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1. • CommentRowNumber1.
   • CommentAuthormmuddasani
   • CommentTimeNov 19th 2017
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   Author: mmuddasani
   Format: MarkdownItexI was trying to follow along the proof for the closure properties [here](https://ncatlab.org/nlab/show/weak+factorization+system#ClosureProperties). Part 5 of the proof (for closure under (co-)products) assumes that the arrow category $\mathcal{C}^{\Delta[1]}$ is a presheaf category whose (co-)limits are computed componentwise. Is this a valid assumption? IIUC, it is folklore (if not already documented in several textbooks) that the collection of evaluation functors out of a $\mathcal{C}$-valued presheaf category jointly reflect (co-)limits. Is is possible to come up with a counterexample category $\mathcal{C}$ and a coproduct diagram in $\mathcal{C}^{\Delta[1]}$ such that both evaluation functors do not jointly preserve the coproduct?

   I was trying to follow along the proof for the closure properties here. Part 5 of the proof (for closure under (co-)products) assumes that the arrow category $\mathcal{C}^{\Delta[1]}$ is a presheaf category whose (co-)limits are computed componentwise. Is this a valid assumption?

   IIUC, it is folklore (if not already documented in several textbooks) that the collection of evaluation functors out of a $\mathcal{C}$-valued presheaf category jointly reflect (co-)limits.

   Is is possible to come up with a counterexample category $\mathcal{C}$ and a coproduct diagram in $\mathcal{C}^{\Delta[1]}$ such that both evaluation functors do not jointly preserve the coproduct?

2. • CommentRowNumber2.
   • CommentAuthorMike Shulman
   • CommentTimeNov 19th 2017
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   Author: Mike Shulman
   Format: MarkdownItexYour question is whether the proof works even if $C$ is not assumed to have coproducts but $C^{\Delta[1]}$ happens "by accident" to have some coproduct that isn't componentwise? One thing to say is that as soon as $C$ has a terminal object, the two evaluation functors $C^{\Delta[1]}\to C$ are left adjoints, hence preserve all colimits. I think this is beside the point, though; generally whenever people talk about (co)limits in a functor category they mean componentwise ones even if they forget to say it. I don't think I've ever seen any use for "accidental" limits in a functor category that aren't componentwise, just like how all interesting Kan extensions are pointwise.

   Your question is whether the proof works even if $C$ is not assumed to have coproducts but $C^{\Delta[1]}$ happens “by accident” to have some coproduct that isn’t componentwise?

   One thing to say is that as soon as $C$ has a terminal object, the two evaluation functors $C^{\Delta[1]}\to C$ are left adjoints, hence preserve all colimits.

   I think this is beside the point, though; generally whenever people talk about (co)limits in a functor category they mean componentwise ones even if they forget to say it. I don’t think I’ve ever seen any use for “accidental” limits in a functor category that aren’t componentwise, just like how all interesting Kan extensions are pointwise.

3. • CommentRowNumber3.
   • CommentAuthormmuddasani
   • CommentTimeNov 21st 2017
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   Author: mmuddasani
   Format: MarkdownItexThanks for the helpful reminder regarding the evaluation functors being adjoints! If we call both evaluation functors $cod, dom$, then they participate in an adjoint triple $cod \dashv id \dashv dom : \mathcal{C}^{\Delta[1]} \to \mathcal{C}$. So this means $cod$ is always a left adjoint and so always preserves coproducts. And I think that's all we need in order to finish the proof, since the proposed lift remains a well-defined candidate. I think that showing that both of the triangles commute involves regarding each triangle as a square (i.e. arrow of the arrow category) with an identity for the newly introduced (co-)domain and with the left edge of each square being a coproduct. The only part I'm still taking time to verify the details is regarding whether the top arrow of the top triangle is what we expect it to be as we switch from the coproduct P.O.V. versus the sequence of arrows P.O.V. I'll try to add inline diagrams for the above paragraph when I can spend time to learn how to do so. Regarding (co-)limits usually being pointwise in functor categories, I'll acknowledge that note. I'm still curious about the counterexamples, but perhaps less so if it is generally accepted to usually work with pointwise (co-)limits.

   Thanks for the helpful reminder regarding the evaluation functors being adjoints! If we call both evaluation functors $cod, dom$, then they participate in an adjoint triple $cod \dashv id \dashv dom : \mathcal{C}^{\Delta[1]} \to \mathcal{C}$. So this means $cod$ is always a left adjoint and so always preserves coproducts. And I think that’s all we need in order to finish the proof, since the proposed lift remains a well-defined candidate.

   I think that showing that both of the triangles commute involves regarding each triangle as a square (i.e. arrow of the arrow category) with an identity for the newly introduced (co-)domain and with the left edge of each square being a coproduct. The only part I’m still taking time to verify the details is regarding whether the top arrow of the top triangle is what we expect it to be as we switch from the coproduct P.O.V. versus the sequence of arrows P.O.V.

   I’ll try to add inline diagrams for the above paragraph when I can spend time to learn how to do so.

   Regarding (co-)limits usually being pointwise in functor categories, I’ll acknowledge that note. I’m still curious about the counterexamples, but perhaps less so if it is generally accepted to usually work with pointwise (co-)limits.