Equivalence between $\textbf{2Cat}$ and $\textbf{2Cat}_{psd}$

Question

Is there an equivalence between category $\textbf{2Cat}$ of 2-categories and 2-functors and the category $\textbf{2Cat}_{psd}$ of 2-categories and pseudofunctors?

Answer 1

These categories aren't equivalent, because $\mathbf{2Cat}$ has all limits, while $\mathbf{2Cat}_{psd}$ does not. The problem is, basically, that the limits would have to agree with those in $\mathbf{2Cat}$, but strict equalizers of pseudofunctors aren't generally closed under composition.

For instance, let $[2]$ denote the category $0\stackrel{a}{\to} 1\stackrel{b}{\to} 2$, viewed as a 2-category, and let $C$ denote the 2-category generated by morphisms $f:0\to 1,g:1\to 2, h:0\to 2$, and an isomorphism $\alpha: g\circ f\cong h$. Then there are two identity-on-objects pseudofunctors $F,G:[2]\to C$ strictly preserving identity 1-morphisms, sending $a$ to $f$ and $g$ to $b$: $F$ sends $b\circ a$ to $g\circ f$ while $G(b\circ a)=h$. Now $F$ and $G$ are an example of a pair of pseudofunctors admitting no equalizer, essentially because $F$ and $G$ agree on $b$ and on $a$ but not on $b\circ a$.

Indeed, suppose there were such an equalizer $e:E\to C$. Then by the universal property, denoting by $[0]$ the one-point category and by $[1]$ the category generated by a single non-identity arrow, we'd have $$\mathbf{2Cat}_{psd}([0],e)\cong \mathrm{eq}\left(F_*,G_*:\mathbf{2Cat}_{psd}([0],[2])\to \mathbf{2Cat}_{psd}([0],C\right)=\{0,1,2\}$$ and $$\mathbf{2Cat}_{psd}([1],e)\cong \mathrm{eq}\left(F_*,G_*:\mathbf{2Cat}_{psd}([1],[2])\to \mathbf{2Cat}_{psd}([1],C\right)=\{\mathrm{id}_0,\mathrm{id}_1,\mathrm{id}_2,a,b\}$$ where the equalizers on the right are calculated in $\mathbf{Set}$. But this is absurd: if $E$ were a 2-category, it would contain a composite mapping $0\to 2$, but $b\circ a$ is not equalized by $F$ and $G$.

This is part of a larger story. As was remarked in the comments, it's much more natural to ask whether the 2- or 3-categories $\mathbf{2Cat}$ and $\mathbf{2Cat}_{psd}$ are equivalent. If we take the 2-categories in which 2-morphisms are the strictly 2-natural transformations, then the inclusion $\mathbf{2Cat}\to \mathbf{2Cat}_{psd}$ admits a left 2-adjoint: for every 2-category $K$, there exists a 2-category $K'$ together with a strict 2-functor $K'\to K$ which is an equivalence in $\mathbf{2Cat}_{psd}$ but not, generally, in $\mathbf{2Cat}$, such that pseudofunctors out of $K'$ are naturally identified with 2-functors out of $K$. But this 2-adjunction is not an equivalence, and indeed the same argument as above, using strict 2-dimensional limits, shows that the 2-categories $\mathbf{2Cat}$ and $\mathbf{2Cat}_{psd}$ cannot be equivalent. (In case you're wondering, passing to 3-categories won't help. There exist many nontrivial 2-categories $K$ such that there are no nonconstant 2-functors $K\to K'$, because $K'$ contains no strictly invertible 1-morphisms other than the identities.)

In general, this kind of question is thoroughly studied in 2-categorical universal algebra, specifically, the study of 2-monads and the various flavors of morphism between their algebras. Your specific question is actually almost answered by Maclane's coherence theorem, stating that every monoidal category is equivalent to a strict monoidal category; the $K'$ construction in the previous paragraph is essentially the same as Maclane's. The general framework, though, is due to Blackwell, Kelly, and Power's paper Two-Dimensional Monad Theory. A friendly approach to all this, and much more, is in Steve Lack's A 2-Categories Companion.