal{F}(g)$$ and, since $$\mathcal{F}$$ preserves epimorphisms, it directly implies (4).

\end{itemize}