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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.)

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  • $\begingroup$ 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). $\endgroup$
    – R. van Dobben de Bruyn
    Sep 25 at 11:53
  • $\begingroup$ 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. $\endgroup$
    – Federico Cantero
    Sep 25 at 12:21
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    $\begingroup$ 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. $\endgroup$
    – David Wärn
    Sep 25 at 12:28

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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 $F_{i-1}(X)$.

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