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Let $R$ and $S$ be two rings. We say that $R$ and $S$ are Morita equivalent if the categories of right $R$-modules $\text{Mod}_{R}$ and right $S$-modules $\text{Mod}_{S}$ are equivalent.

We say that an $R$-module $P$ is a progenerator of $\text{Mod}_{R}$ if $P$ is a finitely generated projective module which is a generator of $\text{Mod}_{R}$, and we have the following known result:

Proposition. $R$ and $S$ are Morita equivalent if and only if there is a progenerator $P$ of $\text{Mod}_{R}$ such that $\text{End}(P) \simeq S$.

Now, it is also known that $R$ and $S$ are Morita equivalent if and only if the categories of left $R$-modules $_{R}\text{Mod}$ and left $S$-modules $_{S}\text{Mod}$ are equivalent. Therefore, my question is, what is the correspondent result of the above proposition for left modules?

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The equivalent proposition is:

Proposition. $R$ and $S$ are Morita equivalent if and only if there is a progenerator $P$ of $_{R}\text{Mod}$ such that $\text{End}(P) \simeq S^{\text{op}}$.

While searching for the answer of this question, I found some texts in the internet which stated the above proposition putting $\text{End}(P) \simeq S$ instead of $\text{End}(P) \simeq S^{\text{op}}$, and it is wrong! I thought it was important to share this detail, because it may confuse people, as it confused me. Now let's see the explanation:

The fundamental motive for the appearence of $S^{\text{op}}$ instead of $S$ is the way we compose functions. We have the following (which is easily verified):

If we regard the ring $S$ as a right module ($S_{S}$), then $\text{End}(S_{S}) \simeq S$ and if we regard $S$ as a left module ($_{S}S$), then $\text{End}(_{S}S) \simeq S^{\text{op}}$ (those are ring isomorphisms).

Now, for more details, let us examine the necessity of those propositions closely:

If $F : \text{Mod}_{S} \rightarrow \text{Mod}_{R}$ is an equivalence of categories, then putting $P_{R} = F(S_{S})$, we may show that $P_{R}$ is a progenerator, and to verify that $\text{End}(P_{R}) \simeq S$, we proceed like this: $$\text{End}(P_{R}) = \text{End}(F(S_{S})) \simeq \text{End}(S_{S}) \simeq S.$$

And for the left case we have:

If $F : \text{ }_{S}\text{Mod} \rightarrow \text{ }_{R}\text{Mod}$ is an equivalence of categories, then putting $_{R}P = F(_{S}S)$, we may show that $_{R}P$ is a progenerator, and we have: $$\text{End}(_{R}P) = \text{End}(F(_{S}S)) \simeq \text{End}(_{S}S) \simeq S^{\text{op}}.$$

To prove the sufficiency of the propositions, we proceed as follows:

If $P$ is a progenerator of $\text{Mod}_{R}$ such that $\text{End}(P) \simeq S$, then we may show that $$\text{Hom}_{R}(P,-) : \text{Mod}_{R} \rightarrow \text{Mod}_{S}$$ is an equivalence of categories, and for $M$ in $\text{Mod}_{R}$, we regard $\text{Hom}_{R}(P,M)$ as a right $\text{End}(P)$-module in the usual way, by composing functions, so that it becomes a right $S$-module.

For the left case we have:

If $P$ is a progenerator of $_{R}\text{Mod}$ such that $\text{End}(P) \simeq S^{\text{op}}$, then we may show that $$\text{Hom}_{R}(P,-) : \text{ }_{R}\text{Mod} \rightarrow \text{ }_{S}\text{Mod}$$ is an equivalence of categories, and for $M$ in $_{R}\text{Mod}$, we regard $\text{Hom}_{R}(P,M)$ as a left $\text{End}(P)^{\text{op}}$-module in the usual way, by composing functions and making the necessary adjustments (putting the $^{\text{op}}$) so that everything works, and then it becomes a left $S$-module, since $\text{End}(P)^{\text{op}} \simeq (S^{\text{op}})^{\text{op}} = S$.

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    $\begingroup$ I would like to add that $\operatorname{End}_R(P)\simeq S$ is not necessarily so much wrong as very understandably confusing. It is a pretty common convention when working with non-commutative rings to write $\operatorname{End}_R(P)$ for the opposite of the usual endomorphism algebra of a left module, so that $P$ is a $R$-$\operatorname{End}_R(P)$-bimodule. $\endgroup$ Commented Mar 2, 2020 at 15:46

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