Could any "incoherently involutive" endofunctor be made "coherently involutive"?

Question

Given a category $C$ with an endofunctor $F:C \to C$ and a natural isomorphism $\epsilon:FF \cong 1_C$, call the pair $(F, \epsilon)$ an involutive endofunctor.

Also, call the pair $(F, \epsilon)$ a coherently involutive endofunctor if $(F, F, \epsilon^{-1}, \epsilon)$ forms an adjoint equivalence, or equivalently, if $F\epsilon_X=\epsilon_{FX}$ for all objects $X$ of $C$.

Then, given any involutive endofunctor $(F, \epsilon)$, is there always another natural isomorphism $\epsilon^{\prime}:FF \cong 1_C$ for which $(F, \epsilon^{\prime})$ is coherently involutive?

It is known that incoherent equivalences could always be made into coherent adjoint equivalences by changing one of the involved isomorphisms without also changing the other.

Answer 1

Consider a category with a family of objects $X_n$ with $n \in \mathbb Z$. Let the set of morphisms originating from any particular object be $(p,q,r) \in \mathbb Z^2 \times \mathbb N$. A morphism $(p,q,r)$ starting from $X_n$ has codomain $X_{n+2p+2q+r}$.

Let composition be $$(p, q, r) \circ (p', q', r') = (p + p', q + q', r + r').$$ Now construct an involutive functor $F$ which maps $X_n$ to $X_{n+1}$. Its action on morphisms is $F(p,q,r) = (q,p,r)$. This functor is indeed involutive. We construct $\epsilon_{X_n} = (-1,0,0)$, with inverse $(1,0,0)$.

Finally, we prove that it cannot be coherently involutive. Suppose we have $\epsilon_{X_n} = (p, q, r)$. We need $r$ to be even, and $p+q+r/2 = -1$, so that the codomain is correct. Then we need $\epsilon_{X_{n+1}} = \epsilon_{FX_n} = F\epsilon_{X_n} = (q, p, r)$. But now the square formed by $\epsilon$ on $(0,0,1)$ cannot commute, because we need $$ (p,q,r) + (0,0,1) = (0,0,1) + (q,p,r) $$ implying $p = q$ and $r \ne 0$ by parity. But $r \in \mathbb N$, so it cannot ever have an inverse.

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Basically we start with some free category generated with the required stuff (and one object). We would have $F^n X$ for every $n \in \mathbb Z$, and two isomorphisms $F^{n+2} X \to F^{n} X$ given by $\epsilon_{F^n X}$ and $F\epsilon_{F^{n-1}X}$, respectively. All the other morphisms must be a composition of these. But now there's nothing to relate the "even" objects and "odd" objects, so the naturality condition is too weak. We can obtain a coherent isomorphism by modifying $\epsilon_{F^{\text{odd}} X}$ to $F\epsilon_{F^{\text{even}} X}$, leaving the others unchanged.

To fix this, we further add a morphism $r : X\to FX$. Now the naturality square for $\epsilon$ on $X \to FX$ "locks" the relative position of the two aforementioned isomorphisms. So we cannot switch stuff around to get a coherent involution. It remains to rigorously establish this.

Since our category is created from generators and relations, to prove certain things are distinct, we need to establish invariants. For example, in a morphism, the total number of appearances $F^n r$ is preserved by all our equations. So this number is an invariant. Using these invariants we can show that a coherent $\epsilon$ cannot be chosen.

Finally, to simplify the presentation, we quotient out some stuff not required by our proof above. For instance we can assume the two types of aforementioned isomorphisms commute with each other, which makes them describable with $\mathbb Z^2$ instead of a complicated string of symbols. We can also remove isomorphic copies of objects, leaving only two. This creates my counterexample above.

Answer 2

If $\mathcal C$ is group $G$ considered as a one-object category, the question reduces to whether there exists an automorphism $f\colon G\to G$ with the property that $\{e\in G:f^2(g)=e^{-1}ge$ for all $g\in G\}$ is non-empty and disjoint from $\{e\in G:f(e)=e\}$.

Since any automorphism preserves the identity of the group, it follows that $f^2$ cannot be the trivial automorphism, whence $G$ cannot be abelian.

Moreover, if $f(g)=h^{-1}gh$, then $ff(g)=(h^2)^{-1}gh^2$ and $f(h^2)=h^{-1}h^2h=h^2$, so $f$ cannot be an inner automorphism.

Finally, $ff(g)=ffff^{-1}(g)=f(e^{-1}f^{-1}(g)e)=f(e)^{-1}gf(e)$ shows that $e$ and $f(e)$ define the same inner automorphism. Thus $G$ has to have a non-trivial center.

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The group of automorphisms of the dihedral group $G=(\mathbb Z/3)^\times\lhd\mathbb Z/n$ for $n>2$ is isomorphic to the semi-direct product $(\mathbb Z/n)^\times\lhd\mathbb Z/n=\{bx+c:b\in(\mathbb Z/n)^\times,c\in\mathbb Z/n\}$ with elements $(s,a)\in(\mathbb Z/3)^\times\times\mathbb Z/n$ corresponding to the inner automorphisms $sx-2as$, and action of $bx+c$ given by $(1,a)\mapsto(1,ba)$ and $(-1,a)\mapsto(-1,ba+c)$.

In order for the group to have a non-trivial center, it is necessary that $n=2k$, in which case the pairs of elements determining the same inner automorphisms are $(s,a),(s,a+k)$.

The square of an automorphism $bx+c$ is $(bx+c)\circ(bx+c)=b(bx+c)+c=b^2x+bc+c$.

A necessary condition for the square to be an inner automorphism is that $b^2\equiv\pm1\bmod n$. The condition is also sufficient since, the existence of $a$ such that $bc+c\equiv -2ab^2\bmod n$, i.e. such that $c(b+1)/b^2\equiv-2a\bmod n$, follows from $n$ being even and $b^2$ being odd implying $c/b^2(b+1)$ is even.

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If $b^2\equiv 1\bmod n$, then the property holds if and only if $ba\equiv a+k\bmod n$, i.e. if $(b-1)a\equiv k\bmod n$. In particular, $b\not\equiv1\bmod n$, and $k$ must be even, so $n=4j$. But then $b\pm 1$ being even implies either $b-1$ or $b+1$ is divisible by $4$, whence either $b-1$ is divisible by $4$ or $-2a$ is divisible by $4$. Either way, $k$ is divisible by $4$, so in fact $n=8i$.

A solution is then obtained by taking $n=8$, $k=4$, $b=3$, $a=2$, $c=1$. Indeed, then $b^2\equiv1\bmod 8$, $c(b+1)/b^2\equiv 4\equiv-2a\bmod 8$, $ba\equiv 6\equiv a+k\bmod 8$.