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Let $X$ be a Hausdorff space such that the irrationals $\mathbb P$ (in their usual topology) form a dense subspace of $X$.

Let $C$ be the Cantor set. The set of "non-endpoints" of $C$ is homeomorphic to $\mathbb P$.

Question. If $f:C\to X$ is a continuous surjection such that $f\restriction \mathbb P$ is the identity map, then is $X$ necessarily disconnected?

NOTE: I suppose we could ask the same question with $\mathbb P$ replaced with the rationals $\mathbb Q$ (identifying the endpoints of $C$ with $\mathbb Q$).

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  • $\begingroup$ "The Cantor set" is vague. You probably mean a Cantor subspace of the real line (e.g., the "standard" one), since otherwise the notion of endpoint is senseless. $\endgroup$ – YCor Nov 18 '17 at 20:19
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    $\begingroup$ Yes. Or in the $2^\omega$ sense the "endpoints" would mean the eventually constant sequences... $\endgroup$ – Forever Mozart Nov 18 '17 at 20:20
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    $\begingroup$ So you have the given inclusion $i:P\subset C$, and in addition a dense embedding $j:P\to X$, with $X$ Hausdorff (possibly assuming that $j$ induces a homeomorphism $P\to j(P)$ and that $j(P)$ has countable complement). You suppose the existence of a continuous map $f:C\to X$ such that $f\circ i=j$. Your question seems to be: does it follow that $X$ is disconnected? $\endgroup$ – YCor Nov 18 '17 at 20:26
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    $\begingroup$ Check for instance p30 of eudml.org/doc/172152 (Keane: Interval exchange transformations). Here he doubles all points, but you can glue back all pairs of points outside a countable set of pairs. But again, you have a very natural examples: the map mapping a decimal expansion to the real it defines. It's almost injective, but there are countably manyt fibers of size two (e.g. $734999999\dots$ and $735000000\dots$ which are both mapped to the same real $0.735$), and it maps a Cantor set onto a segment. $\endgroup$ – YCor Nov 18 '17 at 23:35
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    $\begingroup$ @მამუკაჯიბლაძე You may have this intuition, even if it's not necessary to really think of the gaps if you don't care about a real embedding for the resulting Cantor set. $\endgroup$ – YCor Nov 19 '17 at 8:39
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To rephrase and explicitate YCor's example in the comments, consider the devil's staircase function: it maps the non-endpoints of the standard Cantor set $C$ to the non-dyadic reals in $[0,1]$. Now compose (on the left) with the inverse of the question mark function, which is a self-homeomorphism of $[0,1]$ restricting to an increasing bijection between the dyadics to the irrationals. The composition $[0,1]\to[0,1]$ is continuous, nondecreasing, its restriction to $C$ is a surjection $C\to[0,1]$ mapping bijectively the set of non-endpoints of $C$ to the set of irrationals in $[0,1]$.

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