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In https://www.sciencedirect.com/science/article/pii/0001870891900378 section 6 a cotilting module T over an algebra A is said to be strong in case $\hat{add(T)}$ coincides with the subcategory of modules having finite injective dimension.

Here $\hat{add(T)}$ is just the full subcategory of all module $M$ such that there is an exact sequence $0 \rightarrow T_n \rightarrow ... \rightarrow T_0 \rightarrow M \rightarrow 0$ with $T_i \in add(T)$.

Let $\Gamma=End_A(T)$. Proposition 6.5. in this article states that T is a strong cotilting module iff every simple $\Gamma$-module is contained in T as a $\Gamma$-module.

Question: What means "every simple $\Gamma$-module is contained in T as a $\Gamma$-module"?

Im a bit confused here. For example let $A$ be an algebra of finite positive global dimension and $T=A$ (then $\Gamma=A$) with dominant dimension at least one (for example the quiver algebra of the quiver of Dynkin type $A_2$).

Then $T$ is a strong cotilting module but soc(T)=soc(A) does not contain every simple module as a right or left module.

So I think that I have a thinking error or I use the wrong definition of "contained".

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I believe I had this exact same question some time ago. Auslander and Reiten attribute this result to a paper of Auslander and Green. I still have the latter paper, but can no longer find where this result is in that paper. From my old notes, it looks like I concluded that this proposition is misstated in Auslander and Reiten's paper. Most likely, they may have meant to write "strong tilting module", rather than strong cotilting module. In your example of $A=kA_2$, the strong tilting module is $DA$, and this contains all simple $A$ modules (as submodules) on both sides. Alternatively, for strong cotilting modules, I believe the proper statement is the dual that every simple $\Gamma^{op}$-module is a quotient of $T_{\Gamma}$.

If you want to see a proof of the statement for strong tilting modules, see Prop 7.1 in the paper https://arxiv.org/abs/1407.2690

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