Degree and Local Homology

Invariance of Domain

Invariance of Domain

Suppose and are nonempty open sets and are homeomorphic. Then .

Proof Suppose is a homeomorphism. Pick any and let . This homeomorphism induces a map of pairs . This map in turn induces an isomorphism on the relative homology groups: By the Excision Theorem, we can excise the complement of a neighborhood around (and ) to get: The long exact sequence for the pair and the fact that is contractible shows that . Thus, we have:Since these groups are isomorphic for all , we must have .

The Degree of a Map

Degree

The degree of a map is the integer such that the induced map on the -th homology group, , is multiplication by .

Properties of the Degree

Let be continuous maps.

1. .
2. If is not surjective, then .
3. If (i.e., is homotopic to ), then .
4. .
5. If is a reflection across a hyperplane through the origin, then .
6. The degree of the antipodal map, , is .
7. If has no fixed points, then .

Proof Sketch

• (5) Think of as two -simplices, a northern hemisphere and a southern hemisphere , glued along their common boundary (the equator). A generator for is represented by the cycle . A reflection across the equator maps to and vice versa. So, . Thus, .
• (6) The antipodal map can be written as a composition of reflections. For example, in , is the composition of for . Using properties (4) and (5), the result follows.
• (7) If for all , then the line segment from to does not pass through the origin. This allows us to construct a homotopy from to the antipodal map: So . By properties (3) and (6), .

Applications

Hairy Ball Theorem

has a nowhere-vanishing continuous vector field if and only if is odd.

Proof () Suppose is a nowhere-vanishing vector field on . We can normalize it so for all . Since is a vector in the tangent space at , it is orthogonal to . Consider the homotopy: For , , which is the identity map. For , , which is the antipodal map. This shows that . By the properties of degree, this implies . This equality holds only if is even, which means must be odd. () If is odd, say , we can view as the unit sphere in . For , we can define a vector field . This vector field is nowhere zero and is tangent to the sphere.

Free Group Actions on Spheres

If is even, the only nontrivial group (up to isomorphism) that can act freely on is .

Proof Let be a group acting freely on . This means for any with , the map given by has no fixed points. Consider the map given by . This is a group homomorphism. Since the action is free, for any , has no fixed points. By the properties of degree, this implies that . If is even, then is odd, so for all . The kernel of is . Since for all , the kernel is just . This means is an injective group homomorphism. By the first isomorphism theorem, is isomorphic to a subgroup of . The only nontrivial subgroup is itself, which is isomorphic to .

Local Homology

Local Homology

The local homology of a space at a point is defined as: By the Excision Theorem, for any open neighborhood of , we have an isomorphism:

e.g. The local homology of at the origin is:

Induced Maps on Local Homology

Question: Suppose we have a map with . When does this induce a map on local homology groups, ?

Answer: For to be a map of pairs , we need . This is equivalent to the condition that .

However, since local homology only depends on arbitrarily small neighborhoods, a less strict condition is sufficient. We only need there to exist some open neighborhood of such that maps to . This requires , which is equivalent to: This means that is the only preimage of within the neighborhood .

The Local Degree

Local Degree

Let be a map. Suppose for a point , its preimage is a finite set of points . For each , we can find a small enough neighborhood such that . This allows to induce a map on local homology: This map is multiplication by an integer called the local degree of at , denoted .

Remark

If is a local homeomorphism at , then the local degree is either or .

• if is orientation-preserving at .
• if is orientation-reversing at .

Degree Formula

Suppose is a map such that for some , the preimage is a finite set . Then the (global) degree of is the sum of the local degrees at these points:

Proof Choose disjoint open neighborhoods of each . Consider the following commutative diagram which involves maps from the long exact sequences of pairs: By excision, . Let be a generator. The map sends , where is the corresponding generator for . Following the diagram:

• Going down then right: , where is the generator for .
• Going right then down: .

By commutativity, the results must be equal. Thus, .

e.g. Constructing a map of any integer degree. For any and , we can construct a map with . Case : 1. Choose disjoint open disks in . 2. Consider the quotient map that collapses the complement to a single point. This gives a map . 3. Follow this with the “folding” map which is the identity on each sphere in the wedge. 4. Let the resulting map be . If we choose a point (not the collapse point), its preimage will be where each . 5. At each , the map is a local homeomorphism that preserves orientation, so . 6. By the theorem, . A constant map gives degree 0.

Case : First, construct a map of degree as above. Then compose it with a reflection , which has degree . The map will have .

Degree of a Suspension

For any map , its suspension has the same degree.

Proof The proof follows from the naturality of the suspension isomorphism . The following diagram commutes: Since the horizontal maps are isomorphisms and is multiplication by , it follows that must be multiplication by the same integer.

The Splitting Lemma

The Splitting Lemma

For a short exact sequence of abelian groups (or modules): The following are equivalent:

1. There exists a homomorphism such that . (This map is called a retraction).
2. There exists a homomorphism such that . (This map is called a section or a splitting).
3. There is an isomorphism such that .

If any of these conditions hold, the short exact sequence is said to split.

Proof We will show that (1) and (2) are equivalent, and so are all three conditions. : Given a retraction , notice that , we can define by . This is well-defined because if , then , so for some , therefore, . Also, , so . : Given a section , since is injective, , define by . To show this is well-defined, we need to show . By exactness, , and . Also, , so .