Consider the category $\mathbb{N}_{\geq}$ whose objects lie in $\mathbb{N}$, and there is a single morphism $n \rightarrow m$ if $n \leq m$, and otherwise there is no morphism between $n$ and $m$.
A category $C$ is called cofiltered if:
It is non-empty.
For every $i, j \in Ob(C)$ there is $k \in Ob(C)$ with morphisms $k \rightarrow i$, $k \rightarrow j$.
For every $i, j \in Ob(C)$ and pair of morphisms $f_1, f_2: i \rightarrow j$, there is $k \in Ob(C)$ and a morphism $g: k \rightarrow i$ such that $f_1 \circ g = f_2 \circ g$.
Is $\mathbb{N}_{\geq}$ cofiltered? I believe it is, since it is clearly non-empty, and for any $i$, $j \in Ob(C)$ we can just take the lesser of the two which automatically gives unique morphisms to each. For the third requirement we can just do the same again and take the identity morphism on the lesser of $i$ and $j$ since there is only one morphism between each.
A category $C$ is filtered if $C^{\text{op}}$ is cofiltered. I believe $C$ is also filtered by similar reasoning. Can anyone confirm or contradict my claims? My main motivation for asking is that my textbook uses $\mathbb{N}_{\geq}$ as an example of a filtered category and $\mathbb{N}_{\geq}^{\text{op}}$ as an example of a cofiltered category, but I believe both fill both requirements. Is this just a badly written example or am I making a mistake?
Thanks in advance.
You're right, this poset is both filtered and cofiltered. It's a rather degenerate example of a cofiltered poset, though, as it has a least element, or in categorical language an initial object. We are often interested in filtered colimits and cofiltered limits, but the limit of a diagram indexed by a category admitting an initial object is given trivially, by evaluation at that object.