Enough projectives

Remarks

Example: the category of modules over a ring $R$ has enough projectives. For a module $M$, let $F$ be the free module with basis $\set{e_m : m \in M}$. The map $F \to M$ with $e_m \mapsto m$ is surjective, and free modules are projective. Non-example: the category of finite abelian groups has no nonzero projective object. For $n \ge 2$ the quotient map $\Z/n^2\Z \to \Z/n\Z$ does not split, since $\Z/n^2\Z$ is cyclic, so $\Z/n\Z$ is not projective there. Every nonzero finite abelian group has such a cyclic direct summand, and a direct summand of a projective object is projective. So no epimorphism from a projective object onto $\Z/2\Z$ exists. The dual notion is enough injectives: every object admits a monomorphism into an injective object.