Proposition 7.6.4.23 (Rewriting Pullbacks as Equalizers). Let $\operatorname{\mathcal{C}}$ be an $\infty $-category and let $f_0: X_0 \rightarrow X$ and $f_1: X_1 \rightarrow X$ be morphisms of $\operatorname{\mathcal{C}}$. Suppose that $X_0$ and $X_1$ admit a product $X_0 \times X_1$, and let $\pi _0: X_0 \times X_1 \rightarrow X_0$ and $\pi _1: X_0 \times X_1 \rightarrow X_1$ denote the projection maps. For $i \in \{ 0,1\} $, let $g_ i: X_0 \times X_1 \rightarrow X$ denote a composition of $\pi _ i$ with $f_ i$ in the $\infty $-category $\operatorname{\mathcal{C}}$. Then an object of $\operatorname{\mathcal{C}}$ is a pullback of $X_0$ with $X_1$ over $X$ if and only if it is an equalizer of the pair of morphisms $(g_0, g_1)$.
Proof. Let $\operatorname{\mathcal{K}}$ denote the category which is freely generated by a non-commutative square, as indicated in the diagram
Note that the upper right and lower left regions of this diagram determine monomorphisms $\tau _0, \tau _1: \Delta ^2 \hookrightarrow \operatorname{N}_{\bullet }(\operatorname{\mathcal{K}})$. The images of $\tau _0$ and $\tau _1$ are simplicial subsets of $\operatorname{N}_{\bullet }(\operatorname{\mathcal{K}})$, whose union is $\operatorname{N}_{\bullet }(\operatorname{\mathcal{K}})$ and whose intersection is the discrete simplicial set $\{ Y_{01}, Y \} $. It follows that $\tau _0$ and $\tau _1$ induce an isomorphism of simplicial sets $(\tau _0, \tau _1): \Delta ^2 {\coprod }_{ \{ 0,2\} } \Delta ^2 \simeq \operatorname{N}_{\bullet }(\operatorname{\mathcal{K}})$.
For $i \in \{ 0,1\} $, let $\sigma _ i$ be a $2$-simplex of $\operatorname{\mathcal{C}}$ which witnesses $g_ i$ as a composition of $\pi _ i$ with $f_ i$ (in the sense of Definition 1.4.4.1). Then there is a unique morphism of simplicial sets $q: \operatorname{N}_{\bullet }(\operatorname{\mathcal{K}}) \rightarrow \operatorname{\mathcal{C}}$ satisfying $q \circ \tau _ i = \sigma _ i$, which we indicate as a diagram
Let $\operatorname{\mathcal{K}}_{+} \subseteq \operatorname{\mathcal{K}}$ denote the full subcategory spanned by the objects $Y_{01}$ and $Y$. Then the nerve $\operatorname{N}_{\bullet }( \operatorname{\mathcal{K}}_{+} )$ can be identified with the simplicial set $( \bullet \rightrightarrows \bullet )$ of Notation 7.6.4.1, and the restriction $q_{+} = q|_{ \operatorname{N}_{\bullet }( \operatorname{\mathcal{K}}_{+} ) }$ corresponds to the pair of morphisms $g_0, g_1: X_0 \times X_1 \rightarrow X$. Note that the full subcategory $\operatorname{N}_{\bullet }( \operatorname{\mathcal{K}}_{+} ) \subset \operatorname{N}_{\bullet }(\operatorname{\mathcal{K}})$ is coreflective, so the inclusion map $\operatorname{N}_{\bullet }(\operatorname{\mathcal{K}}_{+}) \hookrightarrow \operatorname{N}_{\bullet }(\operatorname{\mathcal{K}})$ is left cofinal (Corollary 7.2.3.7). It follows that an object of $\operatorname{\mathcal{C}}$ is an equalizer of $g_0$ and $g_1$ if and only if it is a limit of the diagram $q$ (Corollary 7.2.2.11).
To complete the proof, it will suffice to show that an object of $\operatorname{\mathcal{C}}$ is a limit of $q$ if and only if it is a fiber product of $X_0$ with $X_1$ over $X$. Let $\operatorname{\mathcal{K}}_{-} \subseteq \operatorname{\mathcal{K}}$ denote the full subcategory spanned by the objects $Y_0$, $Y_1$, and $Y$. By virtue of Corollaries 7.3.8.2 and 7.3.8.14, it will suffice to show that the functor $q$ is right Kan extended from $\operatorname{N}_{\bullet }( \operatorname{\mathcal{K}}_{-} )$. E quivalently, we wish to show that the natural map
is a limit diagram in $\operatorname{\mathcal{C}}$. Unwinding the definitions, we see that $\operatorname{\mathcal{K}}_{-} \times _{\operatorname{\mathcal{K}}} \operatorname{\mathcal{K}}_{ Y_{01}/ }$ can be written as a disjoint union of subcategories having initial objects $Y_0$ and $Y_1$, respectively. In particular, the inclusion map
is left cofinal. The desired result now follows from Corollary 7.2.2.3, together with our assumption that the maps $\pi _0$ and $\pi _1$ exhibit $X_0 \times X_1$ as a product of $X_0$ with $X_1$. $\square$