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Sep 12, 2011 at 6:31 history edited DamienC CC BY-SA 3.0
typo corrected according to GIorgio Mossa's last comment; added 90 characters in body
Sep 12, 2011 at 6:30 comment added DamienC @Giorgio Mossa: yes you suppose well, and you're right about the typo-error in the answer. I'll fix it.
Sep 11, 2011 at 21:04 comment added Giorgio Mossa @DamienC I suppose you meant $Hom_{t\leq0(Cat)}(\mathcal C \times \Delta,\mathcal D)$ in your last comment, but I think I get it. Thanks a lot. Just to be completely correct I think that there are some typo-error in your answer, you say that the natural-transformation $\phi$ should be such that $\phi(-,0)=F$ e $\phi(-,0)=G$, did you meant $\phi(-,1)=G$, right?
Sep 11, 2011 at 19:55 comment added DamienC @Giorgio Mossa: to be precise I should have said that $Hom_{t\leq0}(Cat)}(\mathcal C,\mathcal D)$ is the set of morphisms of a category with objects being functors $\mathcal C\to\mathcal D$.
Sep 11, 2011 at 11:59 comment added Giorgio Mossa @DamienC But $Hom_{t≤0(Cat)}(\mathcal C\times \Delta^1,\mathcal D)$ should be the set of the functors between the categories $\mathcal C \times \Delta^1$ and $\mathcal D$, this should be (in our description) the set of natural transformations, this description seems lacking of the objects of the category, unless you're considering arrow-only categories.
Sep 11, 2011 at 11:32 history edited DamienC CC BY-SA 3.0
added 69 characters in body
Sep 11, 2011 at 11:30 comment added DamienC @Giorgio Mossa: $Hom_{t\leq0}(Cat)}(\mathcal C,\mathcal D)$ is a set but $Hom_{t\leq0}(Cat)}(\mathcal C\times(0\to 1),\mathcal D)$ is a category. About your second question you are right, I should have written that an $n$-functor $\mathcal C\times G^n\to\mathcal D$ is a natural transformation between two $n$-functors. I'm going to fix it.
Sep 11, 2011 at 10:06 comment added Giorgio Mossa @DamienC thanks a lot, this stuff is really cool, though there are some things that I don't get straight, for instance: above you say that $\hom_{t \leq 0 (Cat)}(\mathcal C,\mathcal D)$ should be a category but it seems to me it should be a set, at the same time you say that a functor $\mathcal C \times G^n \to \mathcal D$ should be a $n$-functor, but in this case if we consider $n=2$, then a $2$-functor should be a natural transformation in the sense of the definition in the question, where am I wrong?
Sep 9, 2011 at 22:15 history edited DamienC CC BY-SA 3.0
added 4 characters in body; added 1 characters in body
Sep 9, 2011 at 22:07 history answered DamienC CC BY-SA 3.0