We know that $\mathbb Q$ is not a projective $\mathbb Z$-module, but,
I've read this paper and it says in Example 2.11 that it's easy to check that
"$\mathbb Q$ as $\mathbb Z$-module is projective in $\sigma[\mathbb Q]$"
where $\sigma[\mathbb Q]$ is a full subcategory that contains all modules isomorphic to a submodule of an $\mathbb Q$-generated module.
How can we prove it?
Also, as $\mathbb Z$-module, How to show that $\mathbb Q$ is $\mathbb Z$-projective?
that is for $f:\mathbb Q \rightarrow N$ and $g:\mathbb Z\rightarrow N$ onto, there exists $h:\mathbb Q\rightarrow\mathbb Z$ such that $g\circ h=f$.
Thank you
An answer to your last question: If $\mathbb Z \to N$ is onto, then any homomorphism $f:\mathbb Q \to N$ is easily shown to be trivial. In particular we can always chose $h = 0$ (the only homomorphism $\mathbb Q \to \mathbb Z$ anyway) to satisfy $g \circ h = f$.
This shows (a boring fact, since the involved maps are trivial, so nothing happens here) that $\mathbb Q$ is $\mathbb Z$-projective.