# Reference request: Construction of torsion pair from a class closed under quotients, extensions and coproducts.

The category I'm working in is $$\operatorname{Mod}A$$ for some unitary ring $$A$$.

I'm looking for a reference on when certain subcategories of $$\operatorname{Mod}A$$ give rise to torsion pairs. Let $$\mathcal{T}$$ be a strict full subcategory of $$\operatorname{Mod}A$$ which is closed under quotients, extensions and coproducts. Then when is the pair $$(\mathcal{T}, \mathcal{T}^{\perp_{0}})$$ a torsion pair in $$\operatorname{Mod}A$$?

Edit: $$\mathcal{T}^{\perp_{0}} = \{ X \in \operatorname{Mod}A \: | \: \operatorname{Hom}_{A}(\mathcal{T}, X) = 0 \}$$

Obviously, $$\operatorname{Hom}(X, Y) = 0$$ for all $$X \in \mathcal{T}$$ and $$Y \in \mathcal{T}^{\perp_{0}}$$.

Also, by definition, $$\operatorname{Hom}(\mathcal{T}, Y) = 0$$ if and only if $$Y \in \mathcal{T}^{\perp_{0}}$$.

The other way is less clear to me, namely $$\operatorname{Hom}(X, \mathcal{T}^{\perp_{0}}) = 0$$ if and only if $$X \in \mathcal{T}$$. The 'if' part is obvious, but the 'only-if' part I can't show.

Any help on seeing why this is true? Alternatively, why it's not true and which further conditions I need to make it true? Or just a reference to any of these.

• What does $\perp_0$ mean? Perpendicular with respect to Hom but not necessarily with respect to higher Ext? – Jeremy Rickard May 24 '19 at 13:23
• @JeremyRickard Pardon me, forgot to specify that. Yes, hom-orthogonal. – Auclair May 24 '19 at 20:04

Take $$Y$$ to be the sum of submodules of $$X$$ that are in $$\mathcal{T}$$. By the coproduct and quotient hypothesis, this is in $$\mathcal{T}$$.
Now if $$Z\in\mathcal{T}$$ and $$Z\to X/Y$$, then the image is a submodule of $$X/Y$$ that is in $$\mathcal{T}$$ (as a quotient of $$Z$$), therefore so is its pullback to $$X$$ (it is an extension with $$Y$$), therefore the image is included in $$Y$$; therefore $$Z\to X/Y$$ is $$0$$.
Therefore $$X/Y\in\mathcal{T}^{\bot_0}$$; therefore if $$\hom(X,\mathcal{T}^{\bot_0})=0$$, $$X/Y=0$$ so $$X=Y\in\mathcal{T}$$.