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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? –  Qiaochu Yuan Jul 4 '11 at 2:54
    
@Qiaochu Sorry yeah, I wasn't entirely sure how to show the map I came up with was continuous. –  Harry Stern Jul 4 '11 at 6:00
    
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? –  Harry Stern Jul 4 '11 at 6:08
    
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.) –  Qiaochu Yuan Jul 4 '11 at 11:01
    
Alternately, work in polar coordinates. –  Qiaochu Yuan Jul 4 '11 at 11:02

1 Answer 1

The Riemann mapping theorem states that any non-empty, simply connected open subset of the complex plane, that is not the whole plane, is biholomorphic to a disk.

So the triangle and disk are biholomorphic, which implies homeomorphic.

For a more elementary approach that doesn't use the RMT that is essentially what you proposed above, choose a fixed point in the interior of your triangle and make it the origin. Map any other point $P$ in the interior of the triangle to $P/d$, where $d$ is the length of the segment from the origin to the boundary of the triangle that passes through $P$. This gives a homeomorphism on to the unit disk.

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While the Riemann mapping theorem is nice, I'm looking for a more elementary proof. As you said, the explicit map gives a homeomorphism. I was asking for a proof that it is one. –  Harry Stern Jul 4 '11 at 19:10

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