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1. • CommentRowNumber1.
   • CommentAuthorTodd_Trimble
   • CommentTimeOct 8th 2011
   • (edited Oct 8th 2011)
   • PermaLink
   Author: Todd_Trimble
   Format: MarkdownItexA <a href="http://mathoverflow.net/questions/77498/the-binary-product-preserves-pushouts/77501#77501">question</a> appeared on Math Overflow which I answered, but I'm not completely satisfied with my answer, and I wondered whether any of you had seen this before. In some generality, the question is this. Let $E$ be a cocomplete pretopos (say), and let $F$ be a finitary polynomial endofunctor on $E$ without constant term. Then, if I'm not mistaken, the canonical arrow $\sum_i F(A_i) \to F(\sum_i A_i)$ is monic. Is something like this true for wide pushouts in place of coproducts, i.e., is the canonical arrow $(\sum_i)_{F P} F(A_i) \to F((\sum_i)_P A_i)$ monic, for any cone from $P$ to the collection of the $A_i$? I'm pretty sure that it's going to be true, but I don't think I'm thinking all that clearly this morning. This seems like a handy little lemma to have archived at the Lab...

   A question appeared on Math Overflow which I answered, but I’m not completely satisfied with my answer, and I wondered whether any of you had seen this before.

   In some generality, the question is this. Let $E$ be a cocomplete pretopos (say), and let $F$ be a finitary polynomial endofunctor on $E$ without constant term. Then, if I’m not mistaken, the canonical arrow $\sum_i F(A_i) \to F(\sum_i A_i)$ is monic. Is something like this true for wide pushouts in place of coproducts, i.e., is the canonical arrow $(\sum_i)_{F P} F(A_i) \to F((\sum_i)_P A_i)$ monic, for any cone from $P$ to the collection of the $A_i$? I’m pretty sure that it’s going to be true, but I don’t think I’m thinking all that clearly this morning.

   This seems like a handy little lemma to have archived at the Lab…

2. • CommentRowNumber2.
   • CommentAuthorMike Shulman
   • CommentTimeOct 8th 2011
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   Author: Mike Shulman
   Format: MarkdownItexIt looks to me as though your proof at MO should work in the generality of a cocomplete pretopos, if you do the zigzag thing appropriately internally. Offhand, I don't recall seeing this anywhere before.

   It looks to me as though your proof at MO should work in the generality of a cocomplete pretopos, if you do the zigzag thing appropriately internally. Offhand, I don’t recall seeing this anywhere before.

3. • CommentRowNumber3.
   • CommentAuthorTodd_Trimble
   • CommentTimeOct 9th 2011
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   Author: Todd_Trimble
   Format: MarkdownItexThanks. I think I was worried over nothing. In any case, I did a little [write-up](http://ncatlab.org/toddtrimble/published/Lemma+on+wide+pushouts) on my web which hopefully does the job.

   Thanks. I think I was worried over nothing. In any case, I did a little write-up on my web which hopefully does the job.

4. • CommentRowNumber4.
   • CommentAuthorMike Shulman
   • CommentTimeOct 9th 2011
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexNice! I was wondering last night whether the category has to be an infinitary-pretopos, or whether a cocomplete pretopos suffices -- and also whether every cocomplete pretopos is an infinitary-pretopos, i.e. whether arbitrary coproducts in a pretopos are necessarily stable and disjoint. Is there a nice argument for why finite power functors preserve filtered colimits? Perhaps that is where we have to assume something (like extensivity) about the colimits being constructed in a "set-like way"?

   Nice! I was wondering last night whether the category has to be an infinitary-pretopos, or whether a cocomplete pretopos suffices – and also whether every cocomplete pretopos is an infinitary-pretopos, i.e. whether arbitrary coproducts in a pretopos are necessarily stable and disjoint.

   Is there a nice argument for why finite power functors preserve filtered colimits? Perhaps that is where we have to assume something (like extensivity) about the colimits being constructed in a “set-like way”?

5. • CommentRowNumber5.
   • CommentAuthorTodd_Trimble
   • CommentTimeOct 10th 2011
   • PermaLink
   Author: Todd_Trimble
   Format: MarkdownItexSo I was construing a finite power functor as $$C \stackrel{\Delta}{\to} C^n \stackrel{prod}{\to} C$$ where $\Delta$ preserves filtered colimits because it is left adjoint to $prod$, and $prod$ preserves filtered colimits because of the interchange between finite limits and filtered co... ohhh, dang, you're right, I misapplied the famous theorem. Hm! Back to the old drawing board... I'd love to think about the other questions, too, but today it feels like I have a big mess of confusions of various sorts, all crying for attention. :-)

   So I was construing a finite power functor as

   $$C \stackrel{\Delta}{\to} C^n \stackrel{prod}{\to} C$$

   where $\Delta$ preserves filtered colimits because it is left adjoint to $prod$, and $prod$ preserves filtered colimits because of the interchange between finite limits and filtered co… ohhh, dang, you’re right, I misapplied the famous theorem. Hm! Back to the old drawing board…

   I’d love to think about the other questions, too, but today it feels like I have a big mess of confusions of various sorts, all crying for attention. :-)

6. • CommentRowNumber6.
   • CommentAuthorTodd_Trimble
   • CommentTimeOct 10th 2011
   • (edited Oct 10th 2011)
   • PermaLink
   Author: Todd_Trimble
   Format: MarkdownItexOkay, let me try again. If $C$ is an $\infty$-pretopos, then $c \times -: C \to C$ preserves arbitrary colimits. So now let $J$ be a filtered category, and let $F, G$ be functors $J \to C$. Then $$(colim_{j: J} F(j)) \times (colim_{k: J} G(k)) \cong colim_{k: J} (colim_{j: J} F(j)) \times G(k) \cong colim_{j: J} colim_{k: J} (F(j) \times G(k)) \cong colim_{(j, k): J \times J} F(j) \times G(k) \cong colim_{j: J} F(j) \times G(j)$$ where the last isomorphism comes about because $\Delta: J \to J \times J$ is final. So $prod: C \times C \to C$ preserves filtered colimits. Does that work? (Maybe I should say infinitary pretopos instead of $\infty$-pretopos?)

   Okay, let me try again. If $C$ is an $\infty$-pretopos, then $c \times -: C \to C$ preserves arbitrary colimits. So now let $J$ be a filtered category, and let $F, G$ be functors $J \to C$. Then

   $$(colim_{j: J} F(j)) \times (colim_{k: J} G(k)) \cong colim_{k: J} (colim_{j: J} F(j)) \times G(k) \cong colim_{j: J} colim_{k: J} (F(j) \times G(k)) \cong colim_{(j, k): J \times J} F(j) \times G(k) \cong colim_{j: J} F(j) \times G(j)$$

   where the last isomorphism comes about because $\Delta: J \to J \times J$ is final. So $prod: C \times C \to C$ preserves filtered colimits. Does that work?

   (Maybe I should say infinitary pretopos instead of $\infty$-pretopos?)

7. • CommentRowNumber7.
   • CommentAuthorMike Shulman
   • CommentTimeOct 10th 2011
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexAh, that looks good! So that actually says that any functor of two variables which preserves filtered colimits in each variable separately preserves filtered colimits in both variables together. I knew that for reflexive coequalizers, but I don't recall seeing it for filtered colimits before, although I could easily be misremembering. ("Infinitary pretopos" is an nLabism, intended to express the same thing as the more common "$\infty$-pretopos", but also avoid a terminological collision with "$\infty$-topos", where the $\infty$ denotes the category-level rather than the cardinality of extensivity. You can use it or not, as you like. (-: )

   Ah, that looks good! So that actually says that any functor of two variables which preserves filtered colimits in each variable separately preserves filtered colimits in both variables together. I knew that for reflexive coequalizers, but I don’t recall seeing it for filtered colimits before, although I could easily be misremembering.

   (“Infinitary pretopos” is an nLabism, intended to express the same thing as the more common “$\infty$-pretopos”, but also avoid a terminological collision with “$\infty$-topos”, where the $\infty$ denotes the category-level rather than the cardinality of extensivity. You can use it or not, as you like. (-: )