Properties of an A-module

Question

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.

Answer 1

\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}