Kerodon

Warning 2.2.4.13. Let $\operatorname{\mathcal{C}}$ and $\operatorname{\mathcal{D}}$ be strict $2$-categories, and let $\operatorname{\mathcal{C}}_0$ and $\operatorname{\mathcal{D}}_0$ denote their underlying ordinary categories (obtained by ignoring the $2$-morphisms of $\operatorname{\mathcal{C}}$ and $\operatorname{\mathcal{D}}$, respectively). Every strict functor $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ induces a functor of ordinary categories $F_0: \operatorname{\mathcal{C}}_0 \rightarrow \operatorname{\mathcal{D}}_0$. However, if a functor $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ is not strict, then it need not give rise to a functor from $\operatorname{\mathcal{C}}_0$ to $\operatorname{\mathcal{D}}_0$. If $X \xrightarrow {f} Y \xrightarrow {g} Z$ is a composable pair of $1$-morphisms in $\operatorname{\mathcal{C}}$, then Definition 2.2.4.5 guarantees that the $1$-morphisms $F(g) \circ F(f)$ and $F(g \circ f)$ are isomorphic (via the composition constraint $\mu _{g,f}$), but not that they are identical.