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1. • CommentRowNumber1.
   • CommentAuthorHarry Gindi
   • CommentTimeJun 3rd 2010
   • PermaLink
   Author: Harry Gindi
   Format: MarkdownItexLet $F,G\in [C,D]$ be two functors commuting with small colimits, and let $\alpha: F \to G$ be a natural transformation between them. Then according to Cisinski's book [Astérisque 308](http://www-math.univ-paris13.fr/~cisinski/ast.pdf) Corollaire 1.1.8, the natural transformation $\alpha$ induces a functor $H: D\to Arr(D)$ that commutes with small colimits as well. That's it. That's the only information he gives about $H$. I assume that it's trivial because he doesn't go into more detail, but I have no idea what it should be. If I had to guess, and this is _very_ speculative, I would say that for $X$ in $D$, we can look at the diagram induced by $\alpha$, that is, the diagram defined by $\{\alpha_Z:X\to G(Z) | Z\in F^{-1}(X)\}$ and take $$H(X):=X\to colim_{Z\in F^{-1}(X)}G(Z),$$ the induced map to the colimit.

   Let $F,G\in [C,D]$ be two functors commuting with small colimits, and let $\alpha: F \to G$ be a natural transformation between them.

   Then according to Cisinski’s book Astérisque 308 Corollaire 1.1.8, the natural transformation $\alpha$ induces a functor $H: D\to Arr(D)$ that commutes with small colimits as well. That’s it. That’s the only information he gives about $H$. I assume that it’s trivial because he doesn’t go into more detail, but I have no idea what it should be.

   If I had to guess, and this is very speculative, I would say that for $X$ in $D$, we can look at the diagram induced by $\alpha$, that is, the diagram defined by $\{\alpha_Z:X\to G(Z) | Z\in F^{-1}(X)\}$ and take

   $$H(X):=X\to colim_{Z\in F^{-1}(X)}G(Z),$$

   the induced map to the colimit.

2. • CommentRowNumber2.
   • CommentAuthorDavidRoberts
   • CommentTimeJun 3rd 2010
   • PermaLink
   Author: DavidRoberts
   Format: MarkdownItexThe natural transformation induces a functor $C' \to Arr(C)$ (switching to his notation, C replaced by C' and D replaced by C) - this is standard and trivial. I wonder if it is a typo?... It is a typo, because by the following paragraph, he refers to $F' = H^{-1}(\tilde{F})$, using the notation of the corollary this is a class of arrows in $C'$, hence $H$ is a functor from $C'$ to $Arr(C)$. The only potentially tricky bit is showing that $H$ commutes with inductive limits. I leave it as an exercise.

   The natural transformation induces a functor $C' \to Arr(C)$ (switching to his notation, C replaced by C’ and D replaced by C) - this is standard and trivial. I wonder if it is a typo?… It is a typo, because by the following paragraph, he refers to $F' = H^{-1}(\tilde{F})$, using the notation of the corollary this is a class of arrows in $C'$, hence $H$ is a functor from $C'$ to $Arr(C)$. The only potentially tricky bit is showing that $H$ commutes with inductive limits. I leave it as an exercise.

3. • CommentRowNumber3.
   • CommentAuthorHarry Gindi
   • CommentTimeJun 3rd 2010
   • PermaLink
   Author: Harry Gindi
   Format: MarkdownItexAh, so it was a typo!

   Ah, so it was a typo!

4. • CommentRowNumber4.
   • CommentAuthorHarry Gindi
   • CommentTimeJun 3rd 2010
   • PermaLink
   Author: Harry Gindi
   Format: MarkdownItexHmm... Then in Proposition 1.1.7, how is $f\mapsto \tilde{f}$ an endofunctor of $Arr(C)$? It's sending arrows in $Arr(C)$ to objects in $Arr(C)$. Rather, it seems like a functor $Arr(Arr(C))\to Arr(C)$, which is not an endofunctor. Is this also a mistake, or am I missing the point...

   Hmm… Then in Proposition 1.1.7, how is $f\mapsto \tilde{f}$ an endofunctor of $Arr(C)$? It’s sending arrows in $Arr(C)$ to objects in $Arr(C)$. Rather, it seems like a functor $Arr(Arr(C))\to Arr(C)$, which is not an endofunctor. Is this also a mistake, or am I missing the point…

5. • CommentRowNumber5.
   • CommentAuthorTodd_Trimble
   • CommentTimeJun 3rd 2010
   • (edited Jun 3rd 2010)
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   Author: Todd_Trimble
   Format: MarkdownItex(Comment deleted.)

   (Comment deleted.)

6. • CommentRowNumber6.
   • CommentAuthorHarry Gindi
   • CommentTimeJun 3rd 2010
   • PermaLink
   Author: Harry Gindi
   Format: MarkdownItexAlright, I've concluded that calling it an endofunctor is incorrect. If it is a functor Arr(Arr(C))->Arr(C), this makes sense with the proof, since it turns a retract of squares into a retract of arrows, and it preserves identities blah blah.

   Alright, I’ve concluded that calling it an endofunctor is incorrect. If it is a functor Arr(Arr(C))->Arr(C), this makes sense with the proof, since it turns a retract of squares into a retract of arrows, and it preserves identities blah blah.

7. • CommentRowNumber7.
   • CommentAuthorDavidRoberts
   • CommentTimeJun 3rd 2010
   • PermaLink
   Author: DavidRoberts
   Format: MarkdownItexHmm... indeed!

   Hmm… indeed!

8. • CommentRowNumber8.
   • CommentAuthorHarry Gindi
   • CommentTimeJun 3rd 2010
   • (edited Jun 3rd 2010)
   • PermaLink
   Author: Harry Gindi
   Format: MarkdownItexAlright, I e-mailed Cisinski about the errors. I'm probably the first person who didn't skip through this section.

   Alright, I e-mailed Cisinski about the errors. I’m probably the first person who didn’t skip through this section.