tea.mathoverflow.net - Discussion Feed (Continuous bijective way...) 2018-11-04T23:22:45-08:00 http://mathoverflow.tqft.net/ Lussumo Vanilla & Feed Publisher Noah Stein comments on "Continuous bijective way..." (12347) http://mathoverflow.tqft.net/discussion/877/continuous-bijective-way/?Focus=12347#Comment_12347 2011-01-02T17:06:27-08:00 2018-11-04T23:22:45-08:00 Noah Stein http://mathoverflow.tqft.net/account/480/ @Anton: Very nice, but I believe a continuous map needs to be proper for the induced function on one-point compactifications is also continuous. So you have shown that there is no proper map of the ... Anton Geraschenko comments on "Continuous bijective way..." (12343) http://mathoverflow.tqft.net/discussion/877/continuous-bijective-way/?Focus=12343#Comment_12343 2011-01-02T15:56:26-08:00 2018-11-04T23:22:45-08:00 Anton Geraschenko http://mathoverflow.tqft.net/account/2/ I vote to reopen it (meaning that I'll vote to reopen once some other people do). The original question (where "bijective" read "1-1") was awful because of the (sort of sensible) ... I vote to reopen it (meaning that I'll vote to reopen once some other people do). The original question (where "bijective" read "1-1") was awful because of the (sort of sensible) confusion that "injective" = "1-1 or 0-1" and "bijective" = "1-1".

I think there is no such bijection. A line in the plane is almost the same as a plane through the origin in 3-space (intersecting with the plane at height 1), except there's one plane through the origin that doesn't give you a line (the z=0 plane). So the space of lines in the plane is homeomorphic to $\mathbb{RP}^2$ minus a point: an open mobius strip! So the question is asking if there is a continuous bijection $f:D\to M$ from the open disk $D$ to the open mobius strip $M$.

Suppose there were such a bijection $f$, then $f$ would induce a continuous bijection of one-point compactifications, $\bar f:\bar D\to \bar M$, which would have to be a homeomorphism (a continuous bijection from a compact space to a hausdorff space is a homeomorphism). But $\bar D=S^2$ and $\bar M = \mathbb{RP}^2$ are not homeomorphic.

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Andres Caicedo comments on "Continuous bijective way..." (12337) http://mathoverflow.tqft.net/discussion/877/continuous-bijective-way/?Focus=12337#Comment_12337 2011-01-02T14:58:39-08:00 2018-11-04T23:22:45-08:00 Andres Caicedo http://mathoverflow.tqft.net/account/251/ (Not having thought deeply about it) I think there is a potentially interesting question here, and have voted to reopen: ...
I'm opening this thread in case somebody wants to weigh in opinions one way or the other.]]>