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1. • CommentRowNumber1.
   • CommentAuthorUrs
   • CommentTimeMay 20th 2010
   • PermaLink
   Author: Urs
   Format: MarkdownItexcreated [[exact (infinity,1)-functor]]

   created exact (infinity,1)-functor

2. • CommentRowNumber2.
   • CommentAuthorMike Shulman
   • CommentTimeApr 4th 2017
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   Author: Mike Shulman
   Format: MarkdownItexShould not this notion rather be called "flat $(\infty,1)$-functor"? In 1-category theory it's pretty well established that "left exact" only applies in case the categories in question actually have finite limits.

   Should not this notion rather be called “flat $(\infty,1)$-functor”? In 1-category theory it’s pretty well established that “left exact” only applies in case the categories in question actually have finite limits.

3. • CommentRowNumber3.
   • CommentAuthorUrs
   • CommentTimeApr 4th 2017
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   Author: Urs
   Format: MarkdownItexSure. Since #1 is from 7 years ago, this might have been before we had the corresponding refinement of the entries for the 1-category case.

   Sure. Since #1 is from 7 years ago, this might have been before we had the corresponding refinement of the entries for the 1-category case.

4. • CommentRowNumber4.
   • CommentAuthorMike Shulman
   • CommentTimeApr 4th 2017
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   Author: Mike Shulman
   Format: MarkdownItexOk, done. I also added a remark about "internal flatness". (Where) is it proven that an $(\infty,1)$-functor is flat in Lurie's sense $F:C\to D$ iff $Lan_F : \mathcal{P} C \to \mathcal{P} D$ is left exact?

   Ok, done. I also added a remark about “internal flatness”. (Where) is it proven that an $(\infty,1)$-functor is flat in Lurie’s sense $F:C\to D$ iff $Lan_F : \mathcal{P} C \to \mathcal{P} D$ is left exact?