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• CommentRowNumber1.
• CommentAuthorzskoda
• CommentTimeApr 21st 2011

This is an excerpt I wrote at logical functor:

As far as cartesian morphism there are two different universal properties in the literature, which are equivalent for Grothendieck fibered categories but not in general. In what Urs calls the “traditional definition” (but is in fact a later one) one has for every $x'$, for every $h$, for every $g$ such that … there exist a unique da da da. This way it is spelled in Vistoli’s lectures. This is in fact the strongly cartesian property, stronger than one in Gabriel-Grothendieck SGA I.6. The usual, Grothendieck, or weak property takes for $g$ the identity, and the unique lift is of the identity at $p(x_1)$. Then a Grothendieck fibered category is the one which has cartesian lifts for all morphisms below and all targets, and cartesian morphisms are closed under composition. With the strong cartesian property one does not need to require the closedness under composition. Now a theorem says that in a Grothendieck fibered category, a morphism is strongly cartesian iff it is cartesian.

Now I have made some changes to cartesian morphism, so that the entry is aware of the two variants of the universal property, which are not equivalent in general but are equivalent for Grothendieck fibered categories.

There was also a statement there

In words: for all commuting triangles in Y and all lifts through p of its 2-horn to X, there is a unique refinement to a lift of the entire commuting triangle.

which is too vague and I am not happy with, as it does not involve the essential parameter: the morphism for which we test cartesianess. I made a hack to it, and still it is not something I happy with (I like the idea of horn mentioned, however not the lack of appropriate quantifiers/conditions etc.). It is cumbersome to talk horn. (Maybe we could skip the whole statement in this imprecise form, and just mention please note the filling of the horn in $X$ with prescribed projection in $Y$ or alike). Here is the temporary hack:

In imprecise words: for all commuting triangles in $Y$ (involving $p(f)$ as above) and all lifts through $p$ of its 2-horn to $X$ (involving $f$ as above), there is a unique refinement to a lift of the entire commuting triangle.

• CommentRowNumber2.
• CommentAuthorzskoda
• CommentTimeApr 21st 2011

I have added SGA I.6 and Vistoli among principal references. SGA I.6 is very precise, one of my favorite texts in mathematics.

• CommentRowNumber3.
• CommentAuthorUrs
• CommentTimeApr 21st 2011

Thanks!