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Definition. A function $f:Bbb RtoBbb R$ will be called potentially continuous if there is a bijection $phi:Bbb RtoBbb R$ so that $fcirc phi$ 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 $Bbb RtoBbb R^2$) and only look at the $x$-component $c_x:Bbb RtoBbb R$. This is a continuous function which attains every value uncountably often.


• The question can also be asked this way. Given a family of pairs $(r_i,kappa_i),iin I$ of real numbers $r_i$ and cardinal numbers $kappa_ilemathfrak c$ so that ${r_imid iin I}$ is connected. Can we find a continuous function $f:Bbb RtoBbb R$ 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
  $$f(x)=begin{cases}x-1&text{for $xinBbb N$}\x&text{otherwise}end{cases}$$
  is not p.c., even though its image is all of $Bbb R$.