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Comment by blintz | Hacker News Reader

After trying the last puzzle for a while, I think I've concluded it's a torus - is that right?

Nope.

Added in edit: I'm not going to answer any other questions or suggestions - I'd love to see the working behind the answers. Why should it be a Torus? Why should it be a Klein Bottle? Why should it be what you claim? Show me your working!

FWIW, as and when I get the chance I will upvote every guess that shows its working.

A Mobius strip.

Working?

This is not the best wording, but it's what I have for now: Label the top, left, bottom, and right edges of the square T, L, B, and R, respectively (so T and L are the ones which must be glued to each other, in suitable fashion). Cut the square down the \ diagonal; this produces two pieces: an L\B triangle and a T\R triangle. Mark the \ sides of these pieces as having to be ultimately glued to each other in the appropriate fashion, and now re-attach them in the way in which L and T are already indicated to be glued (in the process, flipping one of them over). Now one has the equivalent of the M diagram from before, and thus a Mobius strip.

I also find a Mobius strip, I made the same reasoning

I believe the bottom of L (BL) must coincide with the left of T (LT) and the top of L (TL) must coincide with the right of T (RT); however, musn't all four points coincide in order to properly reglue the \ diagonal? Wouldn't this appear as a "crimped Möbius strip" (if it is even possible)?

Another way to get to this, with perhaps wrong reasoning:

1. Deform the figure until the non-arrowed edges are a disc, and the arrowed edges are next to one another, so you have something like a Pac man figure where the arrowed edges are the mouth.

2. This figure clearly has 1 edge (the outer edge) but only 1 side -- to get to the 'back' side, go through the mouth

Since all the matters topologically is number of edges and sides, this is identical to a Mobius strip.

  > Since all the matters topologically
  > is number of edges and sides, ...

That turns out not to be the case. A torus with a hole and a sphere with a hole both have two sides and one edge.

A Klein Bottle with a hole also has one side and one edge, and is not a Möbius strip.

So that's not enough, and might not be right.

Yeah I think a problem with this article is that hints at what you can do in topology, but it doesn't really explicitly tell you what all the rules of the game are.

I guess to some extent that's the difference between an article intended for a general audience, and an actual proper course on point-set topology. There are limits to what you can cover.

Speaking of topology, check out this numberphile video from a couple days ago: https://www.youtube.com/watch?v=k8Rxep2Mkp8 https://www.youtube.com/watch?v=6Qpfv5y-7WU

Cliff Stoll is a treasure.

It's a Mobius strip: http://i.imgur.com/1MVkbta.png

Interesting. Are you sure a single point can be stretched out to a line? Certainly lines can be stretched or shrunk, but having a point become a line seems a bit of a stretch (if you'll pardon the pun).

It's a nice argument, but is it valid ... ?

It isn't valid. If it were, the cylinder and the cone would be identical.

I'm pretty sure it's valid as long as you put the line back to a point at the end.

Hmmm, actually now that I think about it I'm starting to believe it.

The idea is that the red point is being identified with the entire line, not stretched. My math is a little rusty, but I think you can make a diagram like http://i.imgur.com/PchSkyI.png and use the fact that Q1 and Q2 are quotient maps to define g and show it's a homeomorphism.

The argument goes something like: there's a universal property on quotient spaces X with quotient map Q which says that if F(a) = F(b) for all a and b in X such that a and b are identified (a ~ b), there's a unique continuous G such that F = G ∘ Q.

So let's take a and b in the top left space. We have a homeomorphism f such that a ~ b if and only if f(a) ~ f(b). Since quotients take equivalent elements to equal elements, Q2(f(a)) = Q2(f(b)). Therefore there's a unique continuous g1 such that Q2 ∘ f = g1 ∘ Q1. We can apply the the same reasoning to Q1 ∘ f^-1 to get a unique continuous g2 such that Q1 ∘ f^-1 = g2 ∘ Q2. Furthermore, g1 and g2 are inverses: g1 ∘ g2 ∘ Q2 = g1 ∘ Q1 ∘ f^-1 = Q2 ∘ f ∘ f^-1 = Q2, and Q2 is surjective. Therefore g = g1 is a homeomorphism.

This argument is almost there. The theorem you mentioned about the quotient requires that you are taking a topological quotient on the left of your diagram, which prescribes a specific topology for the quotient space[1] (which you have not demonstrated is the topology you are using). Let Q1: X -> Y. This is the topology such that the open sets for Y are precisely those for which their preimage via Q1 is open in X.

If you think this sounds a lot like continuity, you're right, but it adds one crucial detail: continuity only requires that whatever open sets we have for Y, their preimage is open (so for example you can make Y have the topology of a point, and Q1 would still be continuous). For quotient maps, we require that any set in Y with an open preimage be open, which fixes exactly one topology. This makes sense, because otherwise the quotient could pick from a laundry list of (non-isomorphic) topologies which keep the quotient map continuous, breaking the topological invariance of the quotient map.

[1]: https://en.wikipedia.org/wiki/Quotient_space_(topology)

I think that can't be because a torus doesn't an edge, and the figure clearly still has one

Correct. If it helps, the given diagram clearly has exactly one edge and exactly one surface.

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