# 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?

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• 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 $2$.
Sorry but your proof does not hold. You make 2 mistakes: 1) $[A, C] = [B, C]= C^1$ is not $\mathrm{Set}(\mathrm{ob} \ C)$ , it is exactly the category $C$ itself. 2) $s\downarrow t= Hom(s, t)$ is the discrete category whose objects are the morphisms from s to t and only has the identity morphisms. When you then rewrite the adjoint iso, you see that it actually holds if we set $F(X) = s$. The category $\mathrm{Arr}(s, t))$ that you mention is actually $C\downarrow C$ also known as $C^2$. – magma Sep 18 '13 at 18:36
@magma: Thanks for catching that, I agree I messed up $[A, C]$ and $S \downarrow T$. But $F(X) = s$ does not make the adjoint iso hold because the right hom set still depends on the choice of $X$ whereas the left does not. I've edited my answer to fix those two mistakes. – Jim Sep 19 '13 at 6:59