Inverse of Natural Projection? May this be a silly question, but can I construct an inverse of a natural projection $p$ from a module $M$ to its quotient module $M/A$? Of course more than one element can be assigned for each coset in $M/A$, but if we limit the inverse's range to direct complement of $A$, I think such inverse could be well-defined.
As far as I know such inverse homomorphism might be useful for various situations but I have never seen any.
There were only isomorphisms from the quotient module $M/A$ itself. There might be some reason for this. Can anyone teach me? Thank you!
 A: You have an exact sequence
$$ 0 \longrightarrow A \overset{i}\longrightarrow M \overset{p}\longrightarrow M/A \longrightarrow 0.$$
By the Splitting lemma, this sequence splits, i.e. $M\cong A\oplus M/A$, exactly if there is a morphism $r:M/A\to M$ such that $p\circ r = \operatorname{id}_{M/A}$. Look at the proof of the splitting lemma to see how to construct $r$ from the isomorphism $M\overset{\sim}\rightarrow A\oplus M/A$ and how to construct this isomorphism from a given right-inverse $r$ of $p$.
A: Are you familiar with the language of exact sequences?  The Wikipedia article is a good introduction.  Given a module $M$ with submodule $A$, there is a natural short exact sequence $0 \to A \to M \to M/A \to 0$.  The first (non-zero) arrow is the inclusion $i$ and the second (non-zero) arrow is the projection $p$.
You are asking if there is a homomorphism $s: M/A \to M$ with the property that $(p \circ s) = id_{M/A}$.  Such a map is not guaranteed to exist.  For a simple example, consider the short exact sequence of abelian groups ($\mathbb{Z}$-modules) $$0 \to \mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}/m\mathbb{Z} \to 0$$ where the first map is multiplication by $m$.   The only homomorphism $\mathbb{Z}/m\mathbb{Z} \to \mathbb{Z}$ is the zero map, so there's no "inverse" to the projection.
The splitting lemma from homological algebra says that such a map exists if and only if
$M$ is the direct sum of $M/A$ and $A$.  If you want an example with a more natural inclusion, consider $\mathbb{Z} \subset \mathbb{Q}$.  Note that $\mathbb{Q} \neq \mathbb{Q}/\mathbb{Z} \oplus \mathbb{Z}$.
