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For discrete opfibrations $A\to D\leftarrow B$ with weakly filtered categories, the map taking a pair to its connected components of a category is a bijection. For surjectivity, send two base objects in the same component to a common target and lift both arrows. For injectivity, choose common targets upstairs for two representatives; a cocone over the resulting finite connected diagram downstairs makes their lifted composites agree. Hence the representatives have a common target in the pullback in a category. Over a filtered category, the component functor on the slice of discrete opfibrations preserves all finite categorical limits: it preserves pullbacks and takes the terminal object to a singleton.

A functor with a unique lift, from each specified object $c$, of every arrow $Fc\to d$. The lifted arrow's target is part of the uniqueness requirement. This is dual to a discrete fibration. Lifting a cocone of a finite connected category gives a cocone upstairs: lifts of its legs have a common target because uniqueness identifies targets along every diagram arrow. Thus a discrete opfibration over a weakly filtered category has weakly filtered domain, including the possibility of an empty domain.

A filtered category is nonempty, any two objects have arrows to a common object, and any parallel pair $u,v:A\rightrightarrows B$ is equalized by some arrow $B\to C$. Equivalently, every finite diagram has a cocone: first choose a common target for its finitely many objects, then successively equalize the finitely many discrepancies along its arrows. A weakly filtered category requires a cocone only for finite nonempty connected diagrams. Equivalently, each connected component of a category is filtered. To see this, two objects in one component are joined by a finite zigzag; a cocone on that zigzag gives a common target, and a parallel-pair diagram gives the equalization condition. Conversely a connected diagram lies in one filtered component. An empty category is weakly filtered, since it has no such diagrams.

The unique outgoing-arrow lifting property is precisely that of a discrete opfibration. Let $F:\mathcal{C}\to \mathcal{D}$ have this property, with $\mathcal{D}$ weakly filtered. Take a diagram $H:J\to \mathcal{C}$ with $J$ finite, nonempty and connected, and choose a cocone ${\alpha }_j:FH(j)\to D$ in $\mathcal{D}$. Lift each ${\alpha }_j$ uniquely from $H(j)$, obtaining ${\tilde{\alpha }}_j:H(j)\to C_j$. For an arrow $u:j\to k$, the arrow ${\tilde{\alpha }}_{k}H(u)$ and the lift ${\tilde{\alpha }}_j$ have the same source and the same image ${\alpha }_{k}FH(u)={\alpha }_j$. Unique lifting gives equality of the arrows and, importantly, $C_j=C_k$. Connectedness of $J$ makes all targets a single object $C_0$. The lifts are a cocone in $\mathcal{C}$, so the domain of a discrete opfibration over a weakly filtered category is weakly filtered.

Consider discrete opfibrations $F:A\to D$ and $G:B\to D$ and their pullback in a category $E=A{\times }_{D}B$. The assumed creation of pullbacks describes its objects as pairs $(a,b)$ with $Fa=Gb$, and its arrows as pairs $(u,v)$ with $Fu=Gv$. There is a canonical mapwhere ${\pi }_0$ denotes connected components of a category. Assume the vertices are weakly filtered categories.

For surjectivity, choose components $[a]$ and $[b]$ whose images belong to the same component of $D$. That component is filtered, so there are arrows $Fa\to d$ and $Gb\to d$. Lifting them from $a$ and $b$ gives $a\to a_1$ and $b\to b_1$ with $F{a}_1=G{b}_1=d$. Thus $(a_1,b_1)\in E$ maps to the chosen component pair.

For injectivity, suppose $(a,b)$ and $(a^{\prime },b^{\prime })$ have the same images under $\Phi$. Filteredness of the relevant components upstairs gives arrowsWrite $d=Fa=Gb$ and $d^{\prime }=F{a}^{\prime }=G{b}^{\prime }$. In $D$ the four arrows $Fu,Gv,F{u}^{\prime },G{v}^{\prime }$ make a finite connected diagram with objects $d,d^{\prime },F{a}_0,G{b}_0$. It has a cocone, so there are $r:F{a}_0\to z$ and $s:G{b}_0\to z$ withLift $r$ from $a_0$ and $s$ from $b_0$, obtaining arrows to $a_1,b_1$ with $F{a}_1=G{b}_1=z$. Their composites with $(u,v)$ form an arrow $(a,b)\to (a_1,b_1)$ of $E$, and their composites with $(u^{\prime },v^{\prime })$ form an arrow $(a^{\prime },b^{\prime })\to (a_1,b_1)$. Hence the two representatives lie in the same component. This proves the connected components of discrete-opfibration pullbacks formulaThe map is the canonical comparison, so this is preservation of the pullback, not just an accidental bijection of its objects.

Now fix a small filtered category $C$. Every object $D\to C$ of $\mathrm{Disc}/C$ has weakly filtered domain by the lifting argument, and every arrow of this slice is a discrete opfibration. Pullbacks in the slice are the created pullbacks above; their resulting domains are weakly filtered as well, since a pullback projection is a discrete opfibration onto a weakly filtered domain. The terminal slice object is $1_C:C\to C$. Because $C$ is nonempty and filtered, it is connected, so ${\pi }_0(C)$ is a singleton. Thus the component functor preserves the terminal object and pullbacks. These construct every finite categorical limit: binary products are pullbacks over the terminal object, and equalizers are pullbacks along a diagonal into a binary product. Therefore the functor from this slice to sets preserves all finite limits.

The statement that limits of shape $\mathcal{I}$ commute with colimits of shape $\mathcal{J}$ means that for every $D:\mathcal{I}\times \mathcal{J}\to \mathcal{C}$, the canonical commutation of limits and colimits mapis an isomorphism whenever the iterated limits and colimits exist.

A filtered category is a nonempty category in which every finite diagram has a cocone. Equivalently, any two objects map to a common object and any parallel pair becomes equal after postcomposition. A weakly filtered category requires cocones only for finite connected diagrams; equivalently, each connected component is filtered.

Write a weakly filtered category as the disjoint union ${\coprod }_s{\mathcal{J}}_s$ of its filtered connected components. Its colimit is the coproduct of the filtered colimits over the ${\mathcal{J}}_s$. In sets, a connected finite limit commutes with coproducts: connectedness forces all coordinates of a compatible tuple to lie in the same coproduct summand. By the assumed theorem, the filtered colimit over each ${\mathcal{J}}_s$ commutes with every finite limit. Applying these two facts successively proves that weakly filtered colimits commute with connected finite limits in $\mathbf{Set}$.

The forgetful functor from abelian group to sets creates finite limits and filtered colimits and reflects isomorphisms. The comparison map for a filtered colimit and a finite limit therefore becomes the corresponding isomorphism of sets, so filtered colimits commute with finite limits in $\mathbf{AbGp}$.

The dual claim fails because inverse limit need not preserve epimorphisms. Take the inverse systemswith identity bonding maps on $A_n$, reduction maps on $B_n$, and levelwise epimorphisms $q_n:A_n\to B_n$. Thenand the induced map $\mathbb{Z}\to {\mathbb{Z}}_2$ is not surjective. Since an epimorphism in abelian groups is a finite-colimit cokernel, cofiltered limits do not commute with finite colimits in $\mathbf{AbGp}$.