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

Remark Let $\operatorname{\mathcal{C}}$ be an $(\infty ,2)$-category. We will refer to vertices of $\operatorname{\mathcal{C}}$ as objects, and to the edges of $\operatorname{\mathcal{C}}$ as morphisms. If $f$ is an edge of $\operatorname{\mathcal{C}}$ satisfying $d_1(f) = X$ and $d_0(f) = Y$, then we say that $f$ is a morphism from $X$ to $Y$ and write $f: X \rightarrow Y$.

Suppose we are given morphisms $f: X \rightarrow Y$, $g: Y \rightarrow Z$, and $h: X \rightarrow Z$ of $\operatorname{\mathcal{C}}$. We will say that a $2$-simplex $\sigma $ witnesses $h$ as a composition of $f$ and $g$ if it is thin and satisfies $d_0(\sigma ) = g$, $d_1(\sigma ) = h$, and $d_2(\sigma ) = f$, as indicated in the diagram

\[ \xymatrix@R =50pt@C=50pt{ & Y \ar [dr]_{g} & \\ X \ar [ur]^{f} \ar [rr]^-{h} & & Z. } \]

Note that:

  • When $\operatorname{\mathcal{C}}$ is an $\infty $-category, this recovers the terminology of Definition (since the $2$-simplex $\sigma $ is automatically thin).

  • If $\operatorname{\mathcal{C}}$ is the Duskin nerve of a $2$-category $\operatorname{\mathcal{E}}$, the $2$-simplex $\sigma $ can be identified with a $2$-morphism $\gamma : g \circ f \Rightarrow h$ of $\operatorname{\mathcal{E}}$, which is invertible if and only if $\sigma $ is thin. In other words, $\sigma $ witnesses $h$ as a composition of $f$ and $g$ if and only if it encodes the datum of an isomorphism $g \circ f \xRightarrow {\sim } h$ in the category $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{E}}}(X,Z)$.

  • Axiom $(1)$ of Definition asserts that the composition of $1$-morphisms in $\operatorname{\mathcal{C}}$ is defined (albeit not uniquely). More precisely, it asserts that for every pair of morphisms $f: X \rightarrow Y$ and $g: Y \rightarrow Z$, there exists a morphism $h: X \rightarrow Z$ and a $2$-simplex which witnesses $h$ as a composition of $f$ and $g$.