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A self-map is a map $f:X\to X$ from a set $X$ to itself. There is an obvious notion of *morphism*, and thus of *isomorphism* and *automorphism*, of self-maps. [A morphism from $f:X\to X$ to $g:Y\to Y$ is a map $\phi:X\to Y$ such that $g\circ\phi=\phi\circ f$.]

A self-map is **rigid** if it has no non-trivial automorphism.

The question is in the title:

Does every set have a rigid self-map?

Clearly, the existence of a rigid self-map of a given set $X$ depends only on the cardinality $|X|$ of $X$.

There is an obvious notion of the *coproduct* $f:X\to X$ of a family $f_i:X_i\to X_i$ of self-maps. [The set $X$ is the disjoint union of the $X_i$ and $f$ coincides with $f_i$ on $X_i$.] Any self-map is the coproduct of its *indecomposable components*, and a self-map is rigid if and only if its indecomposable components are rigid and pairwise non-isomorphic. (In the sequel I use the expression "component" instead of "indecomposable component". Moreover the identity of the empty set does not count as a component.)

We claim:

(1) If $|X|\le2^{2^{\aleph_0}}$, then $X$ has a rigid self-map.

Define $f:\mathbb N\to\mathbb N$ by $f(i)=\max(i-1,0)$. Then $f$ is rigid. Moreover, for each $n\in\mathbb N$ the map $f$ induces a rigid self-map of the set $\{0,1,\ldots,n\}$. This proves that (1) holds for $|X|\le\aleph_0$.

It remains to prove that $X$ has a rigid self-map when $\aleph_0<|X|\le2^{2^{\aleph_0}}$.

This will follow from Lemmas 1, 2 and 3 below.

**Lemma 1.** Let $X$ be an infinite set and $\Sigma$ a set of non-isomorphic rigid surjective indecomposable self-maps of $X$. Assume $|\Sigma|>|X|$. Then the coproduct of the elements of $\Sigma$ is a rigid surjective self-map of a set of cardinality $|\Sigma|$.

This is obvious.

**Lemma 2.** Let $f$ be a rigid surjective self-map of an infinite set $X$, and $Y$ a set satisfying $|X|\le|Y|\le2^{|X|}$. Then $Y$ has a rigid self-map.

*Proof.* Let $X'$ be a set disjoint from $X$ and $\phi:X'\to X$ a bijection. For each subset $S$ of $X'$ put $X_S=X\sqcup S$ (disjoint union) and define $f_S:X_S\to X_S$ by setting $f_S(x)=f(x)$ for $x\in X$ and $f_S(s)=\phi(s)$ for $s\in S$.

It suffices to show that the coproduct $g:Y\to Y$ of the $f_S:X_S\to X_S$ (where $S$ runs over all the subsets of $X'$) is rigid.

Let $h:Z\to Z$ be a component of $g$. Then $h$ is a component of $f_S$ for some $S$. It is easy to see that there is a unique component $f_0:X_0\to X_0$ of $f$ such that, if we set $S_0:=S\cap\phi^{-1}(X_0)$, then $h$ is equal to $$ f_{0,S_0}:X_{0,S_0}\to X_{0,S_0}, $$ where $f_{0,S_0}$ is defined as $f_S$ was defined above (replacing the bijection $\phi:X'\to X$ with the bijection $\phi^{-1}(X_0)\to X_0$ induced by $\phi$).

Let $$ f_{1,T_1}:X_{1,T_1}\to X_{1,T_1} $$ be another component of $g$, corresponding to a subset $T$ of $X'$, and let $$ \psi:X_{0,S_0}\to X_{1,T_1} $$ be an isomorphism from $f_{0,S_0}$ to $f_{1,T_1}$. Since $X_0$ and $X_1$ are the respective images of $f_{0,S_0}$ and $f_{1,T_1}$ by surjectivity of $f$, the isomorphism $\psi$ maps $X_0$ onto $X_1$ and $S_0$ onto $T_1$. By rigidity of $f$ we have $X_0=X_1$ and $\psi(x)=x$ for all $x\in X_0$. Let $s$ be in $S_0$. It suffices to show $\psi(s)=s$. Set $x=\phi(s)\in X_0$. Then $\psi$ maps the fiber of $\phi$ above $x$ to itself, but $s$ is the only point in this fiber. This completes the proof of Lemma 2.

**Lemma 3.** Let $A$ be the set of all increasing self-maps of $\mathbb N$ such that $a(0)\ge1$. Then there is a family of pairwise non-isomorphic rigid surjective indecomposable self-maps
$$
(f_a:X_a\to X_a)_{a\in A},
$$
where each $X_a$ is an infinite subset of $\mathbb N^2$.

*Proof.* Define the subset $X_a$ of $\mathbb N^2$ by the condition that $(i,j)\in X_a$ if $i\in a(\mathbb N)$ or if $j=0$, and define $f_a:X_a\to X_a$ by setting

$\bullet\ f_a(i,j)=(i,j-1)$ if $j\ge1$,

$\bullet\ f_a(i,0)=(i-1,0)$ if $i\ge1$,

$\bullet\ f_a(0,0)=(0,0)$.

Let us fix $a\in A$ and sketch the proof that $f_a$ is a rigid self-map of $X_a$.

The point $(0,0)$ is the only fixed point. The points of the form $(i,0)$ with $i\ge1$ are characterized by the fact that they have ancestors which have two parents, and any two distinct such points are at different distances to $(0,0)$. Therefore the points $(i,0)$ are fixed by any automorphism of $f_a$. The point $(a(n),j)$ with $j\ge1$ has no ancestor with two parents, its first descendent with two parents is $(a(n),0)$, which is fixed by the automorphisms of $f_a$, the point $(a(n),j)$ is at distance $j$ from $(a(n),0)$, and these properties characterize $(a(n),j)$. Thus $(a(n),j)$ is fixed by the automorphisms of $f_a$.

This argument shows also that the $f_a$ are pairwise non-isomorphic. The other statements are clear.

[If $y=f(x)$ we say that $x$ is a *parent* of $y$. If $y=f^n(x)$ for $n\in\mathbb N$ we say that $x$ is an *ancestor* of $y$ and $y$ a *descendent* of $x$.]

set-theory