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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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    @Chris Eagle I didn't realize you could edit the title, thanks.2011-07-04

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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.2011-07-04