# Kerodon

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

Proposition 1.2.5.10. Let $S$ be a Kan complex and let $x$ and $y$ be vertices of $S$. Then $x$ and $y$ belong to the same connected component of $S$ if and only if there exists an edge $e$ of $S$ having source $x$ and target $y$.

Proof. Let $S_0$ denote the set of vertices of $S$. Let $R$ be the collection of pairs $(x,y) \in S_0$ for which there exists an edge $e$ of $S$ having source $x$ and target $y$. Using Remark 1.2.1.23, we can identify $\pi _0( S )$ with the quotient of $S_{0}$ by the equivalence relation generated by $R$. It will therefore suffice to show that $R$ is already an equivalence relation on $S_0$. To prove this, we must verify three things:

• The relation $R$ is reflexive. This follows from the observation that for every vertex $x \in S_0$, the degenerate edge $\operatorname{id}_{x}$ has source $x$ and target $x$.

• The relation $R$ is symmetric. Suppose that $(x,y) \in R$: that is, there exists an edge $e$ of $S$ having source $x$ and target $y$. Then the tuple $(\bullet , \operatorname{id}_ x, e)$ determines a map of simplicial sets $\sigma _0: \Lambda ^{2}_{0} \rightarrow S$ (see Proposition 1.2.4.7), which we depict as a diagram

$\xymatrix@R =50pt@C=50pt{ & y \ar@ {-->}[dr] & \\ x \ar [ur]^{e} \ar [rr]^{\operatorname{id}_ x} & & x. }$

Since $S$ is a Kan complex, we can complete this diagram to a $2$-simplex $\sigma : \Delta ^2 \rightarrow S$. Then $e' = d^{2}_0(\sigma )$ is an edge of $S$ having source $y$ and target $x$, so the pair $(y,x)$ also belongs to $R$.

• The relation $R$ is transitive. Suppose that we are given vertices $x,y,z \in S_0$ with $(x,y) \in R$ and $(y,z) \in R$; we wish to show that $(x,z) \in R$. Let $e$ be an edge of $S$ having source $x$ and target $y$, and let $e'$ be an edge of $S$ having source $y$ and target $z$. Then the tuple $(e', \bullet , e)$ determines a map of simplicial sets $\tau _0: \Lambda ^2_1 \rightarrow S$ (see Proposition 1.2.4.7), which we depict as a diagram

$\xymatrix@R =50pt@C=50pt{ & y \ar [dr]^{e'} & \\ x \ar [ur]^{e} \ar@ {-->}[rr] & & z. }$

Our assumption that $S$ is a Kan complex guarantees that we can extend $\tau _0$ to a $2$-simplex $\tau : \Delta ^2 \rightarrow S$. Then $e'' = d^{2}_1(\tau )$ is an edge of $S$ having source $x$ and target $z$, so that $(x,z)$ belongs to $R$.

$\square$