A category is $M$-complete or $E$-cocomplete if has certain limits or colimits of morphisms in a given class $M$ or $E$.
Not to be confused with an M-category.
Let $C$ be a category and let $M$ be a class of monomorphisms in $C$. (Often, $M$ will be the right class in an orthogonal factorization system.) We say that $C$ is $M$-complete if it admits all (even large) intersections of $M$-subobjects. This means that it admits all (even large) wide pullbacks of families of $M$-morphisms, and such pullbacks are again in $M$. (If $M$ is the right class of an OFS, then any intersection of $M$-morphisms which exists is automatically in $M$.)
If $M$ is the class of all monomorphisms, we may say mono-complete for $M$-complete.
Dually, if $E$ is a class of epimorphisms, we say $C$ is $E$-cocomplete if it admits all cointersection?s of $E$-morphisms, and epi-cocomplete if $E$ is the class of all epimorphisms.
If $C$ is $M$-well-powered, then no large limits are required in the definition of $M$-completeness. Therefore, if $C$ is well-powered and complete, it is $M$-complete whenever $M$ is the right class in an OFS. Dually, if $C$ is well-copowered and cocomplete, it is $E$-cocomplete whenever $E$ is the left class in an OFS.
For similar reasons, the category FinSet is mono-complete and epi-cocomplete—although it is not complete or cocomplete, it is finitely complete and cocomplete, and its subobject lattices and quotient lattices are likewise essentially finite?.
If $C$ is a topological concrete category over a category $D$ which is mono-complete or epi-complete, then $C$ is also mono-complete or epi-complete. For the faithful forgetful functor $U\colon C\to D$ preserves and reflects monos and epis, and so the initial $C$-structure on an intersection of underlying monos in $D$ gives an intersection in $C$ and the final $C$-structure on a cointersection? of underlying epis in $D$ gives a cointersection in $C$.
$M$-completeness is useful for constructing orthogonal factorization systems. The following is Lemma 3.1 in CHK.
Let $M$ be a class of maps in a category $C$, and assume that
Then there is an orthogonal factorization system $(E,M)$, with $E = {}^\perp M$.
Given $f\colon A\to B$, let $m$ be the intersection of all $M$-morphisms $n\colon X \to B$ through which $f$ factors. Then by the universal property of this intersection, we have $f = m e$ for some $e$; thus it suffices to show $e\in E$. Suppose given a commutative square
with $p\in M$. By pulling $p$ back to $Y$ (since pullbacks of $M$-morphisms exist), we may assume that $Y=W$ and $h$ is the identity. But now the composite $m p$ is an $M$-morphism through which $f$ factors, so by definition, $m$ factors through it. Thus $p$ is an isomorphism and so the lifting problem can be solved.
In fact, it is easy to see that the same proof constructs a factorization structure for sinks.
Note that if $M$ is already part of a prefactorization system, then any composite, pullback, or intersection of $M$-morphisms which exists is automatically also in $M$, since $M = E^\perp$.
Let $(E,M)$ be a prefactorization system on a category $C$, and assume that
Then $(E,M)$ is an orthogonal factorization system.
The following is a slight generalization of Theorem 3.3 of CHK. There it is stated only for the case $M=$ strong monomorphisms, in which case a finitely complete and $M$-complete category is called finitely well-complete.
Let $S : A \rightleftarrows C : T$ be an adjunction, and assume that $A$ is finitely complete and $M$-complete for some OFS $(E,M)$, where $M$ consists of monomorphisms and contains the split monics. Define $E_S$ to be the class of maps inverted by $S$, and $M_S = (E_S)^\perp$; then $(E_S,M_S)$ is an OFS on $A$.
First of all, since $M_S$ belongs to a prefactorization system, it is closed under composites, pullbacks, and any intersections which exist. Therefore, if we define $M' \coloneqq M \cap M_S$, then $M'$ satisfies the hypotheses of Theorem , and so we have an OFS $(E',M')$.
Moreover, it is useful to notice that $E_S={}^\perp T(\hom(C))$: this is an easy consequence of the fact that if $S\dashv T$, then $S a\perp b\iff a\perp T b$, since $f\perp T u\iff S f\perp u$ for each $u\in\hom(C)$, so that $S f$ is an isomorphism.
Now suppose given $f\colon A\to B$; we want to construct an $(E_S,M_S)$-factorization. Let $v$ be the pullback of $T S f$ along the unit $\eta_B \colon B \to T S B$. The naturality square for $\eta$ at $f$ shows that $f$ factors through $v$, say $f = v w$.
Since $T S f$ is evidently in $M_S=({}^\perp T(\hom(C)))^\perp\supseteq T(\hom (C))$, so is $v$; thus it suffices to find an $(E_S,M_S)$-factorization of $w$.
Let $w = n g$ be the $(E',M')$-factorization of $w$. Since $M' \subseteq M_S$, it suffices to show that $g\in E_S$. Note also that since $w$ is a first factor of the unit $\eta_A$, by passing to adjuncts we find that $S w$ is split monic: in the former diagram we have $u w=\eta_A$, so that the adjunct $\epsilon_{S A} \cdot S u\cdot S n\cdot S g=1$, hence also $S g$ is a split monic. But $T S g$ is then also split monic, hence belongs to $M$ and thus also to $M'$ (since it obviously belong to $M_S=({}^\perp T(\hom(C)))^\perp\supseteq T(\hom (C))$). Therefore, since $g\in E'$, the naturality square for $\eta$ at $g$ contains a lift: there is an $\alpha\colon X\to T S A$ such that in the diagram
$\alpha\cdot g=\eta_A$ and $T S g \cdot \alpha=\eta_X$. Passing to adjuncts again, we find that $S g$ is also split epic, since we can consider the diagram
and the commutativity
Hence $S g$ is an isomorphism; thus $g\in E_S$ as desired.
This is useful in the construction of reflective factorization systems.
Last revised on April 20, 2014 at 22:08:48. See the history of this page for a list of all contributions to it.