# Is there an infinite chain of endofunctors of finite sets?

We consider the category of endofunctors of finite sets with natural transformations.

Is there an infinite chain $$F_1, F_2, \ldots$$ of endofunctors such that there is a natural transformation from $$F_i$$ to $$F_{i+1}$$ for all $$i\in \mathbb{N}$$ but no natural transformation from $$F_{i+1}$$ to $$F_i$$ for any $$i\in \mathbb{N}$$?

(This problem has a relevance in universal algebra as one can consider for each pair of finite structures $$A, B$$ the functor mapping the set $$\{ 1,2,...,n\}$$ to the set of homomorphisms $$\operatorname{Hom}(A^n,B)$$ with the obvious maps on morphisms. In fact, this was the motivation for the question.)

• One candidate could be $F = \mathcal P$ (the covariant power set) and $F_i = F^i$. The natural transformation $\eta \colon 1 \to F$ taking $a \in A$ to the singleton $\{a\} \in \mathcal P(A)$ induces multiple natural transformations $F_i \to F_{i+1}$ (namely $F^j \eta F^{i-j}$ for $j \in \{0,\ldots,i\}$). It seems to me that there should be no natural transformations in the other direction, but I don't know how to prove this (nor whether this is true, for that matter). Sep 25 at 11:53
• To prove that, let $\alpha: F\to 1$. Then the composition $\eta\circ\alpha$ yields a choice of an element $x_A$ in each set $A$, and this choice is natural with respect to any morphism. Therefore, this element must be fixed by all automorphisms of the set $A$, but this is only possible if the set has at most one element. Sep 25 at 12:21
• Indeed there is no natural transformation $\mathcal P \to \mathrm{id}$, but there is a natural transformation $\mathcal P \mathcal P \to \mathcal P$, which sends everything to the empty set. Sep 25 at 12:28

Yes. Let $$F_i$$ be the subfunctor of the covariant powerset functor given by $$S\in F_i(X)$$ iff $$0<|S|\le i$$.
If $$X$$ has exactly $$i$$ elements, then there is an element of $$F_i(X)$$ fixed by all automorphisms of $$X$$, namely $$X$$ itself; but there is no fixed element of $$F_{i-1}(X)$$.