
Maths in a Twist [pdf] - ColinWright
http://www.ima.org.uk/_db/_documents/Maths%20in%20a%20Twist%20from%20MT%20August%202016.pdf
======
NateyJay
By just stretching, I also think it's a Mobius strip:
[http://i.imgur.com/sshZhxs.png](http://i.imgur.com/sshZhxs.png)

The only way to get the arrows to connect right is if three of the square's
corners are the same point.

~~~
drostie
The cutting solution is a bit more easy to follow, in my opinion...

[http://imgur.com/gallery/eBQq4](http://imgur.com/gallery/eBQq4)

------
JadeNB
Since it's Colin himself posting this, I guess that he linked to where he
intended; but, if you're interested in this article, then you may want to look
at more of "Mathematics Today", the publication in which this article appears:
[http://www.ima.org.uk/activities/publications/mathematics_to...](http://www.ima.org.uk/activities/publications/mathematics_today.cfm.html)
.

------
javitury
MAYBE SPOILER

For the last problem you can bend the sheet such as the two arrows point in to
the same direction(up) and are in the same side, that is a rectangle. On the
side of the arrows(left of the rectangle), the bottom half is an arrow
pointing up and the upper half is an arrow also pointing up.

Then you bend the paper such that the lower side meets the center and the wrap
around the upper side around the previous bend. I got a cilinder

~~~
CarolineW
If I've followed you correctly, I don't think you get a cylinder.

~~~
javitury
You are right. In the last step, the top arrow has to be bent outwards and
then joined against the bottom arrow, instead of overlapping like I previously
did/imagined. The result is a figure with one surface and one edge, like the
one NateyJay points out: its a mobius strip.

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

~~~
ColinWright
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._

~~~
panic
It's a Mobius strip:
[http://i.imgur.com/1MVkbta.png](http://i.imgur.com/1MVkbta.png)

~~~
ColinWright
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 ... ?

~~~
panic
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](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.

~~~
johncolanduoni
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)](https://en.wikipedia.org/wiki/Quotient_space_\(topology\))

