Michael Shulman exact completion of a 2-category

Regular completion

Recall that there is a 2-category $HDC(K)$ of homwise-discrete categories in any finitely complete 2-category $K$. We write $n Cong_s(K)$ for its full sub-2-category spanned by the $n$-congruences (always we take $n=$ 2, (2,1), (1,2), or 1). Recall that there is a functor $\Phi:K\to 2Cong_s(K)$ sending each object to its kernel; if $K$ is an $n$-category then the image of $\Phi$ is contained in $n Cong(K)$.

Lemma

Suppose that $K$ has finite limits. Then:

1. $HDC(K)$ has finite limits.
2. $n Cong_s(K)$ is closed under finite limits in $HDC(K)$.
3. $\Phi$ is 2-fully-faithful (that is, an equivalence on hom-categories) and preserves finite limits.

Proof

It suffices to deal with finite products, inserters, and equifiers. Evidently $\Phi(1)$ is a terminal object. If $D$ and $E$ are homwise-discrete categories, define $P_0 = D_0\times E_0$ and $P_1 = D_1\times E_1$; it is easy to check that then $P_1 \;\rightrightarrows\; P_0$ is a homwise-discrete category that is the product $D\times E$ in $HDC(K)$. Since $(D_0\times E_0) ^{\mathbf{2}} \simeq (D_0) ^{\mathbf{2}} \times (E_0) ^{\mathbf{2}}$, and products preserve ffs, we see that $P$ is an $n$-congruence if $D$ and $E$ are and that $\Phi$ preserves products.

For inserters, let $f,g:C \;\rightrightarrows\; D$ be functors in $HDC(K)$, define $i_0:I_0\to C_0$ by the pullback

$$\array{I_0 & \to & D_1\\ i_0 \downarrow && \downarrow \\ C_0 & \overset{(f_0,g_0)}{\to} & D_0\times D_0,}$$

and define $i_1:I_1 \to C_1$ by the pullback

$$\array{I_1 & \to & X\\ i_1\downarrow && \downarrow\\ C_1 & \overset{(f_1,g_1)}{\to} & D_1\times D_1}$$

where $X$ is the “object of commutative squares in $D$.” Then $I_1 \;\rightrightarrows\; I_0$ is a homwise-discrete category and $i:I\to C$ is an inserter of $f,g$. Also, $I$ is an $n$-congruence if $C$ is, and $\Phi$ preserves inserters.

Finally, for equifiers, suppose we have functors $f,g:C \;\rightrightarrows\; D$ and 2-cells $\alpha,\beta:f \;\rightrightarrows\; g$ in $HDC(K)$, represented by morphisms $a,b:C_0 \;\rightrightarrows\; D_1$ such that $(s,t) a \cong (f_0,g_0)\cong (s,t) b$. Let $e_0:E_0\to C_0$ be the universal morphism equipped with an isomorphism $\phi:a e_0 \cong b e_0$ such that $(s,t)\phi$ is the given isomorphism $(s,t) a\cong (s,t) b$ (this is a finite limit in $K$.) Note that since $(s,t):D_1\to D_0\times D_0$ is discrete, $e_0$ is ff. Now let $E_1 = (e_0\times e_0)^*C_1$; then $E_1 \;\rightrightarrows\; E_0$ is a homwise-discrete category and $e:E\to C$ is an equifier of $\alpha$ and $\beta$ in $HDC(K)$. Also $E$ is an $n$-congruence if $C$ is, and $\Phi$ preserves equifiers.

For any morphism $f:A\to B$ in $K$, $\Phi(f)$ is the functor $ker(A)\to ker(B)$ that consists of $f:A\to B$ and $f^{\mathbf{2}}: A^{\mathbf{2}} \to B^{\mathbf{2}}$. A transformation between $\Phi(f)$ and $\Phi(g)$ is a morphism $A\to B ^{\mathbf{2}}$ whose composites $A\to B ^{\mathbf{2}} \;\rightrightarrows\; B$ are $f$ and $g$; but this is just a transformation $f\to g$ in $K$. Thus, $\Phi$ is homwise fully faithful. And homwise essential-surjectivity follows from the essential uniqueness of thin structures, or equivalently a version of Prop 6.4 in FBMF.

Moreover, we have:

Theorem

If $K$ is an $n$-category with finite limits, then $n Cong_s(K)$ is regular.

Proof

It is easy to see that a functor $f:C\to D$ between $n$-congruences is ff in $n Cong_s(K)$ iff the square

$$\array{C_1 & \to & D_1\\ \downarrow && \downarrow\\ C_0\times C_0 & \to & D_0\times D_0}$$

is a pullback in $K$.

We claim that if $e:E\to D$ is a functor such that $e_0:E_0\to D_0$ is split (that is, $e_0 s\cong 1_{D_0}$ for some $s:D_0\to E_0$), then $e$ is eso in $n Cong_s(K)$. For if $e\cong f g$ for some ff $f:C\to D$ as above, then we have $g_0 s:D_0 \to C_0$ with $f_0 g_0 s \cong e_0 s \cong 1_{D_0}$, and so the fact that $C_1$ is a pullback induces a functor $h:D\to C$ with $h_0=g_0 s$ and $f h\cong 1_D$. But this implies $f$ is an equivalence; thus $e$ is eso.

Moreover, if $e_0:E_0\to D_0$ is split, then the same is true for any pullback of $e$. For the pullback of $e:E\to D$ along some $k:C\to D$ is given by a $P$ where $P_0 = E_0 \times_{D_0} D_{iso} \times_{D_0} C_0$; here $D_{iso}\hookrightarrow D_1$ is the “object of isomorphisms” in $D$. What matters is that the projection $P_0\to C_0$ has a splitting given by combining the splitting of $e_0$ with the “identities” morphism $D_0\to D_{iso}$.

Now suppose that $f:D\to E$ is any functor in $n Cong_s(K)$. It is easy to see that if we define $Q_0=D_0$ and let $Q_1$ be the pullback

$$\array{ Q_1 & \to & E_1 \\ \downarrow && \downarrow\\ Q_0 \times Q_0 & \overset{f_0\times f_0}{\to} & E_0\times E_0}$$

then $f \cong m e$ where $e:D \to Q$ and $m:Q\to E$ are the obvious functors. Moreover, clearly $m$ is ff, and $e$ satisfies the condition above, so any pullback of it is eso. It follows that if $f$ itself were eso, then it would be equivalent to $e$, and thus any pullback of it would also be eso; hence esos are stable under pullback.

Since $m$ is ff, the kernel of $f$ is the same as the kernel of $e$, so to prove $K$ regular it remains only to show that $e$ is a quotient of that kernel. If $C \;\rightrightarrows\; D$ denotes $ker(f)$, then $C$ is the comma object $(f/f)$ and thus we can calculate

$$C_0 = D_0\times_{E_0} E_1 \times_{E_0} D_0 \cong Q_1.$$

Therefore, if $g:D\to X$ is equipped with an action by $ker(f)$, then the action 2-cell is given by a morphism $Q_1=C_0\to X_1$, and the action axioms evidently make this into a functor $Q\to X$. Thus, $Q$ is a quotient of $ker(f)$, as desired.

However, there are three problems with the 2-category $n Cong_s(K)$.

1. It is too big. It is not necessary to include every $n$-congruence in order to get a regular category containing $K$, only those that occur as kernels of morphisms in $K$.
2. It is too small. While it is regular, it is not exact.
3. It doesn’t remember information about $K$. If $K$ is already regular, then passing to $n Cong_s(K)$ destroys most of the esos and quotients already present in $K$.

The solution to the first problem is straightforward. If $K$ is a 2-category with finite limits, define $K_{reg/lex}$ to be the sub-2-category of $2 Cong_s(K)$ spanned by the 2-congruences which occur as kernels of morphisms in $K$. If $K$ is an $n$-category then any such kernel is an $n$-congruence, so in this case $K_{reg/lex}$ is contained in $n Cong_s(K)$ and is an $n$-category. Also, clearly $\Phi$ factors through $K_{reg/lex}$.

Theorem

For any finitely complete 2-category $K$, the 2-category $K_{reg/lex}$ is regular, and the functor $\Phi:K\to K_{reg/lex}$ induces an equivalence

$$Reg(K_{reg/lex},L) \simeq Lex(K,L)$$

for any regular 2-category $K$.

Here $Reg(-,-)$ denotes the 2-category of regular functors, transformations, and modifications between two regular 2-categories, and likewise $Lex(-,-)$ denotes the 2-category of finite-limit-preserving functors, transformations, and modifications between two finitely complete 2-categories.

Proof

It is easy to verify that $K_{reg/lex}$ is closed under finite limits in $2 Cong_s(K)$, and also under the eso-ff factorization constructed in Theorem ; thus it is regular. If $F:K\to L$ is a lex functor where $L$ is regular, we extend it to $K_{reg/lex}$ by sending $ker(f)$ to the quotient in $L$ of $ker(F f)$, which exists since $L$ is regular. It is easy to verify that this is regular and is the unique regular extension of $F$.

In particular, if $K$ is a regular 1-category, $K_{reg/lex}$ is the ordinary regular completion of $K$. In this case our construction reduces to one of the usual constructions (see, for example, the Elephant).

To solve the second and third problems with $n Cong_s(K)$, we need to modify its morphisms.

Exact completion

Recall that 2-congruences in $Cat$ can be identified with certain double categories. As noted in PAPDC, edge-symmetric double categories with a thin structure are essentially the same as 2-categories, and homwise-discreteness makes them the same as 1-categories. Our lack of edge-symmetry means that we really have a 1-category with distinguished subclass of morphisms (the vertical ones), which must be preserved by functors between congruences. (Note that the transformations are “horizontal” and need not have distinguished components. Since every vertical arrow has a horizontal companion, any vertical transformation is represented by a horizontal one.) In order to eliminate the effect of the distinguished vertical morphisms, we can replace functors between congruences by anafunctors.

Definition

Suppose that $K$ is a finitely complete 2-site and that $D$, $E$, and $F$ are 2-congruences in $K$. A functor $g:F\to D$ is a weak equivalence if

1. the square
   $$\array{F_1 &\overset{g_1}{\to} & D_1 \\ \downarrow && \downarrow\\ F_0\times F_0 & \overset{g_0\times g_0}{\to} & D_0\times D_0}$$

   is a pullback, and

2. $g_0:F_0\to D_0$ is a cover (a one-element covering family). An anafunctor $D\to E$ is a span of functors $D \overset{f^s}{\leftarrow} F \overset{f^t}{\to} E$ such that $f^s$ is a weak equivalence.

The primary example we have in mind is when $K$ is a regular 2-category with its regular coverage, but it is useful to consider the general case.

Definition

If $D \leftarrow F \to E$ and $D \leftarrow G \to E$ are anafunctors between 2-congruences, then a transformation $F\to G$ is a transformation between the two induced functors $F\times_D G\;\rightrightarrows\; E$.

(Here $F\times_D G$ denotes the pullback in $2 Cong_s(K)$.)

Theorem

For any subcanonical and finitely complete 2-site $K$ (such as a regular $n$-category with its regular coverage), there is a finitely complete 2-category $2Cong(K)$ of 2-congruences, anafunctors, and transformations in $K$. It contains $2Cong_s(K)$ as a homwise-full sub-2-category (that is, $2Cong_s(K)(D,E)\hookrightarrow 2Cong(K)(D,E)$ is ff) closed under finite limits.

Proof

Composition is, of course, by pullback. Since covers are stable under pullback and composition, the composite of anafunctors is again an anafunctor. The coverage must be subcanonical in order to define the vertical composite of natural transformations. We regard a functor as an anafunctor by taking $f^s$ to be the identity; it is then clear that a transformation between functors is the same as a transformation between their corresponding anafunctors.

It is easy to see that products in $2 Cong_S(K)$ remain products in $n Cong(K)$. Before dealing with inserters and equifiers, we observe that if $A\leftarrow F \to B$ is an anafunctor in $2 Cong(K)$ and $e:X_0\to F_0$ is any eso, then pulling back $F_1$ to $X_0\times X_0$ defines a new congruence $X$ and an anafunctor $A \leftarrow X \to B$ which is isomorphic to the original in $2 Cong(K)(A,B)$. Thus, if $A\leftarrow F\to B$ and $A\leftarrow G\to B$ are parallel anafunctors in $2 Cong(K)$, by pulling them both back to $F\times_A G$ we may assume that they are defined by spans with the same first leg, i.e. we have $A\leftarrow X \;\rightrightarrows\; B$.

Now, for the inserter of $F$ and $G$ as above, let $E\to X$ be the inserter of $X \;\rightrightarrows\; B$ in $2 Cong_s(K)$. It is easy to check that the composite $E\to X \to A$ is an inserter of $F,G$ in $2 Cong(K)$. Likewise, given $\alpha,\beta: F \;\rightrightarrows\; G$ with $F$ and $G$ as above, we have transformations between the two functors $X \;\rightrightarrows\; B$ in $2 Cong_s(K)$, and it is again easy to check that their equifier in $2 Cong_s(K)$ is again the equifier in $2 Cong(K)$ of the original 2-cells $\alpha,\beta$. Thus, $2 Cong(K)$ has finite limits. Finally, by construction clearly the inclusion of $2 Cong_s(K)$ preserves finite limits.

We write $n Cong(K)$ for the full sub-2-category of $2Cong(K)$ on the $n$-congruences, which is a finitely complete $n$-category. Of course, it contains $n Cong_s(K)$ as a homwise-full sub-$n$-category closed under finite limits, and when $K$ is an $n$-category we have $\Phi:K \to n Cong(K)$.

Theorem

If $K$ is a subcanonical finitely complete $n$-site, then the functor $\Phi:K\to n Cong(K)$ is 2-fully-faithful. If $K$ is an $n$-exact $n$-category equipped with its regular coverage, then $\Phi:K\to n Cong(K)$ is an equivalence of 2-categories.

Proof

Since $\Phi:K \to n Cong_s(K)$ is 2-fully-faithful and $n Cong_s(K)\to n Cong(K)$ is homwise fully faithful, $\Phi:K \to n Cong(K)$ is homwise fully faithful. For homwise essential-surjectivity, suppose that $ker(A) \leftarrow F \to ker(B)$ is an anafunctor. Then $h:F_0 \to A$ is a cover and $F_1$ is the pullback of $A ^{\mathbf{2}}$ along it; but this just says that $F_1 = (h/h)$. The functor $F\to B$ consists of morphisms $g:F_0\to B$ and $F_1 = (h/h) \to B ^{\mathbf{2}}$, and functoriality says precisely that the resulting 2-cell equips $g$ with an action by the congruence $F$. But since $F$ is precisely the kernel of $h:F_0\to A$, which is a cover in a subcanonical 2-site and hence the quotient of this kernel, we have an induced morphism $f:A\to B$ in $K$. It is then easy to check that $f$ is isomorphic, as an anafunctor, to $F$. Thus, $\Phi$ is homwise an equivalence.

Now suppose that $K$ is an $n$-exact $n$-category and that $D$ is an $n$-congruence. Since $K$ is $n$-exact, $D$ has a quotient $q:D_0\to Q$, and since $D$ is the kernel of $q$, we have a functor $D \to ker(Q)$ which is a weak equivalence. Thus, we can regard it either as an anafunctor $D\to ker(Q)$ or $ker(Q)\to D$, and it is easy to see that these are inverse equivalences in $n Cong(K)$. Thus, $\Phi$ is essentially surjective, and hence an equivalence.

Note that by working in the generality of 2-sites, this construction includes the previous one. Specifically, if $K$ is a finitely complete 2-category equipped with its minimal coverage, in which the covering families are those that contain a split epic, then $n Cong(K) \simeq n Cong_s(K)$. This is immediate from the proof of Theorem , which implies that the first leg of any anafunctor relative to this coverage is both eso and ff in $n Cong_s(K)$, and hence an equivalence.

We also remark in passing that this allows us to reconstruct 2-exact 2-categories with enough groupoids or discretes from their subcategories of such.

Theorem

If $K$ is a 2-exact 2-category with enough groupoids, then $K\simeq 2 Cong(gpd(K))$. Likewise, if $K$ is 2-exact and has enough discretes, then $K\simeq 2 Cong(disc(K))$.

Proof

Define a functor $K\to 2Cong(gpd(K))$ by taking each object $A$ to the kernel of $j:J\to A$ where $j$ is eso and $J$ is groupoidal (for example, it might be the core of $A$). Note that this kernel lives in $2Cong(gpd(K))$ since $(j/j)\to J\times J$ is discrete, hence $(j/j)$ is also groupoidal. The same argument as in Theorem shows that this functor is 2-fully-faithful for any regular 2-category $K$ with enough groupoids, and essentially-surjective when $K$ is 2-exact; thus it is an equivalence. The same argument works for discrete objects.

In particular, the 2-exact 2-categories having enough discretes are precisely the 2-categories of internal categories and anafunctors in 1-exact 1-categories.

Our final goal is to construct the $n$-exact completion of a regular $n$-category, and a first step towards that is the following.

Theorem

If $K$ is a regular $n$-category, so is $n Cong(K)$. The functor $\Phi:K\to n Cong(K)$ is regular, and moreover for any $n$-exact 2-category $L$ it induces an equivalence

$$Reg(n Cong(K), L) \to Reg(K,L).$$

Proof

We already know that $n Cong(K)$ has finite limits and $\Phi$ preserves finite limits. The rest is very similar to Theorem . We first observe that an anafunctor $A \leftarrow F \to B$ is an equivalence as soon as $F\to B$ is also a weak equivalence (its reverse span $B\leftarrow F \to A$ then provides an inverse.) Also, $A \leftarrow F \to B$ is ff if and only if

$$\array{F_1 & \to & B_1\\ \downarrow && \downarrow \\ F_0\times F_0 & \to & B_0\times B_0}$$

is a pullback.

Now we claim that if $A\leftarrow F \to B$ is an anafunctor such that $F_0\to B_0$ is eso, then $F$ is eso. For if we have a composition

$$\array{ &&&& F \\ &&& \swarrow && \searrow\\ && G &&&& M\\ & \swarrow && \searrow && \swarrow && \searrow\\ A &&&& C &&&& B}$$

such that $M$ is ff, then $F_0\to B_0$ being eso implies that $M_0\to B_0$ is also eso; thus $M\to B$ is a weak equivalence and so $M$ is an equivalence. Moreover, by the construction of pullbacks in $n Cong(K)$, anafunctors with this property are stable under pullback.

Now suppose that $A \leftarrow F \to B$ is any anafunctor, and define $C_0=F_0$ and let $C_1$ be the pullback of $B_1$ to $C_0\times C_0$ along $C_0 = F_0 to B_0$. Then $C$ is an $n$-congruence, $C\to B$ is ff in $n Cong_s(K)$ and thus also in $n Cong(K)$, and $A \leftarrow F \to B$ factors through $C$. (In fact, $C$ is the image of $F\to B$ in $n Cong_s(K)$.) The kernel of $A\leftarrow F\to B$ can equally well be calculated as the kernel of $F\to B$, which is the same as the kernel of $F\to C$.

Finally, given any $A\leftarrow G \to D$ with an action by this kernel, we may as well assume (by pullbacks) that $F=G$ (which leaves $C$ unchanged up to equivalence). Then since the kernel acting is the same as the kernel of $F\to C$, regularity of $n Cong_s(K)$ gives a descended functor $C\to D$. Thus, $A\leftarrow F \to C$ is the quotient of its kernel; so $n Cong(K)$ is regular.

Finally, if $L$ is $n$-exact, then any functor $K\to L$ induces one $n Cong(K) \to n Cong(L)$, but $n Cong(L)\simeq L$, so we have our extension, which it can be shown is unique up to equivalence.

When $K$ is a regular 1-category, it is well-known that $1 Cong(K)$ (which, in that case, is the category of internal equivalence relations and functional relations) is the 1-exact completion of $K$ (the reflection of $K$ from regular 1-categories into 1-exact 1-categories). Theorem shows that in general, $n Cong(K)$ will be the $n$-exact completion of $K$ whenver it is $n$-exact. However, in general for $n\gt 1$ we need to “build up exactness” in stages by iterating this construction.

It is possible that the iteration will converge at some finite stage, but for now, define $n Cong^r(K) = n Cong(n Cong^{r-1}(K))$ and let $K_{n ex/reg} = colim_r n Cong^r(K)$.

Theorem

For any regular $n$-category $K$, $K_{n ex/reg}$ is an $n$-exact $n$-category and there is a 2-fully-faithful regular functor $\Phi:K\to K_{n ex/reg}$ that induces an equivalence

$$Reg(K_{n ex/reg},L) \simeq Reg(K,L)$$

for any $n$-exact 2-category $L$.

Proof

Sequential colimits preserve 2-fully-faithful functors as well as functors that preserve finite limits and quotients, and the final statement follows easily from Theorem . Thus it remains only to show that $K_{n ex/reg}$ is $n$-exact. But for any $n$-congruence $D_1 \;\rightrightarrows\; D_0$ in $K_{n ex/reg}$, there is some $r$ such that $D_0$ and $D_1$ both live in $n Cong^r(K)$, and thus so does the congruence since $n Cong^r(K)$ sits 2-fully-faithfully in $K_{n ex/reg}$ preserving finite limits. This congruence in $n Cong^r(K)$ is then an object of $n Cong^{r+1}(K)$ which supplies a quotient there, and thus also in $K_{n ex/reg}$.