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A cell is any subset of the plane homeomorphic to a disk.

Could someone provide a complete proof that a triangle is homeomorphic to a disk? I have two ideas but I can't seem make them fully rigorous.

One would be via this explicit mapping $f$: Map the boundary of the triangle to the boundary of the disk, and since they are both Jordan curves we're okay so far. Now, map the centroid $C$ (? I think the centroid is always inside the triangle, unlike the orthocenter or circumcenter) of the triangle to the center of the disk and then map radii to radii. Specifically, for each point $P$ on the boundary of the triangle, map the line $PC$ to the line $f(C)f(P)$.

The other way would be showing something like all closed connected subsets of the plane with Euler characteristic 2 are homeomorphic.

Feel free to use any definition of homeomorphism or continuity that you would like.

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    What's wrong with the explicit map $f$? Are you having trouble proving that it's continuous or what?2011-07-04
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    @Qiaochu Sorry yeah, I wasn't entirely sure how to show the map I came up with was continuous.2011-07-04
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    If we use the inverse image of open sets are open definition, then certainly the radii are open sets (if we don't include the endpoints) but if we pick an arbitrary open ball inside the circle...? Is the inverse image open because we can imagine that we're taking the union of all the radii, which are open?2011-07-04
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    If you're having trouble using that definition, it should be a little easier to show that $f$ and its inverse both preserve convergent sequences. (Actually you only have to show that $f$ is continuous to prove that it's a homemorphism, since it's a general fact that a continuous bijection between a compact space and a Hausdorff space is a homeomorphism.)2011-07-04
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    Alternately, work in polar coordinates.2011-07-04
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    @Chris Eagle I didn't realize you could edit the title, thanks.2011-07-04

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