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Proposition 1.3.5.2. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category. Then:

$(1)$

The composition law of Construction 1.3.5.1 is associative. That is, for every triple of composable morphisms $f: W \rightarrow X$, $g: X \rightarrow Y$, and $h: Y \rightarrow Z$ in $\operatorname{\mathcal{C}}$, we have an equality $( [h] \circ [g]) \circ [f] = [h] \circ ([g] \circ [f])$ in $\operatorname{Hom}_{\mathrm{h} \mathit{\operatorname{\mathcal{C}}}}( W, Z)$.

$(2)$

For every object $X \in \operatorname{\mathcal{C}}$, the homotopy class $[\operatorname{id}_ X] \in \operatorname{Hom}_{\mathrm{h} \mathit{\operatorname{\mathcal{C}}}}(X,X)$ is a two-sided identity with respect to the composition law of Construction 1.3.5.1. That is, for every morphism $f: W \rightarrow X$ in $\operatorname{\mathcal{C}}$ and every morphism $g: X \rightarrow Y$ in $\operatorname{\mathcal{C}}$, we have $[\operatorname{id}_ X] \circ [f]= [f]$ and $[g] \circ [ \operatorname{id}_ X]= [g]$.

Proof. We first prove $(1)$. Let $u: W \rightarrow Y$ be a composition of $f$ and $g$, let $v: X \rightarrow Z$ be a composition of $g$ and $h$, and let $w: W \rightarrow Z$ be a composition of $f$ and $v$. Then $( [h] \circ [g] ) \circ [f] = [w]$ and $[h] \circ ([g] \circ [f]) = [h] \circ [u]$. It will therefore suffice to show that $w$ is a composition of $u$ and $h$. Choose a $2$-simplex $\sigma _0$ of $\operatorname{\mathcal{C}}$ which witnesses $v$ as a composition of $g$ and $h$, a $2$-simplex $\sigma _2$ of $\operatorname{\mathcal{C}}$ which witnesses $w$ as a composition of $f$ and $v$, and a $2$-simplex $\sigma _3$ of $\operatorname{\mathcal{C}}$ which witnesses $u$ as a composition of $f$ and $g$. Then the sequence $( \sigma _0, \bullet , \sigma _2, \sigma _3)$ determines a map of simplicial sets $\tau _0: \Lambda ^3_1 \rightarrow \operatorname{\mathcal{C}}$ (Exercise 1.1.2.14), which we depict informally as a diagram

$\xymatrix@C =70pt@R=70pt{ & X \ar [r]^{g} \ar [drr]^{v} & Y \ar@ {-->}[dr]^{h} & \\ W \ar [ur]^{f} \ar@ {-->}[urr]^{u} \ar@ {-->}[rrr]^{w} & & & Z. }$

Using our assumption that $\operatorname{\mathcal{C}}$ is an $\infty$-category, we can extend $\tau _0$ to a $3$-simplex $\tau$ of $\operatorname{\mathcal{C}}$. Then the $2$-simplex $d_1(\tau )$ witnesses $w$ as a composition of $u$ and $h$.

We now prove $(2)$. Fix an object $X \in \operatorname{\mathcal{C}}$ and a morphism $g: X \rightarrow Y$ in $\operatorname{\mathcal{C}}$; we will show that $[g] \circ [\operatorname{id}_ X] = [g]$ (the analogous identity $[\operatorname{id}_ X] \circ [f] = [f]$ follows by a similar argument). For this, it suffices to observe that the degenerate $2$-simplex $s_0(g)$ witnesses $g$ as a composition of $\operatorname{id}_ X$ and $g$. $\square$