# Kerodon

$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ $\newcommand\empty{}$

Proposition 4.4.2.1 (Joyal [MR1935979]). Let $X$ be a simplicial set. The following conditions are equivalent:

$(a)$

The projection map $X \rightarrow \Delta ^0$ is a Kan fibration.

$(b)$

The simplicial set $X$ is a Kan complex.

$(c)$

The simplicial set $X$ is an $\infty$-category and the homotopy category $\mathrm{h} \mathit{X}$ is a groupoid.

$(d)$

The simplicial set $X$ is an $\infty$-category and every morphism in $X$ is an isomorphism.

$(e)$

The projection map $X \rightarrow \Delta ^0$ is a left fibration.

$(f)$

The projection map $X \rightarrow \Delta ^0$ is a right fibration.

Proof of Proposition 4.4.2.1 from Theorem 4.4.2.6. Let $X$ be a simplicial set. By definition, the projection map $X \rightarrow \Delta ^{0}$ is a left fibration if and only if, for every pair of integers $0 \leq i < n$, every morphism of simplicial sets $\sigma _0: \Lambda ^{n}_{i} \rightarrow X$ can be extended to an $n$-simplex $\sigma : \Delta ^ n \rightarrow X$. This condition is automatically satisfied when $n=1$ (we can identify $\sigma _0$ with a vertex $x \in X$, and take $\sigma$ to be the degenerate edge $\operatorname{id}_{x}$), and is satisfied for $0 < i < n$ if and only if $X$ is an $\infty$-category. Assuming that $X$ is an $\infty$-category, it is satisfied for $i = 0$ if and only if every morphism in $X$ is an isomorphism (by virtue of Theorem 4.4.2.6). This proves the equivalence $(d) \Leftrightarrow (e)$, and the equivalence $(d) \Leftrightarrow (f)$ follows by applying the same reasoning to the opposite simplicial set $X^{\operatorname{op}}$. In particular, $(e)$ and $(f)$ are equivalent to one another, and therefore equivalent to $(a)$ (see Example 4.2.1.5). The equivalences $(a) \Leftrightarrow (b)$ and $(c) \Leftrightarrow (d)$ are immediate from the definitions. $\square$