# Kerodon

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

Proposition 1.1.9.10. Let $S_{\bullet }$ be a Kan complex containing a pair of vertices $x,y \in S_0$. Then $x$ and $y$ belong to the same connected component of $S_{\bullet }$ if and only if there exists an edge $e \in S_1$ satisfying $d_0(e) = x$ and $d_1(e) = y$.

Proof. Let $R$ denote the image of the map $(d_0, d_1): S_1 \rightarrow S_0 \times S_0$. According to Remark 1.1.6.23, we can identify $\pi _0( S_{\bullet } )$ 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 map $(d_0, d_1)$ carries the degenerate edge $s_0(x)$ to the pair $(x,x) \in S_0 \times S_0$.

• The relation $R$ is symmetric. Suppose that $(x,y) \in R$: that is, there exists an edge $e \in S_1$ satisfying $d_0(e) = x$ and $d_1(e) = y$. Then the tuple $(e, s_0(x), \bullet )$ determines a map of simplicial sets $\sigma _0: \Lambda ^{2}_{2} \rightarrow S_{\bullet }$ (see Exercise 1.1.2.14), which we depict as a diagram

$\xymatrix { & y \ar [dr]^{e} & \\ x \ar@ {-->}[ur] \ar [rr]^{s_0(x)} & & x. }$

Since $S_{\bullet }$ is a Kan complex, we can complete this diagram to a $2$-simplex $\sigma : \Delta ^2 \rightarrow S_{\bullet }$. Then $e' = d_2(\sigma )$ is an edge of $S_{\bullet }$ satisfying $d_0(e') = y$ and $d_1(e') = x$, which proves that the pair $(y,x)$ 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 $(x,z) \in R$. Choose edges $e, e' \in S_1$ satisfying $d_0(e) = x$, $d_1(e) = y = d_0(e')$, and $d_1(e') = z$. Then the tuple $(e', \bullet , e)$ determines a map of simplicial sets $\tau _0: \Lambda ^2_1 \rightarrow S_{\bullet }$ (see Exercise 1.1.2.14), which we depict as a diagram

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

Our assumption that $S_{\bullet }$ is a Kan complex guarantees that we can extend $\tau _0$ to a $2$-simplex $\tau : \Delta ^2 \rightarrow S_{\bullet }$. Then $e'' = d_1(\tau )$ is an edge of $S_{\bullet }$ satisfying $d_0( e'' ) = x$ and $d_1(e'') = z$, which proves that $(x,z) \in R$.

$\square$