Definition. A function $f:\Bbb R\to\Bbb R$ will be calledpotentially continuousif there is a bijection $\phi:\Bbb R\to\Bbb R$ such that $f\circ \phi$ is continuous.

So one could say a potentially continuous (p.c.) function is "a continuous function with a mixed up domain". I was wondering whether there is an easy way to characterize such functions.

**Some thoughts** $\DeclareMathOperator{\im}{im}$

- If the image $\im(f)$ is not connected (i.e. no interval), then $f$ is not p.c. because even mixing the domain cannot make $f$ satisfy the intermediate value theorem.
- Bijective functions are always p.c. because we can choose $\phi=f^{-1}$. Every injective function with an open connected image is p.c. for a similar reason. However, only having a connected image is
*not*enough, as e.g. there are bijections, but no continuous bijections $f:\Bbb R \to [0,1]$. - Initially I thought a function can never be p.c. if it attains every value (or at least uncountably many values) uncountably often, e.g. like Conways base 13 function. But then I discovered this: take a Peano curve like function $c$ (or any other continuous surjection $\Bbb R\to\Bbb R^2$) and only look at the $x$-component $c_x:\Bbb R\to\Bbb R$. This is a continuous function which attains every value uncountably often.
- The question can also be asked this way. Given a family of pairs $(r_i,\kappa_i),i\in I$ of real numbers $r_i$ and cardinal numbers $\kappa_i\le\mathfrak c$ so that $\{r_i\mid i\in I\}$ is connected. Can we find a continuous function $f:\Bbb R\to\Bbb R$ with $|f^{-1}(r_i)|=\kappa_i$?
- There is no continuous function which attains each real number exactly once
*except*zero which is attained twice. So, e.g. the function $$f(x)=\begin{cases}x-1&\text{for $x\in\Bbb N$}\\x&\text{otherwise}\end{cases}$$ is not p.c., even though its image is all of $\Bbb R$.