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Suppose $f: \mathbb{R} \rightarrow \mathbb{R}$ is continuous.

Is it possible for $f$ to assume each value in its range an even number of times?

To clarify, some values might be taken 0 times, some 2, some 4, etc., but always an even (and therefore finite) number.

I don’t require that there be a value that is assumed any particular number of times.

For example, the function might be surjective, or never take on any value exactly twice.

This question is possibly the same as A continuous function cannot take every value an exact even number of times?.

That question may instead have meant, “If $n$ is even, is it possible for $f$ to assume every value in its range exactly $n$ times?”

In any event, it has not been answered.

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Yes, this is possible. Define $f$ as follows. For $x\leq 0$, $f(x)=-x$. If $n\in\mathbb{N}$, then

$$f(n+x)=\begin{cases} n+3x &\text{ if }0\leq x\leq 1/3 \\ n+2-3x &\text{ if }1/3\leq x\leq 2/3 \\ n-2+3x &\text{ if }2/3\leq x\leq 1 \end{cases}$$

A glance at a graph shows that $f$ is continuous and achieves every positive value four times, achieves the value $0$ twice, and is never negative.

On the other hand, it is impossible for every point in the range of $f$ to have *the same* even number of preimages. For a contradiction, suppose $n$ is an even integer and $f:\mathbb{R}\to\mathbb{R}$ achieves every value in its range exactly $n$ times. Fix some value $a\in f(\mathbb{R})$ and let $x_1<x_2<\dots<x_n$ be the preimages of $a$. Let $p$ be the number of $x_i$ which are local minima of $f$, $r$ be the number of $x_i$ which are local maxima of $f$, and $q$ be the number of $x_i$ which are neither. Then it follows from the intermediate value theorem that if $\epsilon>0$ is sufficiently small, $f$ achieves the value $a+\epsilon$ at least $2p+q$ times near the $x_i$ and $f$ achieves the value $a-\epsilon$ at least $2r+q$ times near the $x_i$. Thus $2p+q\leq n$ and $q+2r\leq n$. But $p+q+r=n$, and so adding these two inequalities together we find that actually $2p+q=n=2r+q$ and thus $p=r$. Since $n$ is even, this implies $q=n-p-r$ is also even.

That is, there are an even number of $x_i$ at which $f(x)-a$ changes sign. Thus $f(x)-a$ has the same sign on both components of $\mathbb{R}\setminus [x_1,x_n]$. Suppose WLOG that $f(x)-a$ is always positive on $\mathbb{R}\setminus [x_1,x_n]$. It follows that $f$ has a global minimum value $b$ which it achieves somewhere on $[x_1,x_n]$.

But now replace $a$ by $b$ and repeat the argument above. Every preimage of $b$ must be a local minimum, so $p=n$, which is clearly impossible. This contradiction means that no such $f$ can exist.

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