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Let $\mathcal C$ be a category which is equivalent to the category $\mathcal C^{\mathcal C}$ of its endofunctors.

Is $\mathcal C$ necessarily equivalent to a category having exactly one object and one morphism?

A particular case of this question was asked here.

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    $\begingroup$ I do not know the answer in the case that the ambient theory is classical set theory, but it is possible to arrange an intuitionistic set theory in which there is a non-trivial set $D$ such that $D \cong D^D$, which would then give you an example (take $\mathcal{C}$ to be the discrete category on $D$). If anyone can answer your question positively, they will have to use excluded middle or choice. $\endgroup$ – Andrej Bauer Jun 18 at 14:00
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    $\begingroup$ I can also produce a counter-example if $\mathcal{C}^\mathcal{C}$ is replaced with the category of endofunctors that preserve filtered colimits. In this case we can find a poset which will do the job. $\endgroup$ – Andrej Bauer Jun 18 at 14:01
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    $\begingroup$ Note that there is a canonical fully faithful functor $const: C \to C^{C}$ sending an object to the constant functor at that object. Moreover, if $C$ is not the terminal category, then the identity functor is not in the essential image of $const$. So if $F: C \to C^{C}$ is an equivalence, then $F^{-1} \circ const: C \to C$ is a fully faithful endofunctor which is not essentially surjective. This rules out, for example, categories with finitely many isomorphism classes of objects. At a more basic level, $C^{C}$ is monoidal under composition, so any such $C$ must be monoidal. $\endgroup$ – Tim Campion Jun 19 at 18:22
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    $\begingroup$ This is an interesting tweak to your other question from last week. I thought about trying to let $D=C^C$ from my answer there, but it doesn't work because the set of objects of $D^D$ has cardinality $End(G)^{End(G)}$, while the set of objects of $D$ has cardinality $End(G)$. $\endgroup$ – Matt Feller Jun 19 at 20:41
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    $\begingroup$ Another small observation: any such $C$ must be connected. For if $C$ has $\kappa$ many connected components, then $C^C$ has at least $\kappa^\kappa$ many connected components, and the only $\kappa$ such that $\kappa = \kappa^\kappa$ is $\kappa = 1$. Since the identity is connected to the constant functors, there is even a bound on the length of zigzag needed to get from one object to another. $\endgroup$ – Tim Campion Jun 20 at 20:55
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Here is a constraint of a different nature from the ones mentioned so far in the comments. I work in ZFC with one universe $V_\kappa =: Set$. Let $CARD$ denote the class of cardinals $<\kappa$. Below all arithmetic is cardinal arithmetic (not ordinal arithmetic).

Proposition: Let $\mathcal C$ be a category with $\mathcal C \simeq \mathcal C^{\mathcal C}$. Assume that

  • The number of isomorphism classes of objects of $\mathcal C$ is $< 2^\kappa$

  • $\mathcal C$ is locally small (its homsets are of size $<\kappa$)

  • $\mathcal C$ has all small powers and all small copowers.

Then $\mathcal C$ is a preorder.

Remarks:

  • There are two important cases where $\mathcal C$ has all small powers and all small copowers:

    1. if $\mathcal C$ has all small products and all small coproducts.

    2. if $\mathcal C$ is a preorder

    So the proposition could be stated as an "if and only if".

  • Of course, the existence of the powers and copowers should be viewed as the"substantive" hypothesis, and the size restrictions as "technical".

  • I think these hypotheses are general enough to make the case that no "nice large category" satisfies $\mathcal C \simeq \mathcal C^{\mathcal C}$, but of course it would be interesting to weaken the hypothesis. That would probably require quite different methods not reducing to the existence of lots of $Set$-functors as below. There's another paper of Koubek on contravariant $Set$-functors, which might allow one to assume only the existence of powers without assuming the existence of copowers (or dually).

Proof: For any $F: Set \to Set$, and $A \in \mathcal C$, we may form a functor $\Phi_{F,A}: \mathcal C \to \mathcal C$, $C \mapsto \amalg_{x \in F(\mathcal C(A,C))} A$. Suppose for contradiction that $\mathcal C$ is not a preorder. Fix $A,B$ such that

$$\lambda := |\mathcal C(A,B)| \geq 2$$

Let $\mathcal A$ be the class of (small) cardinals of the form $\alpha = \lambda^\mu$, and let $\mathcal F$ be the class of all weakly increasing functions $f: \mathcal A \to CARD$ such that

  1. $f(\alpha) \geq 2^\alpha$ for all $\alpha \in \mathcal A$.

  2. For all $\alpha \in \mathcal A$, if $\lambda^{f(\alpha)} = \lambda^\beta$, then $f(\alpha) \leq \beta$.

Clearly $|\mathcal F| = 2^\kappa$. By a theorem of Koubek [1], for every $f \in \mathcal F$, there is a functor $F_f: Set \to Set$ with $|F_f(X)| = f(|X|)$ for all $X$ with $|X| \in \mathcal A$. Because $\mathcal C$, and hence $\mathcal C^{\mathcal C}$, has only $< 2^\kappa$ many isomorphism classes of objects, there are $f_1 \neq f_2 \in \mathcal F$ such that $\Phi_{F_1,A} \cong \Phi_{F_2,A}$ (where $F_i = F_{f_i}$). Since $f_1 \neq f_2$, there is a cardinal $\mu$ such that $f_1(\lambda^\mu) \neq f_2(\lambda^\mu)$. We have that $\Phi_{F_1,A}(\prod_\mu B) \cong\Phi_{F_2,A}(\prod_\mu B)$, i.e. $\amalg_{f_1(\lambda^\mu)} A \cong \amalg_{f_2(\lambda^\mu)} A$. Homming into $B$, we find that $\mathcal C(A,B)^{f_1(\lambda^\mu)} \cong \mathcal C(A,B)^{f_2(\lambda^\mu)}$. But by the second condition in the definition of $\mathcal F$, this implies that $f_1(\lambda^\mu) = f_2(\lambda^\mu)$, contradicting the choice of $\mu$.

[1] Koubek, Václav. "Set functors." Commentationes Mathematicae Universitatis Carolinae 012.1 (1971): 175-195. http://eudml.org/doc/16420, Prop 2.4, bottom of p. 183 (p.10 of the pdf) -- note that bizarrely in this paper that means section 4, second proposition, and is also different from Lemma 2.4.

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This is a partial answer. I tried to mimic the proof of Theorem 3 in

[1] Complete lattices and the generalized Cantor theorem by Roy O. Davies, Allan Hayes and George Rousseau, published in Proc. Amer. Math. Soc. 27 (1971), 253–258.

(The notation used in this answer is different from the one in the question. I'll stick as much as possible to the notation and terminology of [1].)

All categories in this post are small.

Let $A$ and $B$ be categories, and let $b_0,b_1$ be two objects of $B$ such that there is a morphism $b_0\to b_1$ but no morphism $b_1\to b_0$.

We claim

(1) $B^B$ is not equivalent to $B$,

(2) there is no full functor $B^A\to A$ and no essentially surjective functor $A\to B^A$.

It suffices to prove (2).

For any category $C$ write $C_0$ for the set of objects of $C$.

Let $C$ and $D$ be categories. Say that a map $F:C_0\to D_0$ is a weak monomorphism if the existence of morphisms $c\to c'$ and $F(c')\to F(c)$ implies that of a morphism $c'\to c$.

It is clear that the map $C_0\to D_0$ induced by a full functor is a weak monomorphism. It is also clear that the existence of an essentially surjective functor $D\to C$ implies that of a weak monomorphism $C_0\to D_0$.

We claim

(3) There is no weak monomorphism $(B^A)_0\to A_0$.

It suffices to prove (3).

Let $2$ be the ordinal $\{0,1\}$ viewed as a category. We claim

(4) There is no weak monomorphism $(2^A)_0\to A_0$.

Let us show that (4) implies (3).

Let $F:(B^A)_0\to A_0$ be a weak monomorphism, let $J:2\to B$ be a functor mapping $i$ to $b_i$ for $i=0,1$ (such a functor exists by assumption), and define $F':(2^A)_0\to A_0$ by $F'(G):=F(J\circ G)$. It is straightforward to check (using the assumption that there is no morphism $b_1\to b_0$) that $F'$ is a weak monomorphism (details left to the reader). This proves that (4) implies (3).

It suffices to prove (4).

Say that a subset $R$ of $A_0$ is a right ideal if the conditions $a\in R$ and there is a morphism $a\to a'$ imply $a'\in R$. The right ideals form a complete lattice $\mathcal R$ order isomorphic to $(2^A)_0$. (The order on $\mathcal R$ is given by inclusion.)

Thus it suffices to show that there is no weak monomorphism $\mathcal R\to A_0$.

Let $\phi:\mathcal R\to A_0$ be a map. Define the map $f:\mathcal R\to \mathcal R$ by letting $f(R)$ be the right ideal generated by $\phi(R)$. By the corollary to Theorem 1 in [1] there is an $R$ in $\mathcal R$ such that $$ f(R)\le\bigcup_{S>R}f(S). $$ As we have $\phi(R)\in f(R)$, this implies $\phi(R)\in f(S)$ for some $S$ in $\mathcal R$ with

(5) $R<S$,

and thus (by definition of $f(S)$) the existence of a morphism

(6) $\phi(S)\to\phi(R)$.

Now (5) and (6) imply that $\phi$ is not a weak monomorphism.

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I am posting a long comment because in these days I do not have the time to check if this idea actually makes sense. If someone manages to elaborate this poorly written comment into an answer, or someone observes that this point of view is a bad one, I will erase this.

Let me give an unrequired proof that when $\mathsf{C} = \mathsf{Set}$, then $\mathsf{C}$ cannot be equivalent to $\mathsf{C}^\mathsf{C}.$

The proof goes in the following way: when two categories are equivalent, their full subcategory of tiny objects are equivalent, this happens because being tiny is invariant under equivalence of categories. Now, there is only $1$ tiny object in $\mathsf{Set}$, while there is a whole proper class of tiny objects in $\mathsf{Set}^\mathsf{Set}$, namely the image of the Yoneda embedding $y: \mathsf{Set}^\circ \to\mathsf{Set}^\mathsf{Set}$.

Now, assume that $\mathsf{C}$ is cartesian closed so that $\mathsf{C}$ is a $\mathsf{C}$-category. Can't we run kind of the same argument giving an internal definition of tiny object?

The general idea is that in $\mathsf{V}$-Cat one has that $\mathsf{V}$ is equivalent to $\mathsf{V}^\mathsf{V}$ iff the enriched Cauchy completion of $1$ and $\mathsf{V}^\circ$ coincide. This looks very unlikely to happen to me. For example it will never happen when $\mathsf{V}$ is $\mathsf{V}$-complete.

Obviously, one has to be carefull, because I am restricting to $\mathsf{V}$-enriched functors. Is this truly restrictive? I do not know.

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