Consider two categories $\mathbf A$ and $\mathbf B$. Assume that $\mathbf A$ is a small category. Then construct the functor category $[\mathbf A,\mathbf B]$, whose objects are all morphisms from $\mathbf A$ to $\mathbf B$ and whose morphisms are all natural transformations between such functors.

Then if $\mathbf B$ is a large category, $[\mathbf A,\mathbf B]$ is a proper metacategory in the sense that its objects or morphisms form not a class, but a conglomerate.

And it is known that this functor metacategory is legitimate; i.e., it is isomorphic to some category in usual sense.

But how can this situation be realized since metacategories and categories have different sizes?


Your error is the claim that $[\mathbf{A},\mathbf{B}]$ is not a class.

Let $|A|$ be the small set of all objects and arrows of $\mathbf{A}$, and $|B|$ be the class of all objects and arrows of $\mathbf{B}$.

Let $V$ be the class of all (small) sets, so that $|B| \subseteq V$.

Then every functor $\mathbf{A} \to \mathbf{B}$ can be expressed in the obvious way as a function $|A| \to |B|$. Thus, $\mathrm{Ob}[\mathbf{A},\mathbf{B}] \subseteq |B|^{|A|} \subseteq V$.

A similar argument applies to natural transformations.

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  • $\begingroup$ But if |$A$| is the large class and |$B$| is a two point set, the Ob$[\mathbf A,\mathbf B] \nsubseteq V$? $\endgroup$ – A. Gonus Nov 7 '17 at 12:25
  • $\begingroup$ @A.Gonus: In that case you can have $\not\subseteq$. It's not guaranteed, though; e.g. if B is discrete and A connected, then $[A,B] \cong B$. $\endgroup$ – user14972 Nov 7 '17 at 13:01

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