Typically, when defining the functor category $\mathcal{D}^\mathcal{C}$, where objects are functors $\mathcal{C}\rightarrow\mathcal{D}$ and the morphisms between such objects $F,G$ are the natural transformations $F\rightarrow G$ with the obvious composition and identities, one requires the category $\mathcal{C}$ to be small, i.e. that its objects form a set. $\mathcal{D}$ can be any category.

I've always thought that the reason for the smallness requirement on $\mathcal{C}$ is that "surely you'll get into trouble with the morphisms $F\rightarrow G$ in $\mathcal{D}^\mathcal{C}$ potentially not forming a set otherwise", and I've never given it much thought.

Now, does anyone know of an example which fixes a large category $\mathcal{C}$, a category $\mathcal{D}$ and two functors $F,G:\mathcal{C}\rightarrow\mathcal{D}$ such that that the collection of natural transformations $F\rightarrow G$ does not form a set? I've always thought in the back of my head that this should be easily doable with $\mathcal{C}=\mathcal{D}=\mathrm{Set}$ and some simple functors $F,G:\mathcal{C}\rightarrow\mathcal{D}$ by somehow cooking up at least one natural transformation $F\rightarrow G$ for each object of $\mathcal{C}$, i.e. for each set. Maybe my set theory fu is just weak.

I could of course just be missing something basic here. I'm asking because the books I've checked tend to just state $\mathcal{C}$ must be small. Apologies if this is all trivial!


1 Answer 1



let C be a discrete category (i.e. the only morphisms in C are the identity morphisms) and let D be the category consisting of two objects 0 and 1 and (apart from the identity morphisms) two parallel arrows $0\rightrightarrows 1$. Consider the constant functors F with value 0 and G with value 1. A natural transformation $F\to G$ corresponds to the choice of an element in a set of two elements for every object of C so that the conglomerate of all natural transformations $F\to G$ can be identified with the power conglomerate of Ob(C). This is not a set unless Ob(C) is a set.

  • $\begingroup$ Thanks! I'm glad it was as simple as the back of my head thought it was, although I was obviously a bit off in my actual construction. $\endgroup$
    – gspr
    Sep 17, 2010 at 8:49

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