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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!

Christoph
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generic properties
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  • Such an “inverse” (in general only a left inverse) exists precisely when $A$ is a direct summand of $M$. – egreg Oct 30 '13 at 18:21
  • Now then how can I depict such inverse? Just $$p^-1$$ will not work. – generic properties Oct 30 '13 at 18:27
  • The (two-sided) inverse exists if and only if the canonical projection is an isomorphism, which means $A={0}$; not a very interesting case, is it? – egreg Oct 30 '13 at 18:28

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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 $p$.

Christoph
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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 $\mathbb{Q} \neq \mathbb{Q}/\mathbb{Z} \oplus \mathbb{Z}$.

Adam Saltz
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  • Thank you! Actually my focus was the situations where A is a direct summand of M. So learning about the splitting lemma might help me a lot. – generic properties Oct 30 '13 at 18:44