Let $A$ be a finite‑dimensional algebra over an algebraically closed field $k$. I work with right modules.
We have a duality given by $F=\operatorname{Hom}(_,A):\mathrm{mod}(A)\to\mathrm{mod}(A^{op})$. This means that the same functor, seen as $G=(\mathrm{mod}(A))^{op}\to\mathrm{mod}(A^{op})$, is an equivalence of categories.
Let $eA\in\mathrm{mod}(A)$ where $e$ is a primitive idempotent, i.e. $eA$ is an indecomposable projective in $\mathrm{mod}(A)$. Thus $eA$ is injective in $(\mathrm{mod}(A))^{op}$ and is therefore sent to an injective module through $G$, i.e. $G(eA)$ is injective in $\mathrm{mod}(A^{op})$ (=left module). However we can compute that $\operatorname{Hom}(eA,A)\cong Ae$, which is a projective left module.
Where is the mistake? I know that a projective can also be injective and conversely, but obviously it’s not the case in such generality.
This functor is used to define the transpose of a module $M$. This transpose can be turned into a duality between quotients of $\mathrm{mod}(A)$. This category is called the projectively stable category and is defined to be $\underline{mod}(A)=mod(A)/I$, where $I$ is the ideal defined as $I(M,N)={\text{maps from }M\text{ to }N\text{ such that they factor through a projective}}$.
I manage to show that every projective is isomorphic to $0$ in this category. I would like to prove the following result (which I would like to be true):
Let $M\not\cong _{\mathrm{mod}(A)}N$ be two non‑projective, non‑isomorphic modules. Then $M\not\cong_{\underline{\mathrm{mod}}(A)}N$
I.e., I would like to show that in the projectively stable category we don’t kill or identify anything except for the projective modules.
