Final and strongly final objects in Higher Topos Theory

Question

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:

1. 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)

2. 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?

Answer 1

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.)

1. 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.

2. 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 $F \colon \Delta^{n+1} \to \mathcal C$. $\square$