Notation 4.1. Throughout this section, we assume the following:

1. (1) ${\mathcal {M}}$ will denote an abelian category of R-modules that has enough projectives.

2. (2) ${\mathcal {P}}$ will denote a set of pairs of R-modules $(N,P)$ with $N \subseteq P$ , $N,P \in {\mathcal {M}}$ , and P projective.

3. (3) ${\mathcal {P}}'$ will denote a set of pairs such that ${\mathcal {P}} \subseteq {\mathcal {P}}'$ and for every $(L,M) \in {\mathcal {P}}'$ with $L \subseteq M$ , there is a projective module P $\in {\mathcal {M}}$ and a surjection $\pi :P \to M$ in ${\mathcal {M}}$ with $(\pi ^{-1}(L),P) \in {\mathcal {P}}$ .

4. (4) In addition, if $(L,M),(N,M) \in {\mathcal {P}}'$ with $L \subseteq N$ , then there exists $\pi :P \twoheadrightarrow M$ such that $(\pi ^{-1}(L),P),(\pi ^{-1}(N),P) \in {\mathcal {P}}$ .

5. (5) We further assume:

   • (*) Given $L,M,P,\pi $ as above, if we have a commutative diagram

     
     in ${\mathcal {M}}$ satisfying

     1. (a) $\varphi :M \to N$ is surjective,

     2. (b) $q:Q \to N$ is a surjection from a projective module onto N,

     3. (c) $(q^{-1}(\varphi (L)),Q) \in {\mathcal {P}}$ ,

     4. (d) $\tilde {\varphi }:P \to Q$ is any map such that $\varphi \circ \pi = q \circ \tilde {\varphi }$ (such a $\tilde {\varphi }$ always exists by the projectiveness of P),

     then $(\tilde {\varphi }(\pi ^{-1}(L)),Q) \in {\mathcal {P}}$ .

Proof. We first show that $p_c(L,M)$ is well defined, that is, independent of our choice of $\pi $ and P. Accordingly, suppose that $\pi : P {\twoheadrightarrow } M$ and $\pi ': P' {\twoheadrightarrow } M$ in ${\mathcal {M}}$ , where $P, P'$ are projective and $(\pi ^{-1}(L),P),((\pi ')^{-1}(L),P') \in {\mathcal {P}}$ . Let $z\in {p}_{\operatorname {\mathrm {c}}}(L, M)$ with respect to $\pi $ . That is, there is some $y \in p(\pi ^{-1}(L),P)$ with $\pi (y) = z$ .

Since $\pi '$ is surjective and P is projective, there is a map $\varphi : P {\rightarrow } P'$ in ${\mathcal {M}}$ with $\pi = \pi ' \circ \varphi $ . By the cofunctoriality hypothesis,

$$\begin{align*}p(\pi^{-1}(L),P)=p((\pi' \circ \varphi)^{-1}(L),P)=p(\varphi^{-1}((\pi')^{-1}(L)),P) \subseteq \varphi^{-1}(p((\pi')^{-1}(L),P')).\end{align*}$$

So $\varphi (y) \in p((\pi ')^{-1}(L),P')$ . Further, $\pi '(\varphi (y))=\pi (y)=z$ . So $z \in {p}_{\operatorname {\mathrm {c}}}(L,M)$ with respect to $\pi '$ . Hence, ${p}_{\operatorname {\mathrm {c}}}(L,M)$ with respect to P and $\pi $ is contained in ${p}_{\operatorname {\mathrm {c}}}(L,M)$ with respect to $P'$ and $\pi '$ , and by symmetry, they are equal.

Next, we show that ${p}_{\operatorname {\mathrm {c}}}$ satisfies the isomorphism criterion. Suppose $\varphi :M \rightarrow M'$ is an isomorphism, $(L,M) \in {\mathcal {P}}'$ , and $\pi :P {\twoheadrightarrow } M$ with P projective and $(\pi ^{-1}(L),P)= ((\varphi \circ \pi )^{-1}(\varphi (L)),P) \in {\mathcal {P}}$ . Then

$$ \begin{align*} \varphi({p}_{\operatorname{\mathrm{c}}}(L,M))&=\varphi(\pi(p(\pi^{-1}(L),P)))=\varphi(\pi(p(\pi^{-1}(\varphi^{-1}(\varphi(L))),P)))\\ &=\varphi\circ\pi(p((\varphi\circ\pi)^{-1}(\varphi(L)),P))={p}_{\operatorname{\mathrm{c}}}(\varphi(L),M'), \end{align*} $$

which shows that ${p}_{\operatorname {\mathrm {c}}}$ is invariant under isomorphisms.

Now, we show that ${p}_{\operatorname {\mathrm {c}}}$ is cohereditary. Let $L \subseteq N \subseteq M$ with

$$\begin{align*}(N,M),(N/L,M/L) \in {\mathcal{P}}',\end{align*}$$

and let $\pi : P {\twoheadrightarrow } M$ in ${\mathcal {M}}$ with P projective such that $(\pi ^{-1}(N),P) \in {\mathcal {P}}$ . Let $q: M {\twoheadrightarrow } M/L$ be the canonical surjection. Note that $q^{-1}(N/L)=N$ , so that $(q^{-1}(N/L),M) \in {\mathcal {P}}'$ , and $(q \circ \pi )^{-1}(N/L)=\pi ^{-1}(N)$ , so that $((q \circ \pi )^{-1}(N/L),P) \in {\mathcal {P}}$ . Then

$$ \begin{align*} {p}_{\operatorname{\mathrm{c}}}(N/L, M/L) &= (q \circ \pi) (p((q \circ \pi)^{-1}(N/L),P))\\ &= q(\pi p(\pi^{-1}(q^{-1}(N/L)),P))) \\ &= q( {p}_{\operatorname{\mathrm{c}}}(q^{-1}(N/L),M))\\ & = \displaystyle\frac{{p}_{\operatorname{\mathrm{c}}} (q^{-1}(N/L),M)+L}{L}\\ &=\displaystyle\frac{{p}_{\operatorname{\mathrm{c}}} (N,M)+L}{L}.\ \end{align*} $$

Hence, ${p}_{\operatorname {\mathrm {c}}}$ is cohereditary.

We now show that ${p}_{\operatorname {\mathrm {c}}}$ is cofunctorial. Accordingly, let $L \subseteq N$ and let $\varphi : M \to N$ be a homomorphism in ${\mathcal {M}}$ with $(L,N),(\varphi ^{-1}(L),M) \in {\mathcal {P}}'$ . Choose maps $\pi : P {\twoheadrightarrow } M$ and $q: Q {\twoheadrightarrow } N$ in ${\mathcal {M}}$ with $P,Q$ projective such that $(\pi ^{-1}(\varphi (L)),P),(q^{-1}(L),Q) \in {\mathcal {P}}$ . Then, since P is projective and q is surjective, there is some map $\tilde \varphi : P {\rightarrow } Q$ in ${\mathcal {M}}$ with $q \circ \tilde \varphi = \varphi \circ \pi $ .

Now, let $z\in {p}_{\operatorname {\mathrm {c}}} (\varphi ^{-1}(L),M)$ . Then there is some $y \in p(\pi ^{-1}(\varphi ^{-1}(L)), P)$ with $z=\pi (y)$ . Then, by cofunctoriality of p when the ambient modules are projective,

$$ \begin{align*}p(\pi^{-1}(\varphi^{-1}(L)),P)&=p((\varphi \circ \pi)^{-1}(L),P) =p((q \circ \tilde{\varphi})^{-1}(L),P)\\ &=p(\tilde{\varphi}^{-1}(q^{-1}(L)),P)\subseteq \tilde{\varphi}^{-1}(p(q^{-1}(L)),Q).\end{align*} $$

So $\tilde {\varphi }(y) \in p(q^{-1}(L),Q)$ . Hence, $\varphi (z)=q(\tilde {\varphi }(y)) \in {p}_{\operatorname {\mathrm {c}}}(L,N)$ , which shows that ${p}_{\operatorname {\mathrm {c}}}$ is cofunctorial.

To see that $p={p}_{\operatorname {\mathrm {c}}}$ when p is cohereditary, first we show that if P is projective, $K \subseteq L \subseteq P$ , $(L,P) \in {\mathcal {P}}$ , and $(L/K,P/K) \in {\mathcal {P}}'$ , then $p(L/K,P/K)={p}_{\operatorname {\mathrm {c}}} (L/K,P/K)$ . Let $\pi : P \rightarrow P/K$ . Then

$$\begin{align*}{p}_{\operatorname{\mathrm{c}}}(L/K,P/K)=\pi(p(L,P))=\displaystyle\frac{p(L,P)+K}{K}=p(L/K,P/K)\end{align*}$$

since p is cohereditary.

Now, let $(N,M) \in {\mathcal {P}}'$ and $\pi :P \twoheadrightarrow M$ in ${\mathcal {M}}$ with P projective and $(\pi ^{-1}(N),P) \in {\mathcal {P}}$ . Set $\ker (\pi )=K$ so that $\bar {\pi }:P/K \to M$ is an isomorphism. Then, setting $L/K=\bar {\pi }^{-1}(N)$ where $K \subseteq L \subseteq P$ , we have

$$\begin{align*}{p}_{\operatorname{\mathrm{c}}}(N,M)={p}_{\operatorname{\mathrm{c}}}(L/K,P/K)=p(L/K,P/K)=p(N,M).\\[-38pt] \end{align*}$$

Proof. By Proposition 4.3, it suffices to show that ${p}_{\operatorname {\mathrm {c}}}$ is order-preserving on submodules. The final comment follows by Proposition 3.5(c).

Suppose $L \subseteq N \subseteq M$ and $(L,M), (N,M) \in {\mathcal {P}}'$ and suppose P is a projective module such that $\pi :P {\twoheadrightarrow } M$ in ${\mathcal {M}}$ and $(\pi ^{-1}(L),P), (\pi ^{-1}(N),P) \in {\mathcal {P}}$ as in Notation 4.1.

Since $\pi ^{-1}(L) \subseteq \pi ^{-1}(N) \subseteq P$ , and p is order-preserving on submodules, then $p(\pi ^{-1}(L),P) \subseteq p(\pi ^{-1}(N),P)$ . Now, observe that

$$\begin{align*}{p}_{\operatorname{\mathrm{c}}}(L,M)=\pi(p(\pi^{-1}(L),P)) \subseteq \pi(p(\pi^{-1}(N),P))= {p}_{\operatorname{\mathrm{c}}}(N,M), \end{align*}$$

implying that ${p}_{\operatorname {\mathrm {c}}}$ is order-preserving on submodules.

Proof. Suppose $(L,M) \in {\mathcal {P}}'$ and $\pi :P {\twoheadrightarrow } M$ is a surjective homomorphism in ${\mathcal {M}}$ with $(\pi ^{-1}(L),P) \in {\mathcal {P}}$ . Since p is extensive

$$\begin{align*}L = \pi(\pi^{-1}(L)) \subseteq \pi(p( \pi^{-1}(L), P))={p}_{\operatorname{\mathrm{c}}}(L,M),\end{align*}$$

implying that ${p}_{\operatorname {\mathrm {c}}}$ is extensive. Now, by Lemma 3.16, an extensive cohereditary pair operation is residual. Thus, by Proposition 4.3, ${p}_{\operatorname {\mathrm {c}}}$ is residual and cofunctorial. If p is a cofunctorial residual pair operation, then p is a cofunctorial extensive cohereditary pair operation by Lemma 3.16 and again Proposition 4.3 gives us that ${p}_{\operatorname {\mathrm {c}}}=p$ when p is residual.

Notation 4.6. When p is extensive, we will denote ${p}_{\operatorname {\mathrm {c}}}$ by ${p}_{\operatorname {\mathrm {r}}}$ from now on.

When $p={\mathrm {cl}}$ is a closure operation and $L \subseteq M$ , we will denote $p(L,M)=L_M^{{\mathrm {cl}}}$ .

Proof. Since every cofunctorial closure operation is a cofunctorial pair operation, then Corollary 4.5 gives us that ${{\mathrm {cl}}}_{\operatorname {\mathrm {r}}}$ is a cofunctorial residual pair operation on ${\mathcal {P}}'$ and if ${\mathrm {cl}}$ is a residual closure on ${\mathcal {P}}'$ , then ${\mathrm {cl}}={{\mathrm {cl}}}_{\operatorname {\mathrm {r}}}$ . By Proposition 4.4 and Corollary 4.5, we see that ${{\mathrm {cl}}}_{\operatorname {\mathrm {r}}}$ is extensive and order-preserving on submodules. When we have shown ${{\mathrm {cl}}}_{\operatorname {\mathrm {r}}}$ is idempotent, our proof will be complete.

Let $(N, M) \in {\mathcal {P}}'$ and P be a projective R-module with submodule L such that $M \cong P/K$ and $N \cong L/K$ in ${\mathcal {M}}$ , and $(L,P) \in {\mathcal {P}}$ . It will be enough to show that ${{\mathrm {cl}}}_{\operatorname {\mathrm {r}}}$ is idempotent on submodules of $P/K$ . Suppose $\pi :P {\twoheadrightarrow } P/K$ , then by definition,

$$\begin{align*}(L/K)_{P/K}^{{{\mathrm{cl}}}_{\operatorname{\mathrm{r}}}}=\displaystyle\frac{L_P^{{\mathrm{cl}}}+K}{K}=\displaystyle\frac{L_P^{{\mathrm{cl}}}}{K},\end{align*}$$

where the last equality holds because ${\mathrm {cl}}$ is extensive. Now, applying ${{\mathrm {cl}}}_{\operatorname {\mathrm {r}}}$ to $(L/K)_{P/K}^{{{\mathrm {cl}}}_{\operatorname {\mathrm {r}}}}$ and using the above equality and the idempotence of ${\mathrm {cl}}$ , we obtain

$$\begin{align*}((L/K)_{P/K}^{{{\mathrm{cl}}}_{\operatorname{\mathrm{r}}}})_{P/K}^{{{\mathrm{cl}}}_{\operatorname{\mathrm{r}}}}=\displaystyle\frac{(L_P^{{\mathrm{cl}}})_P^{{\mathrm{cl}}}}{K}=\displaystyle\frac{L_P^{{\mathrm{cl}}}}{K}=(L/K)_{P/K}^{{{\mathrm{cl}}}_{\operatorname{\mathrm{r}}}},\end{align*}$$

concluding our proof.

Notation 4.8. When $\operatorname {\mathrm {i}}$ is an interior operation, we denote $L_{\operatorname {\mathrm {i}}}^M:=\operatorname {\mathrm {i}}(L,M)$ .

Proof. First, we show that ${\operatorname {\mathrm {i}}}_{\operatorname {\mathrm {c}}}$ is intensive. Note that if $(L,M) \in {\mathcal {P}}'$ and $\pi :P {\twoheadrightarrow } M$ is a surjective homomorphism in ${\mathcal {M}}$ with $(\pi ^{-1}(L),P) \in {\mathcal {P}}$ , then since $\operatorname {\mathrm {i}}$ is intensive

$$\begin{align*}L_{{\operatorname{\mathrm{i}}}_{\operatorname{\mathrm{c}}}}^M= \pi((\pi^{-1}(L))_{\operatorname{\mathrm{i}}}^P)\subseteq \pi(\pi^{-1}(L))=L,\end{align*}$$

implying that ${\operatorname {\mathrm {i}}}_{\operatorname {\mathrm {c}}}$ is intensive.

Proposition 4.4 implies that ${\operatorname {\mathrm {i}}}_{\operatorname {\mathrm {c}}}$ is order-preserving on submodules. Thus, to show that ${\operatorname {\mathrm {i}}}_{\operatorname {\mathrm {c}}}$ is a cohereditary interior operation, we need only show that ${\operatorname {\mathrm {i}}}_{\operatorname {\mathrm {c}}}$ is idempotent. Suppose P is a projective R-module and $K \subseteq L \subseteq P$ are such that $(L,P) \in {\mathcal {P}}$ and $(L/K, P/K) \in {\mathcal {P}}'$ . By the definition of ${\operatorname {\mathrm {i}}}_{\operatorname {\mathrm {c}}}$ and intensivity, we have

$$\begin{align*}(L/K)_{{\operatorname{\mathrm{i}}}_{\operatorname{\mathrm{c}}}}^{P/K}=\displaystyle\frac{L_{\operatorname{\mathrm{i}}}^P+K}{K} \subseteq L/K.\end{align*}$$

Assuming further that $((L/K)_{{\operatorname {\mathrm {i}}}_{\operatorname {\mathrm {c}}}}^{P/K}, P/K) \in {\mathcal {P}}'$ , then

$$\begin{align*}(L/K_{{\operatorname{\mathrm{i}}}_{\operatorname{\mathrm{c}}}}^{P/K})_{{\operatorname{\mathrm{i}}}_{\operatorname{\mathrm{c}}}}^{P/K} \subseteq (L/K)_{{\operatorname{\mathrm{i}}}_{\operatorname{\mathrm{c}}}}^{P/K}\end{align*}$$

as ${\operatorname {\mathrm {i}}}_{\operatorname {\mathrm {c}}}$ is order-preserving on submodules. We also know that $L_{\operatorname {\mathrm {i}}}^P \subseteq L_{\operatorname {\mathrm {i}}}^P+K$ , so since $(L_{\operatorname {\mathrm {i}}}^P,P), (L_{\operatorname {\mathrm {i}}}^P+K,P) \in {\mathcal {P}}$ by hypothesis, then using the fact that $\operatorname {\mathrm {i}}$ is idempotent, we have

$$\begin{align*}L_{\operatorname{\mathrm{i}}}^P=(L_{\operatorname{\mathrm{i}}}^P)_{\operatorname{\mathrm{i}}}^P \subseteq (L_{\operatorname{\mathrm{i}}}^P+K)_{\operatorname{\mathrm{i}}}^P,\end{align*}$$

which implies that

$$ \begin{align*}(L/K)_{{\operatorname{\mathrm{i}}}_{\operatorname{\mathrm{c}}}}^{P/K}&=\displaystyle\frac{L_{\operatorname{\mathrm{i}}}^P+K}{K}=\displaystyle\frac{(L_{\operatorname{\mathrm{i}}}^P)_{\operatorname{\mathrm{i}}}^P+K}{K}\\ &\subseteq \displaystyle\frac{(L_{\operatorname{\mathrm{i}}}^P+K)_{\operatorname{\mathrm{i}}}^P+K}{K}=((L/K)_{{\operatorname{\mathrm{i}}}_{\operatorname{\mathrm{c}}}}^{P/K})_{{\operatorname{\mathrm{i}}}_{\operatorname{\mathrm{c}}}}^{P/K}. \end{align*} $$

Now, for any $(N,M)\in {\mathcal {P}}'$ , with $M \cong P/K$ and $N \cong L/K$ in ${\mathcal {M}}$ , we have

$$\begin{align*}(N_{{\operatorname{\mathrm{i}}}_{\operatorname{\mathrm{c}}}}^M)_{{\operatorname{\mathrm{i}}}_{\operatorname{\mathrm{c}}}}^M = N_{{\operatorname{\mathrm{i}}}_{\operatorname{\mathrm{c}}}}^M,\end{align*}$$

which implies that ${\operatorname {\mathrm {i}}}_{\operatorname {\mathrm {c}}}$ is idempotent and hence a cohereditary functorial interior operation.

Proof. Suppose $(L, M) \in {\mathcal {P}}'$ , and let $\pi :P {\twoheadrightarrow } M$ be a surjection in ${\mathcal {M}}$ such that $(\pi ^{-1}(L),P) \in {\mathcal {P}}$ . By definition ${p}_{\operatorname {\mathrm {c}}}(L,M)= \pi (p(\pi ^{-1}(L), P))$ . Since p is cofunctorial, then

$$\begin{align*}\pi(p(\pi^{-1}(L), P)) \subseteq \pi(\pi^{-1}(p(L,M)))=p(L,M)\end{align*}$$

since $\pi $ is surjective. Hence, ${p}_{\operatorname {\mathrm {c}}} \leq p$ .

Proof. Since ${\mathrm {cl}}$ is both cofunctorial and extensive, and cohereditary, extensive operations are residual by Lemma 3.16, this is a direct consequence of Proposition 4.10.

Proof. If $p \leq q$ , then for all $(N,P) \in {\mathcal {P}}$ , $p(N,P) \subseteq q(N,P)$ . If $(L,M) \in {\mathcal {P}}'$ , then there exist a projective module P and surjection $\pi :P {\twoheadrightarrow } M$ in ${\mathcal {M}}$ such that ${p}_{\operatorname {\mathrm {c}}}(L,M)=\pi (p(\pi ^{-1}(L),P))$ and ${q}_{\operatorname {\mathrm {c}}}(L,M)=\pi (q(\pi ^{-1}(L),P))$ . Since $p(\pi ^{-1}(L),P) \subseteq q(\pi ^{-1}(L),P)$ , then

$$\begin{align*}{p}_{\operatorname{\mathrm{c}}}(L,M)=\pi(p(\pi^{-1}(L),P)) \subseteq \pi(q(\pi^{-1}(L),P))={q}_{\operatorname{\mathrm{c}}}(L,M),\end{align*}$$

giving us the result.

Proof. Let $y\in P$ with $\pi (y) \in {L}^{{{\mathrm {cl}}}_{\operatorname {\mathrm {r}}}}_{M}$ . Let $q: M {\twoheadrightarrow } M/L=:\bar M$ be the natural surjection. Then, by functoriality of ${{\mathrm {cl}}}_{\operatorname {\mathrm {r}}}$ and by definition, we have

$$\begin{align*}(q \circ \pi)(y) \in q({L}^{{{\mathrm{cl}}}_{\operatorname{\mathrm{r}}}}_{M}) \subseteq {0}^{{{\mathrm{cl}}}_{\operatorname{\mathrm{r}}}}_{M/L} = (q \circ \pi)\left((\ker (q \circ \pi))^{\mathrm{cl}}_P\right) = (q \circ \pi)\left(\pi^{-1}(L)^{\mathrm{cl}}_P\right). \end{align*}$$

Hence, there is some $a \in \pi ^{-1}(L)^{\mathrm {cl}}_P$ with $q(\pi (y)) = q(\pi (a))$ . That is, $\pi (y-a) \in \ker q = L$ . Thus, $y-a \in \pi ^{-1}(L)$ , whence

$$\begin{align*}y=(y-a)+a \in \pi^{-1}(L)+ \pi^{-1}(L)^{\mathrm{cl}}_P = \pi^{-1}(L)^{\mathrm{cl}}_P.\\[-34pt] \end{align*}$$

Proof. Since ${{\mathrm {cl}}}_{\operatorname {\mathrm {r}}}$ is residual, by Lemma 3.18, it is enough to show that if $L \subseteq M$ are finitely generated modules with $L \subseteq ({\mathfrak {m}} L)_M^{{{\mathrm {cl}}}_{\operatorname {\mathrm {r}}}}$ , we have $L \subseteq 0_M^{{{\mathrm {cl}}}_{\operatorname {\mathrm {r}}}}$ . Let P be a finitely generated free module, and let $\pi :P \twoheadrightarrow M$ be a surjection. Let $K=\ker \pi $ . We claim that $\pi ^{-1}({\mathfrak {m}} L)=K+{\mathfrak {m}} \pi ^{-1}(L)$ . To see this, first note that $\pi (K+{\mathfrak {m}} \pi ^{-1}(L)) \subseteq \pi (K)+{\mathfrak {m}} \pi (\pi ^{-1}(L)) \subseteq {\mathfrak {m}} L$ , whence $K+{\mathfrak {m}} \pi ^{-1}(L) \subseteq \pi ^{-1}({\mathfrak {m}} L)$ . For the reverse inclusion, let $x \in \pi ^{-1}({\mathfrak {m}} L)$ . That is, $\pi (x) \in {\mathfrak {m}} L$ . Then there exist $y_j \in {\mathfrak {m}}$ and $z_j \in L$ such that $\pi (x)=\sum _{j=1}^n y_jz_j$ . By the surjectivity of $\pi $ , there exist $z_j' \in \pi ^{-1}(L)$ with $\pi (z_j')=z_j$ . It follows that $x-\sum _{j=1}^n y_jz_j' \in \ker \pi =K$ . Thus, $x \in K+{\mathfrak {m}} \pi ^{-1}(L)$ .

Now, let $z \in \pi ^{-1}(L)$ . Then since $\pi (z) \in L \subseteq ({\mathfrak {m}} L)_M^{{{\mathrm {cl}}}_{\operatorname {\mathrm {r}}}}$ , by Lemma 4.14, we have $z \in (\pi ^{-1}({\mathfrak {m}} L))_P^{{\mathrm {cl}}}=(K+{\mathfrak {m}} \pi ^{-1}(L))_P^{\mathrm {cl}}$ . Thus, we have $K \subseteq \pi ^{-1}(L) \subseteq (K+{\mathfrak {m}} \pi ^{-1}(L))_P^{\mathrm {cl}}$ , so by the Nakayama property for cl $\pi ^{-1}(L) \subseteq K_P^{\mathrm {cl}}=(\ker \pi )_P^{\mathrm {cl}}$ . Then, by the definition of ${\mathrm {cl}}_r$ , $L=\pi (\pi ^{-1}(L)) \subseteq \pi ((\ker \pi )_P^{\mathrm {cl}})=0_M^{{{\mathrm {cl}}}_{\operatorname {\mathrm {r}}}}$ .

Proof. First, we prove that if $(0,M) \in {\mathcal {P}}'$ , then

$$\begin{align*}0_M^{{{\mathrm{cl}}}_{\operatorname{\mathrm{r}}}}=\bigcup_{(0,N) \in {\mathcal{P}}', N \subseteq M \text{ f.g.}} 0_N^{{{\mathrm{cl}}}_{\operatorname{\mathrm{r}}}}.\end{align*}$$

Let $(0,M) \in {\mathcal {P}}'$ , and let $\pi :P \to M$ be a surjection from a projective module such that $(\ker (\pi ),P) \in {\mathcal {P}}$ . Let $x \in 0_M^{{{\mathrm {cl}}}_{\operatorname {\mathrm {r}}}}=\pi ((\ker (\pi ))_P^{{\mathrm {cl}}})$ . Since ${\mathrm {cl}}$ is finitistic, there is some $\ker \pi \subseteq U \subseteq P$ such that $U/\ker (\pi )$ is finitely generated, $(\ker (\pi ),U) \in {\mathcal {P}}$ , and $x \in \pi ((\ker (\pi ))_U^{{\mathrm {cl}}})$ . Since U is a submodule of a projective module, $x \in \pi ((\ker (\pi ))_U^{{{\mathrm {cl}}}_{\operatorname {\mathrm {r}}}})$ . Since ${{\mathrm {cl}}}_{\operatorname {\mathrm {r}}}$ is functorial by Proposition 4.3, and by our hypothesis $(0,\pi (U)) \in {\mathcal {P}}'$ , we have $\pi ((\ker (\pi ))_U^{{{\mathrm {cl}}}_{\operatorname {\mathrm {r}}}}) \subseteq 0_{\pi (U)}^{{{\mathrm {cl}}}_{\operatorname {\mathrm {r}}}}$ . Since $\pi (U) \cong U/\ker (\pi )$ , it is finitely generated and $(0,\pi (U)) \in {\mathcal {P}}'$ , and so $\pi (U)$ is a finitely generated submodule N of M such that $x \in 0_N^{{{\mathrm {cl}}}_{\operatorname {\mathrm {r}}}}$ . This proves our first claim.

Next, we prove that this implies that ${{\mathrm {cl}}}_{\operatorname {\mathrm {r}}}$ is finitistic on ${\mathcal {P}}'$ . Let $(L,M) \in {\mathcal {P}}'$ such that $(0,M/L) \in {\mathcal {P}}'$ . Since ${{\mathrm {cl}}}_{\operatorname {\mathrm {r}}}$ is residual, $L_M^{{{\mathrm {cl}}}_{\operatorname {\mathrm {r}}}}=\pi ^{-1}(0_{M/L}^{{{\mathrm {cl}}}_{\operatorname {\mathrm {r}}}})$ , where $\pi :M \to M/L$ is the quotient map. Let $x \in L_M^{{{\mathrm {cl}}}_{\operatorname {\mathrm {r}}}}$ . By our first claim, there is some finitely generated $N/L \subseteq M/L$ such that $(0,N/L) \in {\mathcal {P}}'$ and $x \in \pi ^{-1}(0_{N/L}^{{{\mathrm {cl}}}_{\operatorname {\mathrm {r}}}})$ . Let $q:N \to N/L$ be the quotient map. We claim that $x \in q^{-1}(0_{N/L}^{{{\mathrm {cl}}}_{\operatorname {\mathrm {r}}}})$ . We know that q agrees with the composition $N \hookrightarrow M \twoheadrightarrow M/L$ . Since $x \in \pi ^{-1}(0_{N/L}^{{{\mathrm {cl}}}_{\operatorname {\mathrm {r}}}})$ , $x \in \pi ^{-1}(N/L) = N$ . So $x \in L_N^{{{\mathrm {cl}}}_{\operatorname {\mathrm {r}}}}=q^{-1}(0_{N/L}^{{{\mathrm {cl}}}_{\operatorname {\mathrm {r}}}})$ , as desired.

Notation 5.1. Throughout this section, we assume the following:

1. (1) ${\mathcal {M}}$ will denote an abelian category of R-modules that has enough injectives.

2. (2) ${\mathcal {P}}$ will denote a set of pairs of R-modules $(C,E)$ with $C \subseteq E$ , $C,E \in {\mathcal {M}}$ and E injective.

3. (3) ${\mathcal {P}}'$ will denote a set of pairs such that ${\mathcal {P}} \subseteq {\mathcal {P}}'$ and for every $(A,B) \in {\mathcal {P}}'$ with $A \subseteq B$ , there is an injective module E and an injection $i:B \to E$ in ${\mathcal {M}}$ with $(i(A),E) \in {\mathcal {P}}$ .

4. (4) In addition, if $(A,B),(C,B) \in {\mathcal {P}}'$ with $A \subseteq C$ , there is some injective module E and an injection $i:B \to E$ in ${\mathcal {M}}$ such that $(i(A),E),(i(C),E) \in {\mathcal {P}}$ .

5. (5) We further assume:

   • (**) Given $A,B,E,i$ as above, if we have a commutative diagram in ${\mathcal {M}}$

     
     satisfying

     1. (a) $\psi :B \to C$ is injective,

     2. (b) $j:C {\hookrightarrow } F$ where F is an injective module,

     3. (c) $(j(\psi (A)),F) \in {\mathcal {P}}$ ,

     4. (d) $\tilde {\psi }:E \to F$ is any map such that $\tilde {\psi } \circ i=j \circ \psi $ (such a $\tilde {\psi }$ always exists by the injectiveness of F),

     then $(\tilde {\psi }^{-1}(j(\psi (A))),E) \in {\mathcal {P}}$ .

Proof. We first show that this is well defined, independent of i and E. Accordingly, let $(A,B) \in {\mathcal {P}}'$ and $i: B {\hookrightarrow } E$ and $i': B {\hookrightarrow } E'$ be inclusions in ${\mathcal {M}}$ , where $E, E'$ are injective and $(i(A),E),(i'(A),E') \in {\mathcal {P}}$ .

Since $i'$ is injective and E is injective, there is some R-linear map $\varphi :E' \to E$ in ${\mathcal {M}}$ with $i=\varphi \circ i'$ . Let $z \in {p}_{\operatorname {\mathrm {h}}}(A,B)$ with respect to $i'$ , so there is some $y \in p(i'(A),E')$ with $i'(z)=y$ . Then $i(z)=(\varphi \circ i')(z)=\varphi (y)$ . We have $\varphi (y) \in \varphi (p(i'(A),E'))$ . By functoriality, this is contained in $p(\varphi (i'(A)),E)=p(i(A),E)$ . But then, $i(z) \in p(i(A),E)$ , so $z \in {p}_{\operatorname {\mathrm {h}}}(A,B)$ with respect to E, as desired. This gives us one inclusion, and the other follows by symmetry.

Next, we show that ${p}_{\operatorname {\mathrm {h}}}$ satisfies the isomorphism criterion. Suppose $\varphi :M \rightarrow M'$ is an isomorphism, $(L,M), (\varphi (L),M') \in {\mathcal {P}}'$ , and $i:M{\hookrightarrow } E$ with E injective and $(i(L),E)=((i\circ \varphi ^{-1})(\varphi (L)),P) \in {\mathcal {P}}$ . Then

$$ \begin{align*} \varphi({p}_{\operatorname{\mathrm{h}}}(L,M))&=\varphi(i^{-1}(p(i(L),E)))=\varphi(i^{-1}(p(i(\varphi^{-1}(\varphi(L))),E)))\\ &=(i\circ \varphi^{-1})^{-1}(p((i\circ\varphi^{-1})(\varphi(L)),E))={p}_{\operatorname{\mathrm{h}}}(\varphi(L),M'), \end{align*} $$

which shows that ${p}_{\operatorname {\mathrm {h}}}$ is invariant under isomorphisms.

Now, we show that ${p}_{\operatorname {\mathrm {h}}}$ is hereditary. Let $A \subseteq C \subseteq B$ with $(A,C),(A,B) \in {\mathcal {P}}'$ , and let $i: B {\hookrightarrow } E$ in ${\mathcal {M}}$ with E injective such that $(i(A),E) \in {\mathcal {P}}$ as in Notation 5.1. Let $j: C {\hookrightarrow } B$ be the canonical inclusion. Then

$$ \begin{align*} {p}_{\operatorname{\mathrm{h}}}(A,C) &= (i \circ j)^{-1} (p((i \circ j)(A),E))\\ &= j^{-1}(i^{-1}( p(i(j(A)),E))) \\ &= j^{-1}( {p}_{\operatorname{\mathrm{h}}}(j(A),B))\\ & = {p}_{\operatorname{\mathrm{h}}} (j(A),B) \cap C\\ &={p}_{\operatorname{\mathrm{h}}}(A,B) \cap C. \end{align*} $$

Note that $j(A)=A$ and $(i \circ j)(A)=i(A)$ , making $(j(A),B) \in {\mathcal {P}}'$ and $((i \circ j)(A),E) \in {\mathcal {P}}$ . Hence, ${p}_{\operatorname {\mathrm {h}}}$ is hereditary.

We now show that ${p}_{\operatorname {\mathrm {h}}}$ is functorial. Accordingly, let $(A,B) \in {\mathcal {P}}'$ and let $\varphi : B {\rightarrow } D$ in ${\mathcal {M}}$ such that $(\varphi (A),D) \in {\mathcal {P}}'$ . Choose maps $i : B {\hookrightarrow } E$ and $i': D {\hookrightarrow } E'$ in ${\mathcal {M}}$ with $E,E'$ injective such that $(i(A),E),(i'(\varphi (A)),E') \in {\mathcal {P}}$ . Then, since $E'$ is injective and i is injective, there is some R-linear map $\tilde \varphi : E {\rightarrow } E'$ in ${\mathcal {M}}$ with $i' \circ \varphi = \tilde \varphi \circ i$ . Now, let $z\in {p}_{\operatorname {\mathrm {h}}} (A,B)$ . Then $y=i(z) \in p(i(A),E)$ . Hence, $\tilde \varphi (y) \in \tilde \varphi (p(i(A),E))$ . By functoriality of p when the ambient modules are injective, this is contained in $p(\tilde \varphi (i(A)),E')=p(i'(\varphi (A)),E')$ . Since $\tilde {\varphi }(y)=\tilde {\varphi }(i(z))=i'(\varphi (z))$ and $i'$ is injective, $(i')^{-1}(\tilde {\varphi }(y))=\{\varphi (z)\}.$ This element must be in $(i')^{-1}(p(i'(\varphi (A)),E'))={p}_{\operatorname {\mathrm {h}}}(\varphi (A),D)$ . Hence, $\varphi (z) \in {p}_{\operatorname {\mathrm {h}}}(\varphi (A),D)$ , so ${p}_{\operatorname {\mathrm {h}}}$ is functorial.

For the final statement, assume that p is hereditary. For $(A,B) \in {\mathcal {P}}'$ and $i:B \to E$ in ${\mathcal {M}}$ such that $(i(A),E) \in {\mathcal {P}}$ , we have ${p}_{\operatorname {\mathrm {h}}} (A,B)=p(A,E) \cap B=p(A,B),$ as desired.

Proof. Since ${p}_{\operatorname {\mathrm {h}}}$ is a functorial hereditary pair operation by Proposition 5.3, it suffices to show that for $A \subseteq C \subseteq B$ with $(A,B),(C,B) \in {\mathcal {P}}'$ , ${p}_{\operatorname {\mathrm {h}}}(A,B) \subseteq {p}_{\operatorname {\mathrm {h}}}(C,B)$ . By Notation 5.1, there is some injective module E containing B, with the inclusion morphism i in ${\mathcal {M}}$ , such that $(i(A),E),(i(C),E) \in {\mathcal {P}}$ , and by definition ${p}_{\operatorname {\mathrm {h}}}(A,B)=p(i(A),E) \cap B$ . Since p is order-preserving on submodules, we have $p(i(A),E) \subseteq p(i(C), E)$ , and intersecting each of these modules with B, we obtain ${p}_{\operatorname {\mathrm {h}}}(A,B)=p(i(A),E) \cap B \subseteq p(i(C), E)\cap B={p}_{\operatorname {\mathrm {h}}}(C,B)$ , which is what we needed to show.