Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $p: C\to D$ be a functor, and let $f:y\to x$ be a morphism of $C$. We say that $f$ is cartesian if the canonical map $Q:(C\downarrow f) \to P:=(C\downarrow x)\times_{(D\downarrow p(x)} (D\downarrow p(f))$ is a surjective (on objects) equivalence of categories. However, if we write out what the (strict 2-) pullback means, the objects are precisely the pairs of morphisms $g: z\to x$ and $h:p(z)\to p(y)$ such that $p(g)=p(f) \circ h$. If we look at the fibres of $Q$ over objects of $A$, we see that that they are contractible groupoids.

Using the more common definition of a cartesian morphism, we must show that any pair of morphisms $(g, h)$ as above uniquely determines an arrow $\ell:z\to y$ such that $f\circ \ell= g$ and $p(\ell)=h$.

I see how the first definition implies the existence of such a map, but how does it determine the map's uniqueness (up to more than a contractible space of choices)?

share|improve this question
add comment

1 Answer

up vote 2 down vote accepted

The only morphisms in the fibers of $Q$ are identity maps, so it is actually an isomorphism of categories. To see this, suppose $\ell,\ell'\colon z\to y$ both induce $g\colon z\to y$. What would a morphism from $\ell$ to $\ell'$ in the fiber of $Q$ be? It would be a morphism $\varphi\colon z\to z$ over $y$ (the first $z$ is over $y$ via $\ell$ and the second via $\ell'$) which induces the identity morphism on $g$ in $(C\downarrow x)$. But $\varphi$ induces the morphism

    φ
z ----> z
 \     /
 g\   /g
   v v
    x

The only way this is the identity morphism of $g$ is if $\varphi=id_z$.

share|improve this answer
    
That's what I thought as well, but the nLab says that it only needs to be a surjective equivalence (a trivial fibration in the natural model structure). ncatlab.org/nlab/show/Cartesian+morphism . In fact, I asked a similar question earlier on the nForum where I raised the same issue math.ntnu.no/~stacey/Vanilla/nForum/… . I agree with you, but maybe someone can explain why the definition on the nForum is so strange? Does it give a good definition for weakly cartesian morphisms? –  Harry Gindi Aug 29 '10 at 8:38
    
I think, as you say, the motivation for the strange phrasing is to weaken well. As Anton shows, the two versions are equivalent in this setting — “surjective equivalence” implies “isomorphism” — so just looking at this case, it'd seem more natural to use the simpler phrasing of “isomorphism”. But if in higher-dimensional/weakened generalisations the general concept needed is “trivial fibration”, that would be an argument for using it here, slightly harder to understand but more principled. –  Peter LeFanu Lumsdaine Aug 29 '10 at 17:58
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.