Proposition 1.2.1.22. Let $S_{\bullet }$ be a simplicial set, and let $u_0: S_0 \rightarrow \pi _0( S_{\bullet } )$ be the map of sets given by the component map of Construction 1.2.1.18. Then $u_0$ exhibits $\pi _0(S_{\bullet } )$ as the coequalizer of the face operators $d^{1}_0, d^{1}_1: S_{1} \rightrightarrows S_0$.

**Proof of Proposition 1.2.1.22.**
Let $I$ be a set and let $f: S_0 \rightarrow I$ be a function satisfying $f \circ d^{1}_0 = f \circ d^{1}_1$ (as functions from $S_1$ to $I$). We wish to show that $f$ factors uniquely as a composition

By virtue of Proposition 1.2.1.19, this is equivalent to the assertion that there is a unique map of simplicial sets $F: S_{\bullet } \rightarrow \underline{I}_{}$ which coincides with $f$ on simplices of degree zero. Let $\sigma $ be an $n$-simplex of $S_{\bullet }$, which we identify with a map of simplicial sets $\sigma : \Delta ^ n \rightarrow S_{\bullet }$. For $0 \leq i \leq n$, we regard $\sigma (i)$ as a vertex of $S_{\bullet }$. Note that if $0 \leq i \leq j \leq n$, then we have $f( \sigma (i) ) = f ( \sigma (j) )$: to prove this, we can assume without loss of generality that $i=0$ and $j = n = 1$, in which case it follows from our hypothesis that $f \circ d^{1}_0 = f \circ d^{1}_1$. It follows that there is a unique element $F(\sigma ) \in I$ such that $F(\sigma ) = f( \sigma (i) )$ for each $0 \leq i \leq n$. The construction $\sigma \mapsto F(\sigma )$ defines a map of simplicial sets $F: S_{\bullet } \rightarrow \underline{I}_{}$ with the desired properties. $\square$