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Mar 26 at 11:44 vote accept Calin Tataru
Mar 26 at 11:44 comment added Calin Tataru This makes sense, thank you! Yeah the proof for $\mathbf{TOrd} \to \mathbf{Pos}$ uses the fact that every partial order has a total order extension.
Mar 26 at 2:54 comment added David Gao A technical point: I glossed over the possibility that $\cup_{j \in J} \mathrm{range}(\psi_j) = \varnothing$ in my proof. But since $\Delta$ does not contain the empty set, this could only happen when $F$ is the empty diagram. But $\Delta$ has no initial object, so in that case $F$ does not admit a colimit in $\Delta$ in the first place.
Mar 26 at 2:43 comment added David Gao I’d say the proof that $\mathbf{TOrd} \to \mathbf{Pos}$ preserves colimits is pretty similar, where one can use the completion of partial orders to total orders to show that $\cup_{j \in J} \mathrm{range}(\psi_j)$ must be totally ordered if a colimit in $\mathbf{TOrd}$ exists.
Mar 26 at 2:34 history answered David Gao CC BY-SA 4.0