Tag Archives: homeomorphism

30 Jul

This talk happened in March and I still remember it (and I was super sleep deprived at the time too).  Immediately after the talk, another grad student and I were chatting in the hallway and marveling at how good it was.  He said something like “I feel like a better person for having gone to that talk.”

A few days later, I ran into the speaker and told her that I had loved her talk, and she said “I’m super unintimidating so feel free to email me or ask me if you have any questions.”

During the talk, at one point she said (again, up to sleep-deprived memory coarseness)

“It’s more important that you learn something than that I get through my talk.  There’s no point in rushing through the material if you don’t take something away from this.”

All of these quotes are to say that this was probably the best talk I’ve seen (and I’ve seen lots of talks).  Particularly because of that last quote above.  Speaker put audience before ego, and that is a rare and beautiful thing (the other contender for best talk I’ve ever seen was by someone who recently won a big award for giving incredible talks).

The good news is that this is something anyone can do – mathematics at this level is a matter of practice and good habits, and not “talent” or “genius”.

OK, done fangirling!  On to the math!

We’ll be talking about a property of groups, so brush up from a previous blog post or wikipedia.  First we need to define a  (total) ordering on a group: a binary relation ≤ that satisfies three properties (which you’d expect them to satisfy):

1. Transitivity: if a≤b and b≤c, then a≤c
2. Totality: for any a, b in the group, a≤b and/or b≤a
3. Antisymmetry: if a≤b and b≤a, then a=b.

A few examples and nonexamples:

• The usual ≤ (less than or equal to) on the real numbers is an ordering.  For the rest of this post, I’ll freely switch between using ≤ to denote being in the group, and being in the real numbers (it should be clear when we’re talking about real numbers).
• Comparing heights of people is not an ordering: it’s not antisymmetric (see picture)
• Ordering words in the dictionary is an ordering: if you’re both before/at the same place and after/at the same place as me, then we must be the same word.
• Consider the group $\mathbb{Z}_2\times\mathbb{Z}_2$, which you can think of as a collection of ordered pairs $\latex \{ (0,0), (0,1), (1,0), (1,1)\}$.  If we define an ordering by (x,y)≤(a,b) if x+a≤y+b, then we’d break antisymmetry.  If we defined it by (x,y)≤(a,b) if x<a and y≤b, then we’d break totality (couldn’t compare (0,1) and (1,0)). Top: reals are good to go. Middle: just because we’re the same height doesn’t mean we’re the same person! Bottom: (0,1) and (1,0) don’t know what to do

• Can you come up with a relation that breaks transitivity but follows totality and antisymmetry?

Notational bit: we say that a<b if a≤b and a does not equal b.

Now we say a group is left orderable if it has a total order which is invariant under left multiplication: this means that a<b implies ga<gb for every g in the group.

Let’s go back to the reals.  If you use multiplication (like 3*2=6) as the group operation, then the usual ordering is not a left(-invariant) order: 2<3, but if you multiply both sides by -2 you get -4<-6, which isn’t true.  However, if you use addition (like 3+2=5) as the group operation, then you see that the reals are left orderable: 2<3 implies 2+x<3+x for every real x.  This is a good example of the fact that a group is a set and a binary operation.  It doesn’t make sense to say the real numbers are left orderable; you need to include what the group operation is.

Here’s an interesting example of a left orderable group: the group of (orientation-preserving) homeomorphisms on the real numbers. (Orientation preserving means that a<b means that f(a)<f(b), all in the reals sense).  If you don’t feel like clicking the link to prev. post, just think of functions.  To prove that the group is left orderable, we just have to come up with a left-invariant order.  Suppose you have two homeomorphisms g and defined on the reals.  If g(0)<f(0), then say g<f.  If f(0)<g(0), then say f<g.  If g(0)=f(0), then don’t define your order yet.  If g(1)<f(1), then say g<f.  And so on, using 2, 3, 4…  Looks like a good left order, right?  WRONG!

If g and f agree on all the integers, they could still be different functions.  So we haven’t defined our order.  We need a better left order.  What can we do?  I know, let’s use a fact!

FACT: the rationals (numbers that can be written as fractions) are countable and dense (roughly, wherever you look in the reals, you’re either looking at a rational or there are a bunch in your peripheral vision).

So now we do the same thing, but using the rationals.  Enumerate them (remember, they’re countable) so use $q_1,q_2,q_3\ldots$ in place of 1,2,3… above.  It’s another fact that if g and f agree on all rationals, then they’re equal to each other.  Let’s make sure we have an ordering:

1. Transitivity: If f≤g and g≤h, then that means there’s some numbers (call them 2 and 3) so that f(2)<g(2) and g(3)<h(3).  But since we had to go to 3 to compare g and h, that means g(2)=h(2).  So f(2)<h(2), so f≤h.
2. Totality: if I have two different homeomorphisms, then there has to be a rational somewhere where they don’t agree, by the second fact.
3. Antisymmetry: We sidestepped this by defining < instead of ≤.  But it works.

Here’s a “classical” THEOREM: If G is a countable group, then it is left orderable if and only if it has an injective homomorphism to $\text{Homeo}_+(\mathbb{R})$.

Remember, injective means that each output matches to exactly one input.  Since we showed that there’s a left order on the group of orientation preserving homeomorphisms on the reals, we’ve already proven one direction: if you have an injection, then take your order on G from the order of the homeomorphisms that you inject onto.  So if h is your injection and g, k are your group elements, say that g<k if h(g)<h(k) in $\text{Homeo}_+(\mathbb{R})$.

One thing Mann does in her paper is come up with an example of an uncountable group that doesn’t do what the theorem says (she also does other stuff).  Pretty cool, huh?  Remarkably, the paper seems pretty self-contained.  If you can read this blog, you could probably do good work getting through that paper (with lots of time), which is more than I can say for most papers (which require lots of background knowledge).

That brings me to the “also”: I’ve been quite tickled to be asked about applying to grad school/what grad school entails a handful of times, some of those times by people who found me via this blog.  So please email me if you’re interested in whatever I have to say on the subject!  I’ve applied to grad school twice and have friends in many different departments and areas.  I hear I can be helpful.

## Good Will Hunting “open” math problem (ha)

30 May

I put an edit in italics to show you where I messed up: maybe you’ll see the mistake too before reading the italics!

Fun little post!  I say “open” problem because that’s what they call it in your favorite Ben Affleck-Matt Damon movie*, but the problem they pose is actually super easy and by the end of this post, you’ll be as good at it as Matt Damon!

Here’s the clip from Good Will Hunting:

If you pause the clip at 2:22, you’ll see what the “open problem” is:

Draw all the homeomorphically irreducible trees with n=10.

And that’s the problem we’ll be dealing with today!

So let’s review/learn what all these words mean.  Remember, a graph is a drawing of dots connected by lines (see these two posts for examples).  A graph is connected if you can follow a path from any dot to any other dot. Left is connected, using the light colored lines (ignoring the three isolated points). The graph on the right is not, since there’s no path from the eye to the mouth.

Now a tree is a connected graph without any loops in it: that is, there’s no way to start heading out from one dot and get back to that same dot without backtracking down the way you came.

One funny thing about graphs: it doesn’t matter which way you draw the graph: all that matters is the way the vertices connect with each other.  So for instance, all of the following are the same graph.

So our GWH problem has to do with drawing a bunch of loop-less graphs, with n=10 dots.  There’s just one last bit of the problem to learn: just like in the topological setting where we learned about it before, a homeomorphism is a function that sends one picture to another picture by wriggling it around continuously.  This is basically what we did in the picture above.

So when I say give me a list of homeomorphically irreducible trees, I want none of the trees you give me to be “reducible” via a homeomorphism.  Rather than a strict definition, I’ll just say that here, homeomorphically irreducible means that none of the dots have exactly two neighbors- if you had such a picture, you could just “erase” that dot and have a picture with about the same amount of information. The left two graphs are homeomorphically reducible. The rightmost one is irreducible, though that doesn’t seem to make it happy.

Instead of diving in to n=10, let’s try something a little smaller, like n=3.  Here are all the homeomorphically irreducible trees with n=3:

Ha ha it’s a joke!  There aren’t any!  If we want to be connected with three dots, there are actually only two possible configurations: a line (maybe kink it around a bit, but it’s homotopic to a straight one), and a triangle.  The triangle is a loop, so we can’t have that on our list.  The line segment automatically has a vertex in the middle with two neighbors, so it’s not homeomorphically irreducible.  So to answer our question, there are no homeomorphically irreducible trees of degree 3.

Okay let’s try something more interesting.  Here are all four trees for n=5: You can tell by the way I track that I’m a tree graph, no time to loop back. Ah, ah ah ah, staying with five, staying with five. Ah ah ah ah, staying with fiiiiiiive.

NOTE!  The blue “F” is homotopic to the blue “4”: try to see how you can swing a leg down and reflect for the homotopy.

Here’s the mistake!  There are only three trees for n=5.  The bottom left one is the same as the blue F and blue 4 by straightening out the kink.  Thanks to Dan in the comments for pointing this out.

Which of these is homeomorphically irreducible?  Just the last one: all the others have a vertex with only two neighbors.

If you want, you can draw all the homotopically distinct trees for n=6, 7, 8, 9 in this way and get to n=10, but let’s try to find a shortcut, shall we?

For this next part, we’ll be talking about the degree of a vertex: the number of other vertices attached to it.  For instance, in our bottom right blue “X” graph, we have four dudes with degree 1, and one vertex with degree 4, since the middle one is connected to all four of the outside guys, but each of them only sees her in the center.

So if I go through all the trees with n=5, I have:

1 graph with 2 of degree 1 and 3 of degree 2 (no good)

2 graphs with 3 of degree 1, 1 of degree 2, and 1 of degree 3 (no good)

1 graph with 4 of degree 1 and 1 of degree 4. (ding ding ding!)

What if I go through all the trees with n=4?

1 graph with 2 of degree 1 and 2 of degree 2 (no good)

1 graph with 3 of degree 1 and 1 of degree 3 (ding ding ding!)

This is just a bunch of data as it is, but maybe if I stare at it long enough I’ll be able to find a pattern that will help me with the problem.  I’ll tell you how I did this problem (literally on the back of a piece of paper right now), and send you to a link as to how to do it using equations.

Looking at the data, I notice that it’s a lot easier to start with the highest degree I allow, and build all the graphs from there.  So for instance, with n=5, I can allow highest degrees of 4, 3, and 2 (and allowing 2 is silly because I don’t want vertices of degree 2).  So I only need to look at graphs with degrees 4 (there’s just the star) and 3 (the upper right and bottom left, which both necessarily include vertices of degree 2).

So for our problem, let’s look at the highest degree possible for a graph, and see how many homeomorphically irreducible trees there are with that degree.

Highest degree 10 is impossible, because you need at least 11 vertices for someone to have 10 neighbors.  So we start with 9, which is just a star.  Add it to our list.

How about 8?  Well, if I start with a vertex and give it 8 neighbors, I’ve got one more vertex to add somewhere.  I can’t add it to the central one because it’s all neighbor’d out.  If I add it to one of the 8 neighbors, I get a vertex of degree 2.  So there are no homeomorphically irreducible trees of size n=10 with highest degree = 8.

However, I can make one with 7.  Use the reasoning above, but since we’re trying to avoid vertices of degree 2, I need to add both of my new vertices to one of the 7 neighbors.  Then do the same with 6- I can’t split the three new vertices up between different neighbors, since that’ll also result in a vertex of degree 2.

Here’s a picture of the cases so far. I bet Kelly Clarkson would be into these graphs: they’re spreading their wings and learning to fly

Not gonna lie, I am getting mighty tired of MS Paint around now.  Phew!  I’ll just tell you the degree sequences for the rest of the graphs, and you draw them, how about that?  That way you’re more like Matt Damon anyway.

For highest degree 5: 5, 3, 3, and seven vertices of degree 1.  5, 5, and eight vertices of degree 1.  5, 3, 3, and seven vertices of degree 1 (there are two graphs with this degree sequence, but they’re different from each other).

There are two graphs with highest degree 4, and two graphs with highest degree 3.

So in total, we have the three trees I drew, and the 3+2+2=7 trees you drew.  That means that we end up with 10 trees, which is exactly what GWH has.  And since we went through all possible combinations of highest degrees and used lots of combinatorial thinking, we’ve proven to ourselves that this is a complete list.

For an answer without proof, see this video from last year (I didn’t know about it when I started this post).

For a more mathy proof, see this link, apparently this answer is due to this Swiss mathematician.

*From googling for the above two links, I realize that in the movie they don’t say this is open, but that it took the MIT math professors two years to solve.  I love it!  Also, hello from Ann Arbor, Michigan!  This is just an explanation for the delay- still traveling, just at a fantastic math conference.

## An open problem in group theory

25 Feb

My last post was about Hee Oh‘s talk at CIRM from that conference I went to last month-it actually covered the first third or so of the first of four lectures she gave.  Étienne Ghys gave seven short talks on his favorite groups, which was a huge blast, so I thought I’d try to share some highlights.  This post is a surprisingly simple open problem in group theory, which talks about functions on a circle.  A circle!  Who would’ve thought we still don’t understand everything there is to know about circles?

If you don’t know what a group is, check out my quick intro post for some examples.  Wikipedia also has a significantly more exhaustive page.

You may remember that I once did a series of posts 1, 2, 4, on the homeomorphisms of the torus.  You don’t need to read all the posts to get this post, I just wanted to point out that at one point I used the notation $Homeo_0(T)$ to indicate the homeomorphisms (continuous functions with continuous inverses) of the torus which are isotopic (wiggle-able) to the identity.  In fact, $Homeo_0(T)$ is a group, very related to the group we’re discussing today.

Instead of homeomorphisms, we can also talk about diffeomorphisms: these are homeomorphisms which are differentiable, whose inverses are also differentiable. Rather than dive into a definition of differentiable here, I’m just going to give you an intuitive definition: differentiable functions are “smooth” instead of chunky. Top is smooth and a differentiable. Bottom isn’t; there are weird kinks in its frown

Some functions are differentiable, and some aren’t (see illustration above).  You can also take second and third and n-th derivatives, and we say functions are n-differentiable if it’s possible to take derivatives.  So in the above example, the red rectangle function is at least 1-differentiable (maybe 2 or 3 or more), but the blue function isn’t differentiable at all.

Notation time: we call a circle $S^1$, the sphere in one dimension.  So a hollow ball would be $S^2$, and so on.  In this post, we’ll be talking about twice-differentiable diffeomorphisms of the circle that preserve the orientation of the circle: so if a point is clockwise from y is clockwise from z, then f(x), f(y), and f(z) are also in clockwise order.  This group is written $Diff^2_+(S^1)$.

Great, now we know the group we’re talking about.  Now let’s get into the nitty-gritty of the problem.  First, a subgroup is a subset of a group which is itself a group.  For instance, a subgroup of the integers, $\mathbb{Z}$ under the operation of addition, is the even integers, $2\mathbb{Z}$.  This is because adding two even numbers gives you an even number (2+2=4).  In contrast, the odd integers are not a subgroup of $\mathbb{Z}$, since adding two odd numbers gives you an even number (3+5=8), which doesn’t lie in the set of odd integers.

Next, we need the concept of a normal subgroup.  FYI, mathematicians really care about normal subgroups: they give us lots of insights about the structure of groups, and they help us cut up groups into smaller, more manageable chunks- lots of times we’ll prove things about normal subgroups in order to say something about the larger group.  We start with a subgroup, call it N.  Then N is normal if for every group elements and in N, $xyx^{-1}$ is also in N.  The $x^{-1}$ means the inverse of with respect to the group operation.  So in the integers under addition, the inverse of 2 is -2, because 2 + (-2) =0.  In the real numbers under multiplication, the inverse of 2 is $\frac{1}{2}$, since $2 \cdot \frac{1}{2} = 1$.

In our example, the even integers is a normal subgroup of the integers (you can convince yourself of this).  It’s pretty easy to find subgroups of most groups, but finding normal subgroups (which aren’t just the identity element or the whole group) can be a little harder.  We say a group is simple if it has no non-trivial normal subgroups.

So here’s the open problem I promised at the beginning: Is $Diff_+^2(S^1)$ simple?

And if you want, here’s another one: is $Homeo (D^2, \partial D^2, area)$ simple?  Those are homeomorphisms of the disk that fix the boundary circle and respect area.

I just find it crazy that we don’t understand everything there is to know about functions of a circle or of a disk!  It’s amazing!

In terms of personal blog time, I did in fact bake last weekend, a lot (we were at a vacation rental house full of bakeware), but I didn’t take photos.  Mostly I baked the cookies, plopped them down in front of our awake and lively friends, and went to bed every night- turns out I’m not great at adjusting to living at 9000 feet.  Day one was those awesome salty shortbread cookies, day two were 3-ingredient peanut butter cookies with added peanuts and chocolate chips, and day three were double chocolate bacon cookies which I totally screwed up on but were still delicious.  Ten people made it through a pound and a half of butter in three days, which is glorious. ## Homeomorphisms of the torus, part IV (topology of the identity component)

30 Dec

See Part I for a definition of homeomorphism and torus and Part II for a bit more linear algebra.

I still owe a Part III for the explanation of the linear algebraic classification of homeomorphisms.  But let’s take a step away from linear algebra and look at shapes (my favorite!)

We know what homeomorphisms are (continuous functions with continuous inverses), with the famous example in the picture below: a coffee cup turns into a donut and vice-versa.

In fact, this pictures doesn’t just show us the homeomorphism (which says where each point of the coffee cup gets sent to in the torus and vice-versa).  It also shows us a homotopy (remember the definition from this post)- essentially, because we can see it traveling through time and back, it’s a homotopy.  And in fact, this homotopy is an isotopy– a type of homotopy where at each frozen point in time, the image is homeomorphic to what we started with.  An example of a homotopy which is not an isotopy is the map that ends up sending $x\mapsto -x$, where x is a real number.  Homotopies take place over time, so I would actually write this map as $\mathbb{R}\times [0,1] \to \mathbb{R}; (x,t)\mapsto (1-t)x+t(-x)$.  So when t=0, we have x mapping to x, and when t=1, x maps to -x.  One reason this isn’t an isotopy is because when t=1/2, all of the real line gets mapped to the point 0.  And mapping everything to 0 isn’t a homeomorphism (what would the continuous inverse be?)

A big part of geometric group theory is using shapes to come up with algebraic theorems, and using algebra to come up with shapes.  One thing you can do (which we’re doing RIGHT NOW!) is take a shape, do some algebra, and then make a new shape.  To be specific, our first shape is the torus.  Our algebra was figuring out the group of homeomorphisms of the torus, also written as Homeo(T)- T for torus.  Sometimes you’ll see $T^2$, to specify that we’re talking about the 2-torus rather than a higher dimension (more on higher dimensional tori later.  Isn’t it cool that the plural of torus is tori?  Pronounced tor-eye.)  Now we’re going to make a new shape from this group of homeomorphisms.

We’ll only consider homeomorphisms isotopic to the identity, written as $Homeo_0(T^2)$.  Starting in 1962 and finishing in 1965, badass Mary-Elizabeth Hamstrom proved in a series of papers that $Homeo_0(X)$ is contractible (homotopic to a point) if X is a two-manifold with a short list of exceptions [torus, sphere, plane, disk, annulus, disk with a hole in it, plane with a hole in it.] Abstractly, I realize that there are many 10-year olds out there who could make a better picture than this. But I’m still so proud of myself. Exceptions to the theorem that the space of homeomorphisms isotopic to the identity is contractible.

Let’s look at our current favorite from this list, $Homeo_0(T^2)$.  If we start from the identity, which homeomorphisms can we isotope to?  Well, I can rotate my torus around its hole-axis, and that ending homeomorphism is definitely isotopic to the identity (the rotation through time is the isotopy; where the points end up is the ending homeomorphism).

Since I can rotate by all the degrees up to 360, which brings me back to the identity, this means that $Homeo_0(T^2)$ contains a circle- each point on the circle represents rotating the torus by that many degrees.

What else can I do that’ll be isotopic to the identity?  I can rotate the torus around its center circle (running through the middle of the donut), like if I was wringing out a towel.

Again, I can do this for 360 degrees before coming back to the identity.  So there’s another circle, different from the first one, in $Homeo_0(T)$.

I can also do any combination of these two: I can rotate 27 degrees around the hole-axis, and then 78 around the center circle.  This is true for any numbers between 0 and 360, but then 0 and 360 are the same for both.  So far, we have the picture on the left.  I colored it to indicate that 0=360 on both axes.  Look familiar? Notice that all four corners are the same homeomorphism: the identity can be had by rotating by 360 degrees in either direction, or in both directions, or by doing nothing.  So we’ve shown that $Homeo_0(T^2)$ actually contains a torus!  This is cool because all those other $Homeo_0(X)$ were contractible.  In fact, $Homeo_0(T^2)$ is a torus- we’ve actually described all the homeomorphisms of the torus which are isotopic to the identity- combinations of rotations around the center axis and around the center circle.

Personally I find this much more exciting than classifying the homeomorphisms by trace (yeup that’s happening in Part III, nope Part IV is coming out before Part III and you’re going to like it), probably because it involves shapes rather than numbers.

Update on health: I’m taking antibiotics to help with whooping cough.  So that explains why I’ve been sick for a month.  My boyfriend walked by and asked what I was doing an hour ago, and I told him I was very busy feeling sorry for myself.  Then I wrote this blog post to be less lump-ish.

## Homeomorphisms of the torus, Introduction (definition of the title)

23 Sep

Besides understanding the proof of the fundamental theorem of geometric group theory, figuring out how to classify the homeomorphisms of the torus is one of the first exercises grad students in my field do.  It involves lots of matrix multiplication and remembering some facts from linear algebra, which we’ll sorta brush over (mostly because I don’t remember anything from my linear algebra class besides that Stephen Goode always said bee-ta and once told us that if we left his class saying bee-ta then we’ll have learned something.  I’d never heard it before so I had no idea what he was talking about until a few years later when I said bee-ta and everyone was like uh you’re not European/Australian… Long story short, Americans say bay-ta for $\beta$).

So.. what’s a homeomorphism of the torus?  Well we know what the torus is: a donut!  Mmmm donuts.  But sometime’s it’s hard to do math on a three-dimensional donut (it’s easy to forget where things connect up on the backside) so we often unfold the torus into a square and draw things on the square instead of on the torus.  We’ve seen this picture before: And the idea is that it describes the torus.  You glue the red edges together so the arrows line up, and you have a cylinder.  Then glue the blue edges (which are now the ends of the cylinder) together so the arrows line up, and you have a torus.  Here’s a cartoon: And here’s some pictures of me doing this in real life:

Visualizing abstractions is one of the hardest skills in math (I think so anyway) but once you’ve started it, it’s difficult to remember that people don’t look at flat squares and see donuts right away.  So hopefully this helped you!

Why would we rather use a torus drawn as a square rather than using a 3D model?  Well, for one, it’s way easier to draw a square than a torus.  But it’s also much easier to mathify things we’re interested in on a square than on a torus: for instance, I can draw a curve on my torus… Taken from George Hart’s brilliant bagel page: the string shows part of the curve, and the dotted lines show where the string continues on the back side.

Or I can use math to describe it.  Notice that the string goes through the hole 3 times (easy to see) and around the hole twice (a little harder to see).  So it goes around the skinny side (the longitude) three times and around the flatter side (the latitude) two times.  Let’s name this the (3,2)-torus knot.  Just like with the curve complex, we don’t care if we wiggle the string a little bit (isotopy).

There’s another cool thing we can do with a square: tile the plane!

I know it doesn’t seem that cool but it is!  If I identify every single square with the torus I had before, I can actually draw the (3,2) curve.  Here it is drawn on just one square, and then drawn on the big grid: Left: follow the numbers to see the knot. Right: look at the bottom-most green line.

It might take a little while to understand this picture.  On the left, start in the lower left corner.  Follow strand 1 and you hit a point about 1/3 the way down the right side.  Since the blue arrows are identified, this is the same as the point about 1/3 the way down the left side, which means we go to strand 2.  Similarly, the middle of the top and middle of the bottom are identified, so we get to strand 3, and finally strand 4 finishes the loop.

Instead of doing that tracing, we could draw the (3,2)-knot on the tiled plane, so that it goes over 3 times and up 2 times.  So we can see this loop as a line with slope 2/3 in the plane, which is way easier to draw than the picture on the left.  I added in all the lines on the right to show how the left and right pictures are related.

To summarize so far: a torus is a donut, and we can think of it as embedded in real space (an actual donut) or more abstractly as a square with the sides identified.  The plus of the flat picture is that it’s easier to draw, and it can tile the plane, which led to us associating loops of the torus with lines of fractional slope in the plane, just like our bagel loop was the line with slope 2/3.

OK, onto the homeomorphisms part.  A morphism is a map $f: X\to Y$ that assigns points in $X$ to points in $Y$.  For instance, the yellow pages describe a morphism from X=People to Y=Addresses.  Yes, I did just refer to the yellow pages.  No, I don’t know if they still exist.  Deal with it.

A homeomorphism, then, is a special kind of morphism.  It goes from one topological object back to itself, so Y=X, and it has an inverse.  In our example, there’s no inverse because if I give you an address, you might give me five people back, while if you give me a person, I’ll give you just one person back.  To have an inverse, your map has to be one-to-one, which means exactly what it sounds like: each person in X gets exactly one address in Y.

There’s one more property that homeomorphisms have to have: continuity.  Roughly, continuity says that if you wiggle around a little bit in $X$, then you only wiggle a little bit in $Y$.  The yellow pages map is super not-continuous, because if you go to your alphabetical neighbor in $X$ you might have to cross all of town in $Y$ to find his address.  This graph is an example of a function which is continuous: THIS CHART TELLS YOU HOW BIG YOUR PUPPY IS GONNA GET http://goldendoodles.com/care/growth_chart.htm

The topic of this series of posts will be homeomorphisms of the torus: invertible, continuous maps from the donut to itself.  Next time we’ll talk about matrices (ooh goody) and how they have to do with these homeomorphisms, and then eventually we’ll classify all homeomorphisms by what kind of matrix they are.