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> Let $C$ be an abelian category with enough projectives and $\underline{C}$ the stable category of $C$ that is obtained by factoring out projective modules.
>  When is the functor $\Omega^1 : \underline{C} \rightarrow \underline{C}$ an equivalence?
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We can assume $C$ is a module category of a ring $R$ in case that helps. For $R$ a finite dimensional algebra, this should be true iff $R$ is self-injective.



When is $\Omega^1$ an equivalence?
There's an elementary proof that if $\Omega$ is a self-equivalence of $\underline{C}$ then $C$ also has enough injectives, and projectives and injectives coincide. In particular, this shows that if $C$ is $\text{Mod-}R$ for a ring $R$, then $R$ is quasi-Frobenius.


For any $C$ with enough projectives, it's straightforward to prove that if $X$ and $Y$ are isomorphic in $\underline{C}$, then $X$ is a summand of $Y\oplus P$ and $Y$ is a summand of $X\oplus Q$ in $C$ for some projectives $P$ and $Q$. In particular $X$ is a subobject of a projective iff $Y$ is, and the only objects isomorphic in $\underline{C}$ to zero are the projectives.


So if $\Omega$ is a self-equivalence of $\underline{C}$ then every object of $C$ must be a subobject of a projective.


If $P$ is projective and $\alpha:P\to X$ a monomorphism, then composing with a monomorphism from $X$ to a projective $Q$ and taking the cokernel, we get a short exact sequence
$$0\to P\to Q\to Y\to0.$$
Since $\Omega Y\cong P\cong0$ in $\underline{C}$ and we’re assuming $\Omega$ is a self-equivalence, $Y$ is projective, and the short exact sequence splits. Hence $\alpha$ splits, and so $P$ is injective.


Therefore all projectives are injective, and since every object is a subobject of a projective, there are enough injectives.


Every injective is a subobject, and therefore a summand, of a projective, and so every injective is projective.


