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A free module is rather intuitive: there exists a subset satisfying suitable properties which may act as a basis for elements. Taking as example an $\mathbb R$-module, or a vector space, we see "free" as a generalisation of this concept.

However, a module $P$ is said to be projective if for every surjective homomorphism $\epsilon : B \to C$ and homomorphism $\gamma : P \to C$, there exists a homomorphism $\beta: P \to B$ with $\epsilon\beta = \gamma$.

When I say am looking for "intuition" regarding this definition, I mean I am looking for this: since we can say a free module is a generalisation of the notion of a vector space with a basis, then a projective module can be seen as a generalisation of what?

If such a comparison can't be made, is there some alternate way of understanding the definition, perhaps more concretely than a notion in homological algebra?

I am aware of the Serre-Swan theorem, however, I am not sure this sheds much light, since a projective module is not really a generalisation of a vector bundle on a compact manifold. They are separate things, related by the theorem.

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An equivalent characterization of projective modules are those modules $P$ for which any short exact sequence $0 \to A \to B \to P \to 0$ splits; that is, $B \cong A \oplus P$. In this sense, $P$ generalizes free modules (which share this property), and so this is even further a generalization of vector spaces.

This viewpoint is also suggestive of the terminology - $P$ can be viewed as a direct summand of any module from which it admits a surjective homomorphism, so $P$ is in fact a projection of that module onto one of its direct summands. In particular, since any module is the image of a free module, $F\cong K\oplus P$ for some free module $F$, so $P$ can be viewed as a projection of a free module.

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The most intuitive way to think about projective module(s) is probably direct summand(s) of some free module, so in a sense it is a generalisation of "complemented subspace" in the study of Banach spaces, except you couldn't really make that generalisation since, e.g., $c_0$ is not complemented in $\ell^\infty$ despite both being Banach.

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  • $\begingroup$ So a projective module $P$ is one for which there exists a $P'$ such that $P \oplus P'$ is free? $\endgroup$ – JamalS Oct 24 '18 at 23:56
  • $\begingroup$ Yes, and of course this gives, e.g., every module over $\mathbb{Z}/(p_1\dots p_n)$ is projective, where $p_i\in\mathbb{Z}$ are distinct primes. $\endgroup$ – user10354138 Oct 25 '18 at 0:10

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