# Kerodon

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Example 2.2.4.14. Let $\operatorname{\mathcal{C}}$ be a $2$-category and let $\operatorname{\mathcal{D}}$ be an ordinary category, which we regard as a $2$-category having only identity $2$-morphisms. If $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ is lax functor of $2$-categories, then its values on the $1$-morphisms of $\operatorname{\mathcal{C}}$ must satisfy the following conditions:

$(1)$

If $u,v: X \rightarrow Y$ are $1$-morphisms of $\operatorname{\mathcal{C}}$ having the same source and target and $\gamma : u \Rightarrow v$ is a $2$-morphism of $\operatorname{\mathcal{C}}$, then $F(u) = F(v)$ (since $F(\gamma ): F(u) \Rightarrow F(v)$ must be an identity $2$-morphism of $\operatorname{\mathcal{D}}$).

$(2)$

If $u: X \rightarrow Y$ and $v: Y \rightarrow Z$ are composable $1$-morphisms of $\operatorname{\mathcal{C}}$, then $F(v \circ u) = F(v) \circ F(u)$ (since the composition constraint $\mu _{v,u}: F(v) \circ F(u) \Rightarrow F(v \circ u)$ is an identity $2$-morphism of $\operatorname{\mathcal{D}}$).

$(3)$

For every object $X \in \operatorname{\mathcal{C}}$, $F( \operatorname{id}_ X )$ is the identity morphism $\operatorname{id}_{ F(X)}$ in $\operatorname{\mathcal{D}}$ (since the identity constraint $\epsilon _{X}: \operatorname{id}_{ F(X)} \Rightarrow F( \operatorname{id}_ X )$ is an identity $2$-morphism of $\operatorname{\mathcal{D}}$).

Conversely, any specification of the values of $F$ on objects and $1$-morphisms which satisfies conditions $(1)$, $(2)$, and $(3)$ extends uniquely to a strict functor $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ (the coherence conditions appearing in Definition 2.2.4.5 are automatic, by virtue of the fact that every $2$-morphism of $\operatorname{\mathcal{D}}$ is an identity). In particular, every lax functor $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ is automatically strict. Beware that the analogous statement is generally false if the roles of $\operatorname{\mathcal{C}}$ and $\operatorname{\mathcal{D}}$ are reversed.