equivalences in/of $(\infty,1)$-categories
Given a functor $p\colon X \to Y$ between categories one may ask for each morphism $f\colon y_1 \to y_2$ if given a lift of its target
there is a universal lift of $f$
There may also be other lifts of $f$, but the universal one is essentially unique, as usual for anything having a universal property. Specifically, $\hat f$ in $X$ is essentially uniquely determined by its target $\hat y_2$ and its image $f = p(\hat f)$ in $Y$, and is called a cartesian morphism. A morphism which is cartesian relative to $p^{op}\colon X^{op}\to Y^{op}$ is called opcartesian or cocartesian.
If there are enough cartesian morphisms in $Y$, they may be used to define functors
between the fibers of $p$ over $y_1$ and $y_2$.
This way a functor $p : X \to Y$ with enough Cartesian morphisms – called a Cartesian fibration or Grothendieck fibration – determines and is determined by a fiber-assigning functor $Y \to Cat^{op}$.
This has its analog in higher categories.
(cartesian morphism)
Let $p : X \to Y$ be a functor. A morphism $f : x_1 \to x_2$ in the category $X$ is strongly cartesian with respect to $p$ (nowdays often just cartesian), or (strongly) $p$-cartesian if for every $x'\in X$, for every $h:x'\to x_2$ and every $u:p(x')\to p(x_1)$ such that $p(h) = p(f) u$, there exists a unique $v:x'\to x_1$ such that $h = f v$ and $u = p(v)$:
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.
There is a weaker universal property, originally devised by Grothendieck and Gabriel, where one requires above lifting property only for $u = id_{p(x_1)}$, and traditionally also called simply cartesian, or rarely weak cartesian. In Grothendieck’s fibered categories (see below), cartesian in the strong sense and cartesian in the weak sense are equivalent properties of morphisms.
If we pass to the nerve $N(X)$ and $N(Y)$ of the categories, then in terms of diagrams in sSet this means that the morphism $f : x \to y$ is $p$-cartesian precisely if for all horn inclusions
such that the last edge of the 2-horn is the given edge $f$, a unique lift $\sigma$
exists.
(Grothendieck fibration)
If for every morphism in $Y$ and every lift of its target there is at least one lift which has as its target the chosen one and is a $p$-cartesian morphism in the strong sense, one says that $p$ is a fibered category (also called Grothendieck fibration). Equivalently,
and
We discuss equivalent reformulations of the above definition of Cartesian morphism that lend themselves better to generalization to higher category theory.
For the following, we need this notation: let
$X/x_2$ by the overcategory of $X$ over the object $x_2$;
$Y/p(x_2)$ the corresponding overcategory of $Y$ over $p(x_2)$;
$X/f$ the category whose objects
are objects $a$ of $X$ eqipped with morphisms to $x_1$ and $x_2$ such that the obvious triangle commutes, and whose morphisms are morphisms between these tip objects such that all diagrams in sight commute.
similarly for $Y/p(f)$.
The condition that $f \in Mor X$ is a Cartesian morphism with respect to $p : X \to Y$ is equivalent to the condition that the functor
into the (strict) pullback of the obvious projection $X/{x_2} \to Y/p(x_2)$ along the projection $Y/p(f) \to Y/p(x_2)$ induced by the commutativity of
is a surjective equivalence, and this in turn is equivalent to it being an isomorphism of categories.
It is immediate to see that $\phi$ being an isomorphism of categories is equivalent to the condition that $f$ is a Cartesian morphism. We discuss that just the condition that $\phi$ is a surjective equivalence already implies that it is an isomorphism of categories.
So assume now that $\phi$ is a surjective equivalence.
Notice that objects in the pullback category are compatible pairs
We have that $\phi$ being surjective on object means that every such pair is in the image of some object
and hence that every filler exists . Assume two such fillers $g$ and $g'$. Then by the fact that an equivalence of categories is a surjection (even an isomorphism) on corresponding hom-sets, it follows that there exists (even uniquely) a morphism in $X/f$ connecting them
such that this maps under $\phi$ to the identity morphism in the pullback category. But in particular this maps to the morphism
in $X/{x_2}$ and evidently is the identity there if and only if $h$ is the identity. Hence this maps also to the identity in the pullback category if and only if $h$ is the identity. So $h$ must be the identity. So if two lifts of an object through the surjective equivalence $\phi$ exist, they must already be equal. Hence the surjective equivalence $\phi$ is even an isomorphism on objects and hence an isomorphism of categories.
The notion of cartesian morphism generalizes from category theory to (∞,1)-category theory. We describe it for two different incarnations of the notion of (∞,1)-category: quasi-categories and sSet categories.
We formulate a notion cartesian edge or cartesian morphism in a simplicial set $X$ relative to a morphism $p : X \to Y$ of simplicial sets. In the case that these simplicial sets are quasi-categories – i.e. simplicial set incarnations of (∞,1)-categories – this yields a notion of cartesian morphisms in $(\infty,1)$-categories.
Let $p : X \to Y$ be a morphism of simplicial sets. Let $f : x_1 \to x_2$ be an edge in $X$, i.e. a morphism $f : \Delta^1 \to X$.
Recall the notion of over quasi-category obtained from the notion of join of quasi-categories. Using this we obtain simplicial sets $X/f$, $X/{x_2}$, $S/p(f)$ and $S/p(x_2)$ in generalization of the categories considered in the above definition of cartesian morphisms in categories.
(cartesian edge in a simplicial set)
Let $p : X \to Y$ be an inner Kan fibration of simplicial sets.
Then a morphism $f : x \to y$ in $X$ is $p$-cartesian if the induced morphism
into the pullback in sSet is an acyclic Kan fibration.
This is HTT, def 2.4.1.1.
The morphism $f : x \to y$ as above, for $p : X \to Y$ an inner fibration, is $p$-cartesian precisely if for all $n \geq 2$ and all right outer horn inclusions
(with $\Lambda[n]_n$ the $n$th horn of the $n$-simplex) such that the last edge of the horn is the given edge $f$, a lift $\sigma$
exists.
This is HTT remark 2.4.1.4.
This means that an inner fibration $p : X \to Y$ with a collection of $p$-cartesian morphisms in $X$ specified satisfies the same kind of condition as a right fibration , the only difference being that not all right outer horns inclusion are required to have lifts, but only those where the last edge of the horn maps to a cartesian morphism.
In this sense a Cartesian fibration is a generalization of a right fibration.
If $p : X \to Y$ is an inner fibration of quasi-categories then a morphism $f : x \to y$ in $X$ is $p$-Cartesian precisely if for all objects $a$ in $X$ the diagram
of hom-objects in a quasi-category is a homotopy pullback square (in sSet equipped with its standard model structure).
This is HTT, prop. 2.4.4.3.
Let $C$ and $D$ be simplicially enriched categories and $F : C \to D$ a sSet-enriched functor.
A morphism $f : x \to y \in C$ is $F$-cartesian if it is so under the homotopy coherent nerve $N : sSet Cat \to sSet$ in the sense of quasi-categories above, i.e. if
is an acyclic Kan fibration.
If $C$ and $D$ are enriched in Kan complexes and if $F$ is hom-wise a Kan fibration, then
$N(F) : N(C) \to N(D)$ is an inner fibration;
a morphism $f :x \to y$ in $N(C)$ is an $N(F)$-cartesian morphism precisely if for all objects $a$ in $C$ the diagram
is a homotopy pullback square in sSet equipped with its standard model structure.
This is HTT, prop. 2.4.1.10.
For $p : \mathcal{E} \to \mathcal{C}$ a functor, if in a diagram
in $\mathcal{E}$ the two vertical morphisms are vertical with respect to $p$ (meaning that $p(g) = Id_p(A)$ and $p(h) = Id(B)$) and if the two horizontal morphisms are $p$-Cartesian morphisms, then this square is a pullback square.
If
is another cone over $C \to D \leftarrow B$, then its image under $p$ is
Since $p(f) = p(k)$, another lift of the right horn of this is given by
which gives a unique filler $Q \to A$ by the fact that $f$ is Cartesian.
But this produces now two fillers – namely the original $Q \to C$ and the $Q \to A \to C$ just obtained – of the horn
over
Since $k$ is Cartesian, these two fillers must be equal. This means that the morphism $Q \to A$ is a cone morphism and unique as such. Hence the original square is a pullback.
This appears as Elephant, lemma 1.3.3.
For $C$ a category, a morphism in $C$ is cartesian with respect to the terminal functor $C \to *$ precisely if it is an isomorphism.
In particular all identity morphisms are cartesian.
This is trivial to see. The analog statement holds also for quasi-categories, where it is rather more nontrivial and quite useful:
For $C$ a quasi-category, a morphism in $C$ is cartesian with respect to the terminal morphism $C \to *$ precisely if it is an equivalence.
More generally, for $p : X \to Y$ an inner fibration, a morphism $f$ in $X$ is an equivalence precisely if it is $p$-cartesian and $p(f)$ is an equivalence in $Y$.
The first statement is a proposition of Andre Joyal, slightly reformulated in the language of cartesian morphisms. It appears as HTT, prop 1.2.4.3. A proof appears below HTT, corollary 2.1.2.2.
The second statement is HTT, prop. 2.4.1.5.
David Roberts: There would surely be an anafunctor version of this, that would require no choices whatsoever. It is unlikely that I would be able to find time to write this up, so my plea goes out to those in the know…
I imagine that there would then be an $(\infty,1)$-version using whatever passes as anafunctors in that setting (dratted memory, failing at the first gate)
Mike Shulman: Yes, there would surely be such a version. (-: The simplest way would be to take the specifications $|f^*|$ for the anafunctor $f^*$ to be the cartesian morphisms over $f$, with domain and codomain giving the functions $\sigma$ and $\tau$. Unique factorization would give you the values of morphisms.
David Roberts: just stumbled on this old comment - I’m reading Makkai more closely, and I’m convinced that basically anything defined by a universal property is given by a saturated anafunctor. So this is a heads up for posterity, that a map is a fibration iff the fairly obvious span of functors defines a saturated anafunctor?.
The traditional reference is SGA I.6 (written by P. Gabriel and A. Grothendieck)
There are excellent lectures of Vistoli:
For the 1-categorical case see for instance section B1.3 of
The $(\infty,1)$-categorical version is in section 2.4 of
See also the references at Grothendieck fibration.