Proof. (a) Clearly, $\psi \varphi = 0$ for all $\varphi \in {}^\perp (\mathcal {I}^\perp )$ and $\psi \in \mathcal {I}^\perp $ . For $M\in \mathrm{mod}\,A, $ let $L \subseteq M$ be the sum of all images of morphisms in $\mathcal {I}$ ending in M. Then $M \rightarrow M/L$ is contained in $\mathcal {I}^\perp $ . It is left to show that the inclusion $\varphi \colon L \rightarrow M$ is contained in ${}^\perp (\mathcal {I}^\perp )$ . Suppose there exists $\psi \in \mathcal {I}^\perp $ with $\psi \varphi \neq 0$ . Then by the definition of $L,$ there exists $\varphi '\colon X \rightarrow M$ in $\mathcal {I}$ with $\psi \varphi ' \neq 0$ , which is a contradiction. Thus, always $\psi \varphi = 0$ and $\varphi \in {}^\perp (\mathcal {I}^\perp )$ .

(b) Similar to (a).

(c) If $\mathcal {I}^\perp = \mathcal {J}$ and ${}^\perp \mathcal {J} = \mathcal {I}$ , then $({}^\perp (\mathcal {I}^\perp ), \mathcal {I}^\perp )$ equals $(\mathcal {I}, \mathcal {J})$ and is an ideal torsion pair by (a). If $(\mathcal {I}, \mathcal {J})$ is an ideal torsion pair, then $ \psi \varphi = 0$ for all $\varphi \in \mathcal {I}$ and $\psi \in \mathcal {J}$ . Thus, $\mathcal {I}^\perp \supseteq \mathcal {J}$ . Now let $f \colon M \rightarrow N$ be a morphism in $\mathcal {I}^\perp $ and consider a short exact sequence

with $\varphi \in \mathcal {I}$ and $\psi \in \mathcal {J}$ . Then $f \varphi = 0$ implies that f factors through $\psi $ . Hence, $f \in \mathcal {J}$ . It follows that $\mathcal {I}^\perp = \mathcal {J}$ and similarly ${}^\perp \mathcal {J} = \mathcal {I}$ .

Proof. (a) Let $(\mathcal {I}, \mathcal {J})$ be an ideal torsion pair and $\varphi f \in \mathcal {I}$ for a morphism $\varphi $ and an epimorphism f. Then $\psi \varphi f = 0$ for all $\psi \in \mathcal {J}$ . Because f is epic, also $\psi \varphi = 0$ for all $\psi \in \mathcal {J}$ . Now Proposition 3.2(c) implies $\varphi \in \mathcal {I}$ .

For the other implication, let $\mathcal {I}$ be an ideal such that $\varphi f\in \mathcal {I}$ implies $\varphi \in \mathcal {I}$ for all morphisms $\varphi $ and epimorphisms f. We show that $(\mathcal {I}, \mathcal {I}^\perp )$ is an ideal torsion pair. Similar to the proof of Proposition 3.2, for $M\in \mathrm{mod}\,A,$ let $L\subseteq M$ be the sum of all images of morphisms in $\mathcal {I}$ ending in M. Then $M \rightarrow M/L$ is contained in $\mathcal {I}^\perp $ . It is left to show that the inclusion $\varphi \colon L \rightarrow M$ is contained in $\mathcal {I}$ . Because M is of finite length, it follows that L is a finite sum and there exist $\varphi _i' \colon X_i \rightarrow M$ in $\mathcal {I}$ such that the image of the induced morphism $\varphi '\colon \bigoplus _{i=1}^n X_i \rightarrow M$ equals L. Now $\varphi _i' \in \mathcal {I}$ implies $\varphi ' \in \mathcal {I}$ and if $f\colon \bigoplus _{i=1}^n X_i \rightarrow L$ denotes the projection onto the image of $\varphi '$ , then $\varphi f = \varphi ' \in \mathcal {I}$ . Since f is an epimorphism, the morphism $\varphi $ is contained in $\mathcal {I}$ .

(b) Similar to (a).

Proof. (i) Let $f\colon M \rightarrow M'$ be an arbitrary morphism in $\mathrm{mod}\,A$ . The morphism $\varphi \colon L \rightarrow M$ is a monic left $\mathcal {I}$ -approximation by Remark 3.1. Similarly, let $\varphi ' \colon L' \rightarrow M'$ be a monic left $\mathcal {I}$ -approximation. Then $f \varphi \in \mathcal {I}$ must factor through $\varphi '$ . Thus, $f(\mathrm{Im}\, \varphi ) \subseteq \mathrm{Im}\, \varphi '$ and the described assignment defines a subfunctor of .

(ii) First, we show that $\mathcal {I}$ and $\mathcal {J}$ are ideals. Let $\varphi \colon L \rightarrow M$ and $\varphi '\colon L' \rightarrow M$ be in $\mathcal {I}$ . Then $\mathrm{Im}\, \varphi , \mathrm{Im}\, \varphi ' \subseteq tM$ . Hence, $\mathrm{Im}\, (\varphi + \varphi ') \subseteq tM$ and $\varphi + \varphi '$ is contained in $\mathcal {I}$ . For a morphism $f\colon X\rightarrow L, $ clearly, $\mathrm{Im}\, \varphi f \subseteq \mathrm{Im}\, \varphi \subseteq tM$ so $\varphi f \in \mathcal {I}$ . For a morphism $g\colon M \rightarrow Y$ , because t is a subfunctor of , the inclusion $g(tM) \subseteq tY$ holds. Thus, $\mathrm{Im}\, g \varphi = g(\mathrm{Im}\, \varphi ) \subseteq g(tM) \subseteq tY$ . It follows that $g \varphi \in \mathcal {I}$ and $\mathcal {I}$ is an ideal. Similarly, $\mathcal {J}$ is an ideal. By definition of $\mathcal {J}$ and $\mathcal {I}$ , we have $\psi \varphi = 0$ for all $\varphi \in \mathcal {I}$ and $\psi \in \mathcal {I} = 0$ . Further, every $M\in \mathrm{mod}\,A$ admits the short exact sequence

with $\varphi \in \mathcal {I}$ and $\psi \in \mathcal {J}$ . It follows that $(\mathcal {I}, \mathcal {J})$ is an ideal torsion pair. Since the short exact sequences are also given as above in (i), it follows that the assignments are mutually inverse.

Proof. The natural isomorphism is given by $\mathrm{Hom}_A(A,M) \ni \varphi \mapsto \varphi (1)$ for $M\in \mathrm{mod}\,A$ . Now if $\varphi \in \mathcal {I}(A,M)$ , then $\mathrm{Im}\, \varphi \subseteq tM$ and in particular $\varphi (1) \in t M $ . Conversely, for $x\in t M, $ let $\varphi \colon A\rightarrow M$ be defined by $\varphi (1) = x$ . Then $\mathrm{Im}\, \varphi \subseteq t M $ and $\varphi \in \mathcal {I}(A,M)$ . Hence, the natural isomorphism restricts to $\mathcal {I}(A,-) \cong t$ .

Proof. For $D \varphi \in D \mathcal {I}$ and $D \psi \in D \mathcal {J,}$ the equality $D \varphi D \psi = D (\psi \varphi ) = 0$ holds. For $M\in \mathrm{mod}\,A^{\mathrm{op}}$ , there exists a short exact sequence

$$ \begin{align*} 0 \longrightarrow D (DM/tDM) \longrightarrow M \longrightarrow D (t DM) \longrightarrow 0 \end{align*} $$

with the first morphism contained in $D \mathcal {J}$ and the second one contained in $D \mathcal {I}$ . Thus, $(D \mathcal {J}, D\mathcal {I})$ is an ideal torsion pair. Further, by the above short exact sequence, the corresponding subfunctor of is equivalent to .

Proof. (a) $\Rightarrow $ (b): By Remark 3.1, the ideal $\mathcal {I}$ is always contravariantly finite. It is left to show that $\mathcal {I}$ is covariantly finite. Let $M\in \mathrm{mod}\,A$ and $\pi \colon A^n \rightarrow M$ an epimorphism. Given a left $\mathcal {I}$ -approximation $\varphi \colon A\rightarrow C_A$ consider the following pushout diagram:

We will show that $\psi $ is a left $\mathcal {I}$ -approximation. By commutativity of the diagram, $\psi \pi \in \mathcal {I}$ so $\mathrm{Im}\, \psi = \mathrm{Im}\, \psi \pi \subseteq tP$ . Hence, $\psi \in \mathcal {I}$ . Now let $f\colon M \rightarrow X$ be a morphism in $\mathcal {I}$ . Then $f \pi \in \mathcal {I}$ must factor through $\varphi ^n$ . By the universal property of the pushout diagram, there exists a morphism $g \colon P \rightarrow X$ with $g \psi = f$ . It follows that $\psi $ is a left $\mathcal {I}$ -approximation.

(b) $\Rightarrow $ (c): By Lemma 3.6, we can show that $\mathcal {I}(A,-) \subseteq \mathrm{Hom}_A(A,-)$ is finitely presented. Because $\mathrm{Hom}_A(A,-)$ is finitely presented, it is enough to show that $\mathcal {I}(A,-)$ is finitely generated. Let ${\varphi \colon A \rightarrow C_A}$ be a left $\mathcal {I}$ -approximation. We show that $\mathrm{Im}\, \mathrm{Hom}_A(\varphi ,-) = \mathcal {I}(A,-)$ . Because $\varphi \in \mathcal {I}$ , the inclusion “ $\subseteq $ ” holds. Now let $f\colon A\rightarrow X$ be a morphism in $\mathcal {I}$ . Then f factors through $\varphi $ and so $f\in \mathrm{Im}\, \mathrm{Hom}_A(\varphi , X)$ . Hence, also “ $\supseteq $ ” holds.

(c) $\Rightarrow $ (a): By Lemma 3.6, the functor $\mathcal {I}(A,-) \subseteq \mathrm{Hom}_A(A,-)$ is finitely presented. Hence, there exists an epimorphism $\mathrm{Hom}_A(M,-) \rightarrow \mathcal {I}(A,-)$ with $M\in \mathrm{mod}\,A$ . Now, by the Yoneda lemma, there exists a morphism $\varphi \colon A\rightarrow M$ with $\mathrm{Im}\, \mathrm{Hom}_A(\varphi ,-) = \mathcal {I}(A,-)$ . We show that $\varphi $ is a left $\mathcal {I}$ -approximation. First, $\varphi = \mathrm{Hom}_A(\varphi , M) (1_M)$ so $\varphi \in \mathcal {I}$ . Now let $f\colon A\rightarrow X$ be a morphism in $\mathcal {I}$ . Then $f\in \mathcal {I}(A,X) = \mathrm{Im}\, \mathrm{Hom}_A(\varphi , X)$ so f factors through $\varphi $ . It follows that $\varphi $ is a left $\mathcal {I}$ -approximation.

(a)’ $\Leftrightarrow $ (b)’ $\Leftrightarrow $ (c)’: If $(\mathcal {I}, \mathcal {J})$ is an ideal torsion pair in $\mathrm{mod}\,A$ , then $(D\mathcal {J}, D\mathcal {I})$ is an ideal torsion pair by Lemma 3.7 with the corresponding subfunctor of equivalent to . By duality, $ DA^{\mathrm{op}}$ admits a right $\mathcal {J}$ -approximation iff $A^{\mathrm{op}}$ admits a left $D\mathcal {J}$ -approximation and the ideal $\mathcal {J}$ is functorially finite iff $D\mathcal {J}$ is functorially finite. Hence, the equivalences follow from (a) $\Leftrightarrow $ (b) $\Leftrightarrow $ (c).

(c) $\Leftrightarrow $ (c)’: Because , the functor t is finitely presented if and only if is finitely presented. By Lemma 2.6, the functor is finitely presented if and only if is finitely presented.

Proof. Since is an abelian subcategory of $(\mod A, \mathrm{Ab})$ , it follows that the collection of finitely presented subfunctors of is a sublattice of the lattice of subfunctors of . Hence, by Corollary 3.9, the lattice of functorially finite ideal torsion pairs is a sublattice of the lattice of ideal torsion pairs. Now the claim follows by Remark 3.5.

Proof. We can uniquely extend F and G to functors $\overline {F}\colon \mathrm{Mod}\,A \rightarrow \mathrm{Mod}\,A$ and $\overline {G}\colon \mathrm{Mod}\,A \rightarrow \mathrm{Ab}$ such that $\overline {F}$ and $\overline {G}$ commute with direct limits and $\overline {F}, \overline {G}$ coincide with $F,G$ on $\mathrm{mod}\,A,$ respectively, as follows. For $M\in \mathrm{Mod}\,A,$ we have $M = \varinjlim M_i$ for $M_i \in \mathrm{mod}\,A$ and we set

$$ \begin{align*} \overline{F}(M) = \varinjlim F(M_i), \qquad \overline{G}(M) = \varinjlim G(M_i). \end{align*} $$

Then $\overline {F}$ and $\overline {G}$ have the desired properties (see, e.g., [Reference Prest23]). Because F and G are finitely presented, it follows from [Reference Krause18, Theorem 9,1] that $\overline {F}$ and $\overline {G}$ commute with products. Now $\overline {G}\,\overline {F}$ is the unique extension, in the above sense, of $GF$ as the functors coincide on $\mathrm{mod}\,A$ and $\overline {G}\,\overline {F}$ commutes with direct limits, since $\overline {F}$ and $\overline {G}$ commute with direct limits. Further, $\overline {G}\,\overline {F}$ commutes with products, since so do $\overline {G}$ and $\overline {F}$ . By [Reference Krause18, Theorem 9,1], it follows that $GF$ is finitely presented.

Proof. (a) For $M\in \mathrm{mod}\,A,$ the inclusion $t'M \rightarrow M$ is in $\mathcal {I}'$ and the inclusion $tt'M \rightarrow t' M$ is in  $\mathcal {I}$ . Thus, the inclusion $tt'M \rightarrow M$ is in $\mathcal {I}' \mathcal {I}$ . Now an arbitrary morphism in $\mathcal {I}' \mathcal {I}$ factors as $\varphi ' \varphi $ for $\varphi \colon L \rightarrow N$ in $\mathcal {I}$ and $\varphi ' \colon N \rightarrow M$ in $\mathcal {I}'$ . Further, $\varphi '$ factors as $N\rightarrow t'M \rightarrow M$ and the composition $L \rightarrow N \rightarrow t'M$ factors as $L \rightarrow t t'M \rightarrow t'M$ . We conclude that $\mathrm{Im}\, \varphi ' \varphi \subseteq t t'M$ and the torsion ideal corresponding to $tt'$ must be equal to $\mathcal {I}' \mathcal {I}$ . It is left to show that $(\mathcal {I}' \mathcal {I})^\perp = \mathcal {J} \diamond \mathcal {J}'$ .

For $f\colon X \rightarrow Y$ in $\mathcal {J} \diamond \mathcal {J}'$ , there exists a commutative diagram of morphisms

with $f = gh$ and $\psi \in \mathcal {J}, \psi ' \in \mathcal {J}'$ , where the horizontal sequence is exact. Now for $\varphi \in \mathcal {I},\varphi ' \in \mathcal {I}'$ , we have $\pi h \varphi ' = \psi ' \varphi ' = 0$ . Hence, $h \varphi '$ must factor through $\iota $ , that is, $h \varphi ' = \iota \alpha $ . Thus, $f \varphi ' \varphi = gh \varphi ' \varphi = g \iota \alpha \varphi = \psi \alpha \varphi = 0$ and so $f \in (\mathcal {I}' \mathcal {I})^\perp $ .

Let $f\colon X \rightarrow Y$ in $(\mathcal {I}' \mathcal {I})^\perp $ and consider the commutative diagram

Since $t' X \rightarrow t'X/tt'X$ is contained in $\mathcal {J}$ and $X\rightarrow X/t'X$ in $\mathcal {J}'$ , we conclude that $X \rightarrow X/tt'X$ is contained in $\mathcal {J}\diamond \mathcal {J}'$ . Because $tt'X \rightarrow X$ is contained in $\mathcal {I}' \mathcal {I}$ , the morphism f factors through $X \rightarrow X/tt'X$ . Thus, also $f\in \mathcal {J} \diamond \mathcal {J}'$ .

(b) Similar to (a).

Proof. Let t and $t'$ be the subfunctors of corresponding to the ideal torsion pairs $(\mathcal {I}, \mathcal {J})$ and $(\mathcal {I}', \mathcal {J}')$ . By Corollary 3.9, the functors are finitely presented. Now $tt'$ and $t"$ , defined by , correspond to $(\mathcal {I}'\mathcal {I}, \mathcal {J} \diamond \mathcal {J}' )$ and $(\mathcal {I} \diamond \mathcal {I}', \mathcal {J}' \mathcal {J}),$ respectively, by Proposition 3.12. By Lemma 3.11, the functors $tt'$ and $t"$ are finitely presented. Hence, the corresponding ideal torsion pairs are functorially finite.

Proof. Let $f \colon X \rightarrow Y$ be in $\langle \mathcal {C} \diamond \mathcal {C}' \rangle $ . Then f factors as $gh$ with g starting in $M \in \mathcal {C} \diamond \mathcal {C}'$ . Hence, there exists a short exact sequence

$$ \begin{align*} 0 \longrightarrow L \longrightarrow M \longrightarrow N \longrightarrow 0 \end{align*} $$

with $L\in \mathcal {C}$ and $N\in \mathcal {C}'$ . We extend this to a commutative diagram

with $\varphi \in \langle \mathcal {C} \rangle $ and $\varphi ' \in \langle \mathcal {C}' \rangle $ . Thus, $f \in \langle \mathcal {C} \rangle \diamond \langle \mathcal {C}' \rangle $ .

Let $f\colon X\rightarrow Y$ be in $\langle \mathcal {C} \rangle \diamond \langle \mathcal {C}' \rangle $ . Then there exists a commutative diagram

with exact middle row, $f = gh$ and $\varphi \in \langle \mathcal {C} \rangle $ , $\varphi ' \in \langle \mathcal {C}' \rangle $ . In particular, $\pi h = \varphi '$ factors through some $C'\in \mathcal {C}'$ and taking a suitable pullback P, we obtain a commutative diagram

with $h = h' \alpha $ . Now $gh' \beta = g\iota = \varphi $ factors through some $C\in \mathcal {C}$ . Taking a suitable pushout Q, we obtain a commutative diagram

with $\gamma g' = gh'$ . Hence, $f = gh = g h' \alpha = \gamma g' h' \alpha $ factors through $Q \in \mathcal {C} \diamond \mathcal {C}'$ . We conclude that $\langle \mathcal {C} \rangle \diamond \langle \mathcal {C}' \rangle = \langle \mathcal {C} \diamond \mathcal {C}' \rangle .$

Proof. (a) Let $\mathcal {I}$ be a torsion ideal and $\varphi $ a morphism such that $\varphi f \in \mathcal {I}$ for all f starting in A. We can choose f as an epimorphism starting in $A^n$ such that $\varphi f\in \mathcal {I}$ . By Lemma 3.3, it follows that $\varphi \in \mathcal {I}$ . Hence, $\mathcal {I}$ is right A-determined.

Let $\mathcal {I}$ be right A-determined and $\varphi $ a morphism such that $\varphi f \in \mathcal {I}$ for an epimorphism f. Then for an arbitrary morphism $f'$ starting in A (such that $\varphi f'$ is defined), $f'$ factors through the epimorphism f. Thus, $ \varphi f' \in \mathcal {I}$ and because $\mathcal {I}$ is right A-determined, we conclude that $\varphi \in \mathcal {I}$ . By Lemma 3.3, it follows that $\mathcal {I}$ is a torsion ideal.

(b) Similar to (a).

Proof. (a) By Remark 4.2, we have $\mathcal {I}(A,-) = \mathrm {I}(\mathcal {I})(A,-)$ . Thus, A admits a left $\mathcal {I}$ -approximation if and only if A admits a left $\mathrm {I}(\mathcal {I})$ -approximation. Now the claim follows by Theorem 3.8.

(b) Similar to (a).

Proof. (a) Let $\mathcal {I}$ be left C-determined. We denote by $\mathcal {J}_C$ the ideal consisting of all morphisms $\psi f$ with $\psi \in \mathcal {J}$ starting in $C^n$ for some $n>0$ and f arbitrary. Clearly, $\mathcal {J}_C \subseteq \mathcal {J}$ and thus $\mathrm {J}(\mathcal {J}_C) \subseteq \mathcal {J}$ . To deduce equality, we show ${}^\perp \mathcal {J}_C \subseteq \mathcal {I}$ . Let $\varphi $ such that $\psi f \varphi = 0$ for all $\psi \in \mathcal {J}$ starting in $C^n$ and arbitrary f. Then $ f \varphi \in {}^\perp \mathcal {J} = \mathcal {I}$ and because $\mathcal {I}$ is left C-determined, also $\varphi \in \mathcal {I}$ . We conclude that ${}^\perp \mathcal {J}_C \subseteq \mathcal {I}$ . Hence, $\mathrm {J}(\mathcal {J}_C) = \mathcal {J}$ . By Corollary 4.3, it suffices to show that $DA^{\mathrm{op}}$ admits a right $\mathcal {J}_C$ -approximation for $\mathcal {J}$ to be functorially finite. Let $\psi _i, 1\leq i \leq m$ generate $\mathcal {J}(C, DA^{\mathrm{op}})$ as a k-module. Then all $\psi _i$ induce a morphism $\psi \colon C^m \rightarrow DA^{\mathrm{op}}$ . Clearly, $\psi \in \mathcal {J}_C$ . Now let $\alpha \colon M \rightarrow DA^{\mathrm{op}}$ be in $\mathcal {J}_C$ , so $\alpha = \beta f$ for some f and for some $\beta \in \mathcal {J}$ starting in $C^n$ with . Then by construction of $\psi $ , the morphism $\beta $ factors through $\psi $ . Hence, $\alpha $ factors through $\psi $ and $\psi $ is a right $\mathcal {J}_C$ -approximation.

Let $\mathcal {J}$ be functorially finite and $\psi \colon C\rightarrow DA^{\mathrm{op}}$ a right $\mathcal {J}$ -approximation. Further, let $\varphi \colon M \rightarrow N$ be a morphism such that $f \varphi \in \mathcal {I}$ for all morphisms f ending in C. We show $\varphi \in {}^\perp \mathcal {J} = \mathcal {I}$ . Let $\alpha \colon N \rightarrow L $ in $\mathcal {J}$ and $\iota \colon L \rightarrow (DA^{\mathrm{op}})^n$ a monomorphism. Then $\iota \alpha $ factors through $\psi ^n$ , that is, $\iota \alpha = \psi ^k f$ . Now $f \varphi \in \mathcal {I}$ implies $\iota \alpha \varphi = \psi ^k f \varphi = 0$ . Thus, $\alpha \varphi = 0$ and we conclude that $\varphi \in {}^\perp \mathcal {J} = \mathcal {I}$ . Hence, $\mathcal {I}$ is left C-determined.

(b) Similar to (a).

Proof. (a) This follows by Lemma 4.4 and Theorem 3.8.

(b) By the functoriality of t, the submodule $tC$ of C is invariant under $\mathrm{End}_A(C)$ . Hence, $tC$ is a submodule of the bimodule ${}_AC_{\mathrm{End}_A(C)^{\mathrm{op}}}$ . To show that the assignment $C \mapsto tC$ is one to one, we construct an inverse assignment: For a submodule X of ${}_AC_{\mathrm{End}_A(C)^{\mathrm{op}}}$ , let $\mathcal {I}_X$ be the collection of all morphisms $\varphi \colon M \rightarrow N$ such that for all $f\colon A\rightarrow M$ and $g\colon N \rightarrow C,$ we have $g \varphi f (1) \in X$ . Clearly, $\mathcal {I}_X$ is an ideal and right A-determined as well as left C-determined by construction. By Proposition 4.1, the ideal $\mathcal {I}_X$ is a left C-determined torsion ideal. Further, $X\rightarrow C$ is contained in $\mathcal {I}_X$ and $C \rightarrow C/X$ in $\mathcal {I}_X ^\perp $ . Hence, if t denotes the subfunctor of corresponding to $\mathcal {I}_X$ , then $tC = X$ . It is left to show $\mathcal {I}_{tC} = \mathcal {I}$ , where t denotes the subfunctor of corresponding to $\mathcal {I}$ . If $\varphi \colon M \rightarrow N$ is in $\mathcal {I}$ , then for all $f \colon A \rightarrow M$ and $g\colon N \rightarrow C,$ we have $\mathrm{Im}\, g \varphi f \subseteq tC$ . Thus, $g\varphi f (1) \in tC$ and $\varphi \in \mathcal {I}_{tC}$ . On the contrary, if $\varphi \colon M \rightarrow N$ is in $\mathcal {I}_{tC}$ , then for all $f\colon A \rightarrow M$ and $g\colon N \rightarrow C,$ we have $ g \varphi f(1) \in tC $ . Thus, always $\mathrm{Im}\, g \varphi \subseteq tC$ and $g \varphi \in \mathcal {I}$ . Because $\mathcal {I}$ is left C-determined, we conclude that $\varphi \in \mathcal {I}$ , so $\mathcal {I}_{tC} = \mathcal {I}$ .

Proof. (a) Let $\mathcal {I}$ be left C-determined and $\varphi ' \colon A \rightarrow M'$ a morphism such that $f \varphi '$ factors through $\varphi $ for all $f \colon M' \rightarrow C$ . Then always $f \varphi ' \in \mathcal {I}$ and because $\mathcal {I}$ is left C-determined, we conclude that $\varphi '\in \mathcal {I}$ . Hence, $\varphi '$ factors through the left $\mathcal {I}$ -approximation $\varphi $ . It follows that $\varphi $ is left C-determined.

Let $\varphi $ be left C-determined and $\varphi ' \colon M' \rightarrow N$ a morphism such that $f \varphi ' \in \mathcal {I}$ for all $f \colon N \rightarrow C$ . Then for all $g \colon A \rightarrow M'$ , the morphism $f \varphi ' g \in \mathcal {I}$ factors through $\varphi $ . Because f is arbitrary and $\varphi $ left C-determined, it follows that $\varphi ' g$ factors through $\varphi $ . Thus, $\varphi ' g\in \mathcal {I}$ and because $\mathcal {I}$ is right A-determined by Proposition 4.1, we conclude that $\varphi \in \mathcal {I}$ . Hence, $\mathcal {I}$ is left C-determined.

(b) Similar to (a).

Proof. This follows by Lemma 4.6 and Proposition 2.5.

Proof. By Lemma 2.3, we can check that $\langle \mathcal {T} \rangle $ is functorially finite if and only if $\langle \mathcal {F} \rangle $ is functorially finite. This is the case by Remark 5.1 and Theorem 3.8.

Proof. (a) Let $\varphi $ be a morphism and f an epimorphism with $\varphi f \in \langle \mathcal {C} \rangle $ . Then $\varphi f$ factors as $\varphi _1 \varphi _2$ with $\varphi _1$ starting in $C\in \mathcal {C}$ . Because $\mathcal {C}$ is an epi-closed class, the image of $\varphi _1$ is contained in $\mathcal {C}$ . Now $\mathrm{Im}\, \varphi = \mathrm{Im}\, \varphi f \subseteq \mathrm{Im}\, \varphi _1 $ implies $\varphi \in \langle \mathcal {C}\rangle $ . By Lemma 3.3, it follows that $\langle \mathcal {C} \rangle $ is a torsion ideal.

Let $M\in \mathrm{ob}\, \mathcal {I}$ and $N \leq M$ . Consider the projection $\pi \colon M \rightarrow M/N$ . Then $\pi = \pi 1_M \in \mathcal {I}$ . Further, $\pi = 1_{M/N} \pi $ . By Lemma 3.3, it follows that $1_{M/N} \in \mathcal {I}$ . Hence, $M/N \in \mathrm{ob}\, \mathcal {I}$ and $\mathrm{ob}\, \mathcal {I}$ is an epi-closed class.

(b) Similar to (a).

Proof. (a) Let $\mathcal {C}$ be a functorially finite epi-closed class and $\varphi \colon A \rightarrow C_A$ a left $\mathcal {C}$ -approximation. Then $C_A \in \mathcal {C}$ and for $M\in \mathcal {C}$ an epimorphism $A^n \rightarrow M$ must factor through $\varphi ^n$ for some . We conclude that $\mathrm{gen}\, C_A = \mathcal {C}$ .

For the other implication, consider the ideal $\mathcal {I} = \langle \mathrm{add}\, M \rangle $ . If $\varphi _1, \dots , \varphi _n$ denotes a basis of the k-module $\mathrm{Hom}_A(A,M)$ , then the induced morphism $\varphi \colon A \rightarrow M^n$ is a left $\mathcal {I}$ -approximation. Thus, by Corollary 4.3, the ideal $\mathrm {I}(\mathcal {I})$ is functorially finite. By Lemma 4.3, the ideal $\mathrm {I}(\mathcal {I})$ equals $\langle \mathrm{gen}\, M \rangle $ , as $\mathrm{gen}\, M$ is the smallest epi-closed class containing M. Hence, $\mathrm{gen}\, M$ is functorially finite.

(b) Similar to (a).

Proof. (a) Let $\varphi \colon M \rightarrow N$ be in $\mathcal {I}$ . An epimorphism in $\mathrm{mor}\,A$ starting in $\varphi $ is given by a commutative square of morphisms

with f and g surjective. Then $\varphi ' f = g \varphi \in \mathcal {I}$ . Hence, $\varphi ' \in \mathcal {I}$ by Lemma 3.3. It follows that $\mathcal {I}$ is an epi-closed class in $\mathrm{mor}\,A$ .

Let $\mathcal {C}$ be an epi-closed class in $\mathrm{mor}\,A$ and $\mathcal {I}$ the collection of all morphisms factoring through a morphism in $\mathcal {C}$ . Clearly, $\mathcal {I}$ is an ideal. To see that $\mathcal {I}$ is a torsion ideal, we make use of Lemma 3.3. Let $\varphi \colon L \rightarrow N$ be a morphism and $f\colon M \rightarrow L$ an epimorphism with $\varphi f\in \mathcal {I}$ . Then $\varphi f $ factors as $\varphi _1 \alpha \varphi _2$ for some $\alpha \colon M' \rightarrow N'$ in $\mathcal {C}$ , $\varphi _1\colon N' \rightarrow N$ and $\varphi _2 \colon M \rightarrow M'$ . Now consider the commutative square

where $\pi _1 \colon M'\rightarrow \mathrm{Im}\, \varphi _1\alpha , \pi _2 \colon N'\rightarrow \mathrm{Im}\, \varphi _1$ are the canonical projections and $\iota \colon \mathrm{Im}\, \varphi _1\alpha \rightarrow \mathrm{Im}\, \varphi _1$ the canonical inclusion. Because $\mathcal {C}$ is epi-closed and $\alpha \in \mathcal {C}$ , we conclude that $\iota \in \mathcal {C}$ . Now $\mathrm{Im}\, \varphi = \mathrm{Im}\, \varphi f = \mathrm{Im}\, \varphi _1 \alpha \varphi _2 \subseteq \mathrm{Im}\, \varphi _1 \alpha $ . Hence, $\varphi $ factors through $\iota \in \mathcal {C}$ and so $\varphi \in \mathcal {I}$ . Thus, $\mathcal {I}$ is a torsion ideal.

(b) Similar to (a).

Proof. (a) Let $\mathcal {I}$ be a functorially finite torsion ideal and $\varphi \colon A \rightarrow C_A$ a left $\mathcal {I}$ -approximation. We show that $\mathcal {I}$ equals the collection of all morphisms factoring through a morphism in $\mathrm{gen}\, \varphi $ . First, $\mathrm{gen}\, \varphi \subseteq \mathcal {I}$ by Lemma 5.7. Let $f\colon M \rightarrow N$ be in $\mathcal {I}$ and $\pi \colon A^n \rightarrow M$ an epimorphism for some . Then $f \pi $ factors through $\varphi ^n$ . Hence, there exists $g \colon C_A^n \rightarrow N$ with $f \pi = g \varphi ^n$ . It follows that $\mathrm{Im}\, f = \mathrm{Im}\, f\pi = \mathrm{Im}\, g \varphi ^n \subseteq \mathrm{Im}\, g$ . Thus, there exists a commutative diagram of morphisms

where $\pi '\colon C_A^n \rightarrow \mathrm{Im}\, g$ denotes the canonical projection and $f = f_2 f_1$ . Clearly, $f_1 \in \mathrm{gen}\, \varphi $ . It follows that f factors through a morphism in $\mathrm{gen}\, \varphi $ and $\mathcal {I} = \mathrm{gen}\, \varphi $ .

Let $\mathcal {I}$ equal the collection of all morphisms factoring through a morphism in $\mathcal {C} = \mathrm{gen}\, \varphi $ for some $\varphi \in \mathrm{mor}\,A$ . Then $\mathcal {I}$ is a torsion ideal by Lemma 5.7. Further, $\mathcal {C}$ is functorially finite in $\mathrm{mor}\,A$ by Proposition 5.6. Hence, for $M\in \mathrm{mod}\,A,$ there exists a left $\mathcal {C}$ -approximation of $1_M$ , given by a commutative square of morphisms

We show that f is a left $\mathcal {I}$ -approximation. First, $f\in \mathcal {I}$ since $f_2 \in \mathcal {C}$ and $f = f_2f_1$ . Now let $g\colon M \rightarrow N$ be a morphism factoring through a morphism $\alpha \colon M' \rightarrow N'$ in $\mathcal {C}$ . Then $g = g_2 \alpha g_1$ for $g_1 \colon M \rightarrow M'$ and $g_2 \colon N' \rightarrow N$ . We consider the commutative square

It must factor through the left $\mathcal {C}$ -approximation of $1_M$ from above, so there exists a commutative diagram of morphisms

with $g_1 = h_1 f_1$ . In total, $g = g_2 \alpha g_1 = g_2 \alpha h_1 f_1 = g_2 h_2 f$ . It follows that f is a left $\mathcal {I}$ -approximation, so $\mathcal {I}$ is functorially finite.