# Verifying 'standup maths' homeomorphism claim

In his newest video, Matt Parker claims that a sphere with three holes (a pair of trousers) and a torus with one hole (a pair of trousers with the legs sewn) are homeomorphic. I assume he meant removing closed discs, since he wanted them to be manifolds.
As justification, he shows that they're homotopy equivalent (to $$S^1\vee S^1$$) and then claims 'because they have thickness' they are homeomorphic.
I'm not fully convinced, but cannot show either way. My thoughts are that they are homeomorphic, but that it's not as simple as he suggested (he's essentially suggesting 'rotating' the perpendicularly glued annuli so that they are glued parallel to each other, in some sort of projection, I'm not convinced this is well-defined).

Thoughts?

• "Because they have thickness" is non-mathematician-speak for "I'm not actually talking about the punctured sphere, but instead a three-dimensional solid which looks like the punctured sphere when you immerse it in R3 just so." Aug 2, 2021 at 19:17

They are not homeomorphic. They are merely homotopy equivalent.

A way to see they are not homeomorphic is that they have different numbers of boundary components (three versus one).

A fancier way (using homology) is to consider the fact that a sphere with three holes can be embedded in the plane, which implies that the algebraic intersection number of any pair of closed loops is $$0$$ modulo $$2$$. But on a torus with one hole, it's easy to come up with a pair of curves that intersect in exactly one point, which means the algebraic intersection number is $$1$$ modulo $$2$$.

Addressing "they have thickness": if we are considering (as Parker does in the video) thickened surfaces, which you might formalize as products of a surface with an interval (and thus are 3-dimensional manifolds), then the thickened surfaces are indeed homeomorphic. They are both genus-2 handlebodies.

In general, if an orientable surface deformation retracts onto a wedge of $$g$$ circles (like $$S^1\vee S^1$$ for $$g=2$$) then their thickenings are homeomorphic to a genus-$$g$$ handlebody.

There is another way you can formalize a thickening that depends on the way a surface is embedded in $$\mathbb{R}^3$$, which is to thicken it into the ambient space (i.e., take a product with the normal bundle, rather than with a trivial $$I$$-bundle like above). You can drop the dependence on orientability with this notion of thickening. An "ambiently thickened" Mobius strip is homeomorphic to an "ambiently thickened" annulus, where both are homeomorphic to a genus-$$1$$ handlebody (a solid torus). But thickened in the first way, they are non-homeomorphic. The Mobius strip gives a non-orientable 3-manifold, but handlebodies are orientable.

• I'm not sure I fully understand (this isn't really my area), surely as stated they are manifolds without boundary? Jul 31, 2021 at 17:14
• @Oddlyasymmetric There are a few ways to deal with that issue. For manifolds without boundary, you can consider the ends of the space, and there will be one end per removed disk. There's also a compactification of the surface where the surfaces become manifolds with boundary. Jul 31, 2021 at 17:18
• @Oddlyasymmetric mentioned in their question the key observation that the shapes Parker is talking about "have thickness", which means he's treating them as 3-manifolds. Doetoe's answer emphasises this, and may be what the question was looking for. On the other hand, Kyle Miller seems to have understood the question to be about 2-manifolds - which is reasonable from its opening sentence. Aug 1, 2021 at 13:16
• @RobinSaunders Thanks for pointing this out (I had minterpreted "have thickness" in the question to mean the thickness of an annulus vs a circle!) I've added some things about thickened surfaces. Aug 1, 2021 at 17:46
• I like your observation about "ambient thickening". Note that while closed unorientable surfaces can't be embedded in $\mathbb{R}^3$, they can be immersed; the ambient thickening still makes sense and gives an immersion of the corresponding 3-manifold. Aug 1, 2021 at 19:16

Watching the video, he seems to be aware that they are not homeomorphic as 2-manifolds.

He says that to justify the crucial step of going from two perpendicularly glued cylinders to two annuli joined on the side within a plane, you have to consider the thickness of the fabric, so that they are 3D objects, rather than the thickness in going from 1D (2 joined circles) to 2D.

• Indeed, the thickness is crucial here to avoid this being what the audience would surely dub a Parker homeomorphism ... Aug 1, 2021 at 9:42
• @HagenvonEitzen I like that. From now on I'll keep using the phrase "Parker homeomorphism" for stuff like this hahaha Aug 4, 2021 at 13:57

There is a loop in the punctured torus whose complement is connected but the complement of a loop in a sphere with three discs removed is never connected, so they cannot be homeomorphic.