Show that there is an $R$-module homomorphism $\bar{h}$ such that $g \circ \bar{h} = h$.

Question

Let $R$ be a unital ring. Suppose that a finitely generated free $R$-module is the internal direct sum of submodules $P$ and $Q$. Let $g:A \rightarrow B$ be a surjective $R$-module homomorphism. Let $h: P \rightarrow B$ be an $R$-module homomorphism.

Show that there is an $R$-module homomorphism $\bar{h}: P \rightarrow A$ such that $g \circ \bar{h} = h$.

Here's my approach:

We can let $A$ be the internal direct sum of the given submodules $P$ and $Q$; then by definition $A \cong P \oplus Q$, and any element of $a \in A$ can be written as $a = p + q$ for $p \in P, q \in Q$.

Then since $g$ is surjective, any $b \in B$ is the image of some $a \in A$, so any $b \in B$ is $b = g(p+q) = g(p) + g(q)$ for $p\in P, q \in Q$.

So let $\bar{h}: P \rightarrow A; p \mapsto (p+q)$ for some $q \in Q$. Then $g \circ \bar{h}: P \rightarrow B; p \mapsto g(p+q)$ = $h$, because $h: P \rightarrow B; p \mapsto g(p+q)$.

Am I on the right track? Also, what more do I need to show?

Answer 1

As already stated in the comments, you are not on the right track.

You are given:

Let $R$ be an unital ring, and $P$ be a $R$-module such that there is another $R$-module $Q$ so that $F := P \oplus Q$ is a free $R$-module.

You need to show:

For any $R$-modules $A$ and $B$ and any surjective $R$-module homomorphism $g\colon A → B$ and any $R$-module homomorphism $h \colon P → B$ , there is a lifted $R$-module homomorphism $\bar h \colon P → A$ such that $h = g ∘ \bar h$.

So here’s a hint: Start showing this with the special case $Q = 0$ (meaning $P$ is free) and then work your way up. Free modules have bases and you will need the axiom of choice if you look closely.

Oh, and draw a triangle diagram with $P$ on the top and $A \overset{g}{→} B → 0$ on the bottom. If you are still stuck or when you are finished, look up projective modules.