Filtered category

A category is filtered when every finite diagram in it admits a cocone. Equivalently, it is nonempty, every pair of objects maps to a common object, and every pair of parallel arrows becomes equal after postcomposition.

A filtered colimit in a category is a colimit whose indexing category is a small filtered category. In the Category of sets, filtered colimits commute with finite limits in sets. A filtered colimit of modules is an example of this general categorical construction.

For a diagram of abelian groups indexed by a nonempty directed poset, take the set colimit as classes $[i,x]$, where $[i,x]=[j,y]$ when their images agree at some common later index. Define $[i,x]+[j,y]=[k,D_{ik}x+D_{jk}y]$ at any common upper bound $k$. Directedness and functoriality prove independence of choices. Negation is $-[i,x]=[i,-x]$ and zero is the class of any zero. This is the unique group structure making the canonical maps homomorphisms, and every compatible group cocone factors by a unique homomorphism. Hence the forgetful functor creates these colimits, even when transition maps are not injective.

A category is weakly filtered when every finite connected diagram in it admits a cocone. Equivalently, each of its connected components is a filtered category.

In the Category of sets, filtered colimits commute with finite limits. Elements in a finite limiting diagram involve only finitely many representatives and finitely many equalities, all of which can be realized at one common stage of a filtered diagram.

In a filtered colimit in a category of sets, any finite zigzag of generating identifications can be moved to one common stage. The filtered category conditions supply common objects and equalize the finitely many competing arrows. Consequently finitely many representatives and finitely many equalities can be realized simultaneously at one stage. This is the elementwise reason filtered colimits commute with finite limits in sets.

Let $A_n=\{m\in \mathbb{N}:m\ge n\}$ with the inverse-system inclusions $A_{n+1}\hookrightarrow A_n$, and let $B_n=C_n=1$. Each pushout of $1\leftarrow A_n\to 1$ is a singleton because $A_n$ is nonempty. Yet $\underleftarrow{\lim }{A}_n=\varnothing$, so the pushout after taking limits is $1{\bigsqcup }_{\varnothing }1$, a two-element set. Thus cofiltered limits do not in general preserve pushouts.