Does the comma category functor have an adjoint? Let $A\xrightarrow{S}C\xleftarrow{T}B$ be a diagram of categories; the classical definition of the comma category $(S\downarrow T)$ leads to the definition of a functor
$$
(S\downarrow-)\colon [B,C]\to\bf Cat
$$
sending $T\mapsto (S\downarrow T)$.
Now I wonder if this functor admits a left adjoint: does it commute with limits?
 A: In general, no.  Here is an example:


*

*Let $A = B = \{\ast\}$, the category with one object and one morphism (the identity of that object), $C = \mathbb Z/2$, the category with one object ($\ast'$) and $\hom(\ast', \ast') = \mathbb Z/2$.

*Check that $[\{\ast\}, \mathbb Z/2] = \mathbb Z/2$ (there is a unique functor that sends $\ast \mapsto \ast'$ and a natural transformation from this functor to itself corresponds to an element of $\mathbb Z/2$ by taking the component at $\ast$).

*Let $S\colon A \to C$ and $T\colon B \to C$ be that unique functor.  Check that $S \downarrow T = \mathrm{Set}(\mathbb Z/2)$, the category with objects $\mathbb Z/2$ and only identity morphisms.


Now if $F\colon\mathbf{Cat} \to [B, C] = \mathbb Z/2$ is a left adjoint that means that for every category $X$ and object $t \in C$ there is a bijection
$$\mathbb Z/2 = \hom_{\mathbb Z/2}(F(X), T) \simeq \hom_\mathbf{Cat}(X, \mathrm{Set}(\mathbb Z/2))$$
But if we choose $X = \mathrm{Set}(\{1, 2\})$ then $\hom_\mathbf{Cat}(X, \mathrm{Set}(\mathbb Z/2))$ is just the set maps $\{1, 2\} \to \mathbb Z/2$, and there are $4$ of these, not $2$.
