Lurie introduced in subchapter 1.2.12 of his Higher Topos Theory
the notion of final and strongly final objects:
Definition 1.2.12.1. let $\mathcal{C}$ be a topological category (e.g. simplicial cats,
simplicial set).
An object $X \in \mathcal{C}$ is final if for each $Y \in \mathcal{C}$,
the mapping space $\text{Map}_{h\mathcal{C}}(Y,X)$
regarded itself as object in associated homotopy category $h\mathcal{C}$
is weakly contractible, that is a final object (in ordinary category sense) in $h\mathcal{C}$.
Definition 1.2.12.3. Let $\mathcal{C}$ be now a simplicial set.
An object ("vertex") $X$
of $\mathcal{C}$ is strongly
final if the projection $p: \mathcal{C}/X \to \mathcal{C} $ is a
trivial fibration of simplicial sets.
The used slice $\mathcal{C}/X$ for a vertex $X: \Delta^0 \to C$ is a simplicial set
whose $n$-simplices are $f \in \text{Hom}_X(\Delta^n \ast \Delta^0,C)$
where the subscript $X$ says that these are subjected to condition
$f \vert _{\Delta^0} =X$ (Proposition 1.2.9.2)
Two questions about some properties of these definitions:
-
Why is object $X \in \mathcal{C}$ final if there is a retraction
of ($h\mathcal{C}$-enriched)
homotopy categories from $h\mathcal{C} \ast [0]$ to
$h\mathcal{C}$ carrying the unique object of $[0]$ to $X$. In other words
why the existence of such retraction implies that
for each $Y \in \mathcal{C}$,
the mapping space $\text{Map}_{h\mathcal{C}}(Y,X)$ is
weakly contractible in above sense. (this statement
in used in the proof of Corollary 1.2.12.5) -
After Definition 1.2.12.3. of strongly
final vertex $X$ is remarked that it's equivalent to that vertex
$X \in \mathcal{C}$ is strongly final if and only if any map
$f_0: \partial \Delta^n \to \mathcal{C}$ such that $f_0(n) =X$
can be lifted to a map $f: \Delta^n \to \mathcal{C}$.
Why that' true? Applying Definition 1.2.12.3 above $X$ is strongly final if the natural
projection $p: \mathcal{C}/X \to \mathcal{C}$ is a trivial fibration of simplicial sets.
In other words if $p: \mathcal{C}/X \to \mathcal{C}$ has the
lifting property with respect to every inclusion
$\partial \Delta^n \subset \Delta^n $. This means that
the condition should be read as that
a map $f_0: \partial \Delta^n \to \mathcal{C}/X$, such that
$p \circ f_0: \partial \Delta^n \to \mathcal{C}$ extends to
$\overline{f}: \Delta^n \to \mathcal{C}$, extends to
a $f: \Delta^n \to \mathcal{C}/X$ with $p \circ f=\overline{f}$.
Equivalently using adjuncion from defining property of the slice
$\mathcal{C}/X$ we have
$$ \text{Hom}_{sSet}(S,\mathcal{C}/X) =
\text{Hom}_X(S \ast \Delta^0,\mathcal{C}) $$
for any simplicial set $S$, the lifting property can be reformulated
in terms of lifting a
$f_0: \partial \Delta^n \ast \Delta^0 \to \mathcal{C}$
with $f_0 \vert _{\Delta^0} =X$ to a
$f: \Delta^n \ast \Delta^0 \to \mathcal{C}$. But this lifting
property is seemingly also not the same as the remark
after Definition 1.2.12.3 in the book I exposed in point 2. Or is it in some implicit sense?
Best Answer
It's good to ask this kind of questions on a critical reading! (They are also great exercises in unwinding the definitions to learn to work with simplicial sets, although in the first case I couldn't quite work it out.)
I actually think that this argument is not complete: if $\mathcal C$ is a simplicial set such that the inclusion $\mathcal C \to \mathcal C^{\triangleright}$ has a retraction $f \colon \mathcal C^{\triangleright} \to \mathcal C$, then $f([0])$ need not be final in $\mathcal C$. This already happens when $\mathcal C$ is the nerve of a $1$-category $\mathscr C$: the construction $\mathscr C^{\triangleright}$ (adjoin a new final object to $\mathscr C$) commutes with taking the nerve (see §1.2.8), and if $\mathscr C$ already contained a zero object $*$, then we may define $F \colon \mathscr C^{\triangleright} \to \mathscr C$ by taking $[0]$ to any fixed object $Y$ and taking the unique map $X \to [0]$ for $X \in \mathscr C^{\triangleright}$ to the zero map $F(X) \to * \to Y$ (where $F(X) = X$ if $X \in \mathscr C$ and $F(X) = Y$ if $X = [0]$).
But this is easily fixed. Morally what is going on in this argument is that the trivial Kan fibration $\pi \colon \mathcal C_{/X} \to \mathcal C$ is a categorical equivalence: this follows from the existence of the Joyal model structure (Thm. 2.2.5.1) since it has the same class of cofibrations as the Kan–Quillen model structure. Since $\pi$ takes the final object $\{X \to X\}$ of $\mathcal C_{/X}$ to $X$ in $\mathcal C$, we conclude that $X$ is final.
This argument relies on the existence (and explicit description) of the Joyal model structure, so instead Lurie tries to give a direct proof. As I said, I think that the proof is incomplete, and I don't immediately see if there is a low-tech argument that completes this proof.
Recall that $\Delta^n \star \Delta^0$ is isomorphic to $\Delta^{n+1}$, where $\Delta^n \star \varnothing \subseteq \Delta^n \star \Delta^0$ corresponds to $\Delta^{\{0,\ldots,n\}} \subseteq \Delta^{n+1}$ and $\varnothing \star \Delta^0 \subseteq \Delta^n \star \Delta^0$ to $\Delta^{\{n+1\}} \subseteq \Delta^{n+1}$. Under this identification, $(\partial \Delta^n) \star \Delta^0 \subseteq \Delta^n \star \Delta^0$ is the outer horn $\Lambda_{n+1}^{n+1} \subseteq \Delta^{n+1}$, and the forgetful functor $\mathcal C_{/X} \to \mathcal C$ is given by restricting $f \colon \Delta^{n+1} \to \mathscr C$ to $\Delta^{\{0,\ldots,n\}}$.
In this language, the condition for trivial fibration that you spell out becomes the following: given $f_0 \colon \Lambda_{n+1}^{n+1} \to \mathcal C$ with $f_0(n+1) = X$ such that the restriction of $f_0$ to $\partial \Delta^{\{0,\ldots,n\}}$ extends to a map $\bar f \colon \Delta^{\{0,\ldots,n\}} \to \mathcal C$, there exists $f \colon \Delta^{n+1} \to \mathcal C$ with $f|_{\Delta^{\{0,\ldots,n\}}} = \bar f$ and $f|_{\Lambda_{n+1}^{n+1}} = f_0$. But $$\partial \Delta^{n+1} = \Lambda_{n+1}^{n+1} \underset{\partial \Delta^{\{0,\ldots,n\}}}\amalg \Delta^{\{0,\ldots,n\}},$$ so this lifting criterion is exactly the condition that any map $F_0 \colon \partial \Delta^{n+1} \to \mathcal C$ with $F_0(n) = X$ extends to a map $F \colon \Delta^{n+1} \to \mathcal C$. $\square$