As discussed here, Using the universal property of spaces, the category of spaces has a universal property: it is the free cocompletion of the terminal category $*$. That is, for any $(\infty,1)$-category $C$ with all colimits, there is an equivalence $$ Fun^{cocontinuous}(Spaces, C) \cong Fun(*,C) \cong C $$ I am wondering if there is a similar universal property for $Cat^{(\infty,1)}$, the $(\infty,2)$-category of $(\infty,1)$-categories.

Note that $Cat^{(\infty,1)}$ has all lax colimits and any category $C$ is the lax colimit of the functor $C \rightarrow * \rightarrow Cat^{(\infty,1)}$. So one guess is that $Cat^{(\infty,1)}$ is the lax cocompletion of the terminal category $*$ in the sense that for any $(\infty,2)$-category $C$ with lax colimits, there is an equivalence $$ Fun^{Lax \ cocontinuous}(Cat^{(\infty,1)}, C) \cong Fun(*, C) \cong C $$ Does this proposed equivalence hold and, if so, has it been studied somewhere?

As discussed here, Using the universal property of spaces, the $(\infty,1)$-category of spaces has a universal property: it is the free $\infty$-categorical cocompletion of the terminal category $*$. That is, for any $(\infty,1)$-category $C$ with all colimits, there is an equivalence $$ Fun^{cocontinuous}(Spaces, C) \cong Fun(*,C) \cong C $$ I am wondering if there is a similar universal property for $Cat^{(\infty,1)}$, the $(\infty,2)$-category of $(\infty,1)$-categories.

Note that $Cat^{(\infty,1)}$ has all lax colimits and any category $C$ is the lax colimit of the functor $C \rightarrow * \rightarrow Cat^{(\infty,1)}$. So one guess is that $Cat^{(\infty,1)}$ is the lax cocompletion of the terminal category $*$ in the sense that for any $(\infty,2)$-category $C$ with lax colimits, there is an equivalence $$ Fun^{Lax \ cocontinuous}(Cat^{(\infty,1)}, C) \cong Fun(*, C) \cong C $$ Does this proposed equivalence hold and, if so, has it been studied somewhere?