Mayer-Vietoris sequence

[PWrong]

Ok, I'm finally going to attempt to explain the Mayer Vietoris sequence.

It's basically a tool for working out homology groups of X, given the homology groups of smaller shapes.

The first step is to find two sets that combine to give the whole set, but whose intersection is not empty. That is, choose A and B, such that AuB = X and AnB is nonempty, where u is union and n is intersection.

We have to choose these so that we already know the homology groups of A, B and AnB.

The Mayer-Vietoris sequence is a sequence of maps between the homology groups of all the groups involved. It looks like this
http://en.wikipedia.org/wiki/Mayer%E2%80%93Vietoris_sequence#Unreduced_version

By working out the what the inclusion maps do to elements of the homology groups, and then doing a thing called diagram chasing, you can solve the sequence for H_qX. The part that I want to talk about is how to choose A and B. This depends on the kind of shape X is.

[PWrong]

Extrusions:
If X is an extrusion with base Y e.g. cube is an extruded square, cylinder is an extruded circle, torinder is an extruded torus, then we choose:
A = Y x S⁰
B = ∂Y x B¹
AnB = ∂Y x S⁰

So one shape is a pair of Y's, the other shape consists of all the lines between them, but only on the boundary of Y. In the frame notation I made up yesterday, to find F_kX = F_kYI we define

A = F_kY x S⁰
B = F_k-1Y x B¹

Example: wireframe cube
The base Y is a wireframe square.
A is a pair of parallel wireframe squares, one above the other.
B is 4 parallel lines arranged in a square.
AnB is 8 points arranged in a cube.

You can see this is a problem for minimum and maximum frames. If k is the maximum frame of X, e.g. X is a solid cube, then
A = F_kY x S⁰ = empty set
B = F_k-1Y x B¹ = solid cube
AnB = empty set.
But this means we can't use Mayer-Vietoris since the intersection is empty.

Similarly if k is the minimum frame, e.g. X is a wireframe cylinder and Y is a circle, then
A = F₁Y x S⁰ = wireframe cylinder
B = F₀Y x B¹ = empty set, since a circle doesn't have a 0-frame.

[PWrong]

Hyperspheres:

The trick for solving an n-sphere (or sphere in n+1 dimensions) is to let
A = X_N = the northern hemisphere, including the equator
B = X_S = the southern hemisphere, including the equator
AnB = the equator.

Now A can be deformed into an n-ball, which can be deformed to a point.
Now A can also be deformed into an n-ball, which can be deformed to a point.
AnB is an (n-1)-sphere.

By solving Mayer-Vietoris for this sets, you can get a recurrence equation:
H_qSⁿ = H_q-1S^n-1
for q > 1, n > 1.
For 0 and 1 the groups are easy to work out, and you end up with the equation for homology groups of n-spheres.

[PWrong]

Torus, duocylinder:

Here, we can think of the shape as a square with opposite edges identified.
http://en.wikipedia.org/wiki/File:TorusAsSquare.svg

Draw a small square in the middle. Let A be the inside of the square, including the square itself. Let B be the outside of the square, including the square itself.
Then A can be deformed to a point, B can be deformed to a wedge sum of two circles, and AnB can be deformed to a circle.

We can solve Mayer-Vietoris for these subsets and get the results we know.

(22), ((21)1), 222:
We take a similar approach except with cubes. These shapes are topologically the same as a cube with opposite faces identified. A is a small solid cube, B is everything outside the cube, and the intersection is the boundary of the cube.

(31), (211):
The method I used for these is to cut the shape in half just like hyperspheres.

So for (31), circle # sphere, we get
A = circle spherated by a northern hemisphere, which deforms into circle x disk, which deforms into circle
B = circle spherated by a southern hemisphere, which deforms into circle x disk, which deforms into circle
AnB = circle spherated by circle, which deforms into duocylinder or torus.

For (211), sphere # circle, we get
A = northern hemisphere # circle, which deforms into disk x circle, which deforms into circle
A = southern hemisphere # circle, which deforms into disk x circle, which deforms into circle
AnB = circle # circle, which deforms into duocylinder or torus.

So these shapes should have the same homology groups, unless the inclusion maps are radically different. I'm going to redo all the toratopes later, to check that I had the maps right the first time. I might get some ideas for how to do other shapes.

[PWrong]

Does any of this make any sense?

[Keiji]

Yes, that all makes sense.

But how do you now find the homology group from those of A, B and AnB?

[PWrong]

Well those are always simpler than the original shape. So you start with the simplest things, like the homology groups of a point, and work your way up using Mayer-Vietoris repeatedly.

[Keiji]

I meant how do you find the homology group you want after you've chosen A, B, AnB and know their homology groups.

[PWrong]

Oh. Well that's really complicated, but it can be fun solving them. I'll have a go at explaining it, using the sphere as an example. I'll have to explain each of the following unless you know them already:

groups
homomorphisms
injective or "1-1" maps
surjective or "onto" maps
image of a map
kernel of a map

I write:
--> for a homomorphism
>--> for an injection
-->> for a surjection
>-->> for a bijection (both 1-1 and onto)

X = sphere
A = northern hemisphere including equator
B = southern hemisphere including equator
AnB = equator

We have:
H₀A = H₀B = Z
H_qA = H_qB = 0 for q > 0

H₀AnB = Z
H₁AnB = Z
H_qAnB = 0 for q > 1


The Mayer-Vietoris sequence looks like this (http://en.wikipedia.org/wiki/Mayer%E2%8 ... ed_version )
... --> H₃X --> H₂AnB --> H₂A⊕H₂B --> H₂X --> H₁AnB --> H₁A⊕H₁B --> H₁X --> H₀AnB --> H₀A⊕H₀B --> H₀X --> 0

By subbing in what we know, we get this:
... --> H₃X --> 0 --> 0 --> H₂X --> Z --> 0 --> H₁X --> Z --> Z⊕Z --> H₀X --> 0

How are you following so far?

[Keiji]

Gosh, the last time I looked at this post, it confused me so much I fell asleep!

I understand the substitution you performed. But how does one look at "H₃X --> 0 --> 0 --> H₂X --> Z --> 0 --> H₁X --> Z --> Z⊕Z --> H₀X --> 0" and derive the values of H₃X, H₂X, H₁X and H₀X?

[PWrong]

Sorry, it's a bit like that. I nearly fell asleep a few times in class learning this thing. It is difficult, and I guess you don't really have the background for it . Anyway explaining it should be a good exercise for me even if you don't understand anything, but I think you'll get it. I'll just give you the bare minimum of background you need understand the next step in this post. I'll define the 6 things I mentioned and let you ask me about those before I go any further with the sequence.

Here's the wiki articles in case my definitions don't help.

http://en.wikipedia.org/wiki/Group_%28mathematics%29
http://en.wikipedia.org/wiki/Group_homomorphism
The wiki articles about kernel seem to talk about slightly different things. This is probably the best one:
http://en.wikipedia.org/wiki/Kernel_%28algebra%29
http://en.wikipedia.org/wiki/Image_%28mathematics%29
http://en.wikipedia.org/wiki/Injective_function
http://en.wikipedia.org/wiki/Surjection

1. A group is a set with a binary operation like "x" or "." on it. In this case the groups are abelian (commutative) so we use "+". A group has the following properties:
a) There is an identity element (in this case 0) such that x+0 = x for every x.
b) Every element x has a unique inverse, (-x) with the property that x + (-x) = 0
c) "+" is associative.

In this case all the groups are either the trivial group 0, or the group of integers Z with addition, or the direct sum of many Z's.
An element of Z+Z looks like (a,b) where a and b are integers. We define "+" on Z+Z by (a,b) + (c,d) = (a+c, b+d).

2. Homomorphisms are maps between groups that preserve the addition structure. If (A,+) and (X,'+') are groups and f: A --> X then
f(a + b) = f(a) '+' f(b)

3. Image:
If you have a homomorphism f: X --> Y, then maybe not everything in Y comes from X. We say y is in the image of Y if there is some x in X with f(x) = y. So

For example you can have a map x2: ℤ --> ℤ that multiplies every integer by 2. But 3 is an element of the ℤ on the right but it doesn't come from multiplying something by 2. So 3 is not in the image of f, but 4 is. We write "Im f = f(ℤ) = the even numbers"

Another example: define f: ℤ --> ℤ⊕ℤ by f(x) = (x, -x). The image of f is a diagonal line. It's the set {(x,-x) : x in ℤ} which is a subset of ℤ⊕ℤ

4. Kernel:
If you have a homomorphism f: X --> Y, often a lot of things in X get sent to zero by f. The kernel of f is the set of elements x in X such that f(x) = 0.

Examples:
Define f: ℤ --> ℤ by f(x) = x + 2. The only element in ℤ that gets sent to 0 is -2. So we write
ker f = {-2} as a subset of the first ℤ.

Define f: ℤ⊕ℤ --> ℤ by f(x,y) = x+y.
The kernel of f is the set of pairs of integers that add to zero. So
ker f = {(x,-x) : x in ℤ} which is a subset of ℤ⊕ℤ

5. A homomorphism is injective or 1-1 if "every unique argument has a unique result". So if f(a) = f(b), the only way this can happen is if a=b.
For homomorphisms, f is injective if and only if the kernel is 0.
I write A >--> B or A >-f-> B if f is an injection from A to B.

Example: the "x2" function is 1-1 since every even number can be divided by 2 to get the original. Also notice that the kernal is 0 since only 0x2 = 0.

Example: f: ℤ⊕ℤ --> ℤ by f(x,y) = x+y. This f is not 1-1 since different pairs of integers can add up to the same thing.


6. A homomorphism f: X --> Y is surjective or "onto" if every element of Y comes from somewhere in X. Also, f is onto if and only if the image of f is Y.

Example: Define f: ℤ⊕ℤ --> ℤ by f(x,y) = x+y.
This is onto since every integer can be found by adding up two integers. The image is ℤ

Example: Define f: ℤ --> ℤ⊕ℤ by f(x) = (x, -x)
This is not onto. The point (1,2) is in ℤ⊕ℤ but you can't reach it with the map f.
The image is not all of ℤ⊕ℤ, but rather the subset {(x,-x) : x in ℤ}

I write A -->> B or A -f->> B if f is an injection from A to B.

7. Finally, a bijection is a homomorphism that is both 1-1 and onto. These maps have an inverse map that works nicely. We write A >-->> B or A >-f->> B.
If A >-->> H_qX, then A and H_qX are effectively the same group. If we find a bijection like this for every q then the Mayer-Vietoris sequence is solved.

There's a lot here to take in, so take your time and let me know if there's anything I can explain better.

[Keiji]

Well, I understood all of that, so feel free to continue...

[PWrong]

Awesome

Ok so what we have is this:
... --> H₃X --> 0 --> 0 --> H₂X --> Z --> 0 --> H₁X --> Z --> Z⊕Z --> H₀X --> 0


The first thing to do is work out what some of these maps are. We don't know anything about maps going in or out of H_qX, but we can figure out the others. In some ways this is the hardest part; it involves a lot of visualisation and guesswork. Lets look at the first map

phi₁: H₀AnB --> H₀A⊕H₀B
phi₁: Z --> Z⊕Z

Now remember what we're looking at. The first group is related to linear combinations of points on the equator. An element of the second group is a pair of points, one on the northern hemisphere and one on the southern hemisphere.

We define two inclusion maps i₁ and i₂. These are just identity maps.
i₁: equator --> northern hemisphere: every point goes to itself.
i₂: equator --> southern hemisphere: every point goes to itself.
Clearly these maps are 1-1, but not onto.

Now the map phi₁: H₀AnB --> H₀A⊕H₀B is related to the function (i₁ - i₂). In fact we call it (i₁ - i₂)_*

Let's take the number 1 from H₀AnB = Z and see where it goes. 1 corresponds to a single point on the equator. Since we're using the positive i₁ and the negative i₂, we have
phi₁(1) = (1,-1)
This is an element of H₀A⊕H₀B = Z+Z.

Now we know phi is a homomorphism, so phi₁(2) = phi₁(1 + 1) = (1,-1) + (1,-1) = (2,-2).
Similarly, for any x in Z, we have phi₁(x) = (x,-x).

So that defines the first map.
phi₁:
Z --> Z⊕Z
x|--> (x,-x)

Following so far?

[PWrong]

Oh, there's a couple more very important things I forgot to mention. The Mayer-Vietoris sequence is an exact sequence.

This means two things.
1. If you follow one map and then the next one, you always get zero.
2. The image of one map and the kernal of the next map are the same.

So if I have:
A -^f-> B -^g-> C

then I always have g(f(a)) = 0 for any a, and ker g = im f.
Note that ker g and im f are both subsets of B so this makes sense. Also, the second fact implies the first.


One more thing:
Often in large MV sequences you get something like the following, where you know A and C and want to calculate B.
A >--> B -->> C

Here the first map is 1-1 and the second is onto. In these cases I cheat a bit. I just assume that B = A⊕C. This always works since you can define the two maps as follows.

f: A >--> A ⊕ C
f(a) = (a, 0)

g: A⊕C -->> C
f(a,c) = c

It's easy to see that f is 1-1 and g is onto. Unfortunately I haven't been able to prove that this is the only solution, and I suspect it isn't. I have to ask my lecturer at some point if this is the right approach.

[PWrong]

Ok the other two maps that we can figure out both have to be the 0 map. The next step is to replace all the arrows with the arrows representing 1-1 or onto maps wherever possible. I can't really write out the whole thing on here, but you could have a go if you like. Remember the image of one map is the kernal of the next.

[PWrong]

My computer's working now, so I can finish this. The following explanation isn't great, mostly because the sphere isn't the best example. I'll do something more general in my next post.


... --> H3X --> 0 --> 0 --> H2X --> Z --> 0 --> H1X --> Z --> Z⊕Z --> H0X --> 0

Clearly a map from anything to 0 has to be the 0 map, that sends everything to 0. This map is clearly onto.

Also, a homomorphism from 0 to anything is also the 0 map. Proof:
f(0) = f(a - a) = f(a) - f(a) = 0
This map is clearly 1-1 because the kernel is {0}.

So wherever we see 0's we can put 1-1 and onto signs nearby.

... --> H₃X -->> 0 >-->> 0 >--> H₂X --> Z -->> 0 >--> H₁X --> Z --> Z⊕Z --> H₀X -->> 0

This is a good start, but we don't have any bijections yet. Let's just look at the map between Z and Z⊕Z. I called this map phi₁ before, but I should have called it phi₀.


phi₀: H0AnB --> H0A⊕H0B
phi₀: Z --> Z⊕Z
phi₀: x|--> (x,-x)

Now as we saw earlier, the image of this map is {(x,-x) : x in ℤ}. This is isomorphic to just ℤ.
The kernel of phi₀ is just {0}.

Now look at the map after Z⊕Z. Call this tau₀. Because the sequence is exact, the kernel of tau₀ is the image of phi₀.

Everything on the diagonal line goes to 0. So the group of things that don't go to zero is the complement of this, which is the other diagonal line. The other diagonal line is also the same as Z, and it's mapped 1-1 to H₀X.

So we can write a new short sequence:
Z >--> H₀X -->> 0

Now we can conclude that H₀X is Z⊕0 = Z

[PWrong]

Before I do the rest of the sphere I'll do a general rule for this kind of thing. Suppose we have:

H_q+1 --> A --> B --> H_q

We'll call the maps in between f,g and h respectively. At this point we've figured out what g is but we don't know f and h. So

H_q+1 -f-> A -g-> B -h-> H_q


Now we know that
im f = ker g
ker h = im g

So if we replace A with the kernel of g, then f will be onto. If we replace B with the complement of the image of g (that is, B\ im g), then h will be 1-1.
We can forget about the map between the two groups.

H_q+1 -->> (ker g)

(B\ im g) >--> H_q

If we do this at every step, then we can work out all the homology groups.

[PWrong]

So at the moment we have
... 0 >--> H₃X -->> 0 >-->> 0 >--> H₂X --> Z -->> 0 >--> H₁X --> Z --> Z⊕Z --> H₀X -->> 0

Doing the stuff in the previous post, we have


...0 >--> H₃X -->> 0 >-->> 0 >--> H₂X -->> Z -->> 0 >--> H₁X -->> 0 --> Z >--> H₀X -->> 0

Now we clearly have

H₀X = Z
H₁X = 0
H₂X = Z
H₃X = 0

and all the rest are 0 because they're surrounded by 0's.

[Keiji]

I still can't really figure out how you're making those conclusions.

[PWrong]

Unfortunately some of it is based on guesswork and some is based on "this is what the lecturer did so it must be right". But I can explain some of it. What's the first thing I said that you don't get?

[Keiji]

...never mind. I re-read it, and now it magically makes sense

[PWrong]