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Given a small abelian category $\mathcal{A}$, the Freyd-Mitchell Embedding Theorem gives me a fully faithful exact functor $F:\mathcal{A}\rightarrow R$-$\mathsf{Mod}$, for some unital ring $R$, so that, in particular, $\mathcal{A}$ is equivalent to a full subcategory of $R$-$\mathsf{Mod}$.

Wikipedia states

However, projective and injective objects in $\mathcal{A}$ do not necessarily correspond to projective and injective $R$-modules.

Given that the definition of projective objects is entirely categorical in nature, how could it be that the notions of projective in each category do not correspond to each other in equivalent categories? Is this because the projective objects in a full subcategory of $R\mathsf{Mod}$ are sometimes different than the projective objects in all of $R$-$\mathsf{Mod}$?

(Of course, you can everywhere replace "projective" with "injective" in the above.)

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    $\begingroup$ As you noted, $\mathcal{A}$ is not equivalent to $R$-Mod but rather to a full subcategory, where injectives need not be the same. $\endgroup$ – Seth Sep 13 '14 at 19:56
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From the definitions, there's no reason to expect projective/injective objects of a subcategory to be projective/injective in the ambient category. If $P, B \in \mathcal{A}$, then any map $P \to B$ factors through any epi $A \twoheadrightarrow B$ with $A \in \mathcal{A}$, but if $A, B$ are not in $\mathcal{A}$, there is no reason to expect a lift.

For a concrete example where a full subcategory has (essentially) no injectives, while the ambient category has enough injectives, see this answer.

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An object $P$ is projective when, given any morphism $f: P \to A$, and any epimorphism $g: B \to A$, there is a unique morphism $h: P \to B$ such that $g \circ h = f$.

Even if an object $P$ is projective in a full subcategory of R-mod, then, it might not be projective in R-mod itself, because in that case the objects $A$ and $B$ might be outside of the subcategory over which $P$ is projective.

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