The question is in the title. Let $X$ be an arbitrary infinite set. Is there always a surjective function $f:X\rightarrow X$ which is two-to-one?
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Yes, at least assuming axiom of choice. What do you know about cardinal numbers? For example, this is fairly straightforward by transfinite induction if you know the well-ordering principle, and also almost immediate from of Tarski's theorem (that $X\sim X\times X$ if $X$ is infinite) and the Cantor-Bernstein-Schroeder theorem. – tomasz Feb 03 '22 at 00:48
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Impose order type $\lambda + \lambda$ on $X$ (where $\vert X \vert = \lambda$) to see a fairly straightforward $2$-$1$ mapping. – Robert Shore Feb 03 '22 at 01:47
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Let $Y$ be a set of the same cardinality as $X$ but disjoint from $X$, $g\colon Y\to X$ be a bijection Then $Z=X\sqcup Y$ has the same cardinality as $X$ since $X$ is infinite. Let $f\colon X\to Z$ be a bijection.Then the map $h\colon X\to X$ defined as $h(x)=f(x)$ if $f(x)\in X$ and $h(x)=g(f(x))$ if $f(x)\in Y$ is a $2$ to $1$ surjective map.
For the ZF issues with that, see this question and answer.
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