proof of Borsuk-Ulam theorem

proof of Borsuk-Ulam theorem

Proof of the

Borsuk-Ulam theorem

: I’m going to prove a

stronger

statement than the one given in
the statement of the Borsak-Ulam

theorem

here, which is:

Every odd (that is, antipode-preserving) map

f

:

S

n

→

S

n

has odd degree.

Proof: We go by

induction

on

n

. Consider the pair

(

S

n

,

A

)

where

A

is the equatorial sphere.

f

defines a map

f

~

:

ℝ

⁢

P

n

→

ℝ

⁢

P

n

. By cellular approximation, this may be
assumed to take the hyperplane at infinity (the

n

-

1

-cell of the standard cell

structure

on

ℝ

⁢

P

n

) to itself. Since whether a map lifts to a

covering

depends only on its

homotopy
class

,

f

is

homotopic

to an odd map taking

A

to itself. We may assume that

f

is such a map.

The map

f

gives us a

morphism

of the

long exact sequences

:

H

n

⁢

(

A

;

ℤ

2

)

→

i

H

n

⁢

(

S

n

;

ℤ

2

)

→

j

H

n

⁢

(

S

n

,

A

;

ℤ

2

)

→

∂

H

n

-

1

⁢

(

A

;

ℤ

2

)

→

i

H

n

-

1

⁢

(

S

n

,

A

;

ℤ

2

)

f

*

↓

f

*

↓

f

*

↓

f

*

↓

f

*

↓

H

n

⁢

(

A

;

ℤ

2

)

→

i

H

n

⁢

(

S

n

;

ℤ

2

)

→

j

H

n

⁢

(

S

n

,

A

;

ℤ

2

)

→

∂

H

n

-

1

⁢

(

A

;

ℤ

2

)

→

i

H

n

-

1

⁢

(

S

n

,

A

;

ℤ

2

)

Clearly, the map

f

|

A

is odd, so by the

induction hypothesis

,

f

|

A

has odd degree.
Note that a map has odd degree if and only if

f

*

:

H

n

⁢

(

S

n

;

ℤ

2

)

→

H

n

⁢

(

S

n

,

ℤ

2

)

is an

isomorphism

. Thus

f

*

:

H

n

-

1

⁢

(

A

;

ℤ

2

)

→

H

n

-

1

⁢

(

A

;

ℤ

2

)

is an isomorphism.
By the commutativity of the

diagram

, the map

f

*

:

H

n

⁢

(

S

n

,

A

;

ℤ

2

)

→

H

n

⁢

(

S

n

,

A

;

ℤ

2

)

is
not

trivial

. I claim it is an isomorphism.

H

n

⁢

(

S

n

,

A

;

ℤ

2

)

is generated by cycles

[

R

+

]

and

[

R

-

]

which are the

fundamental classes

of the upper and lower hemispheres, and the antipodal
map exchanges these. Both of these map to the fundamental class of

A

,

[

A

]

∈

H

n

-

1

⁢

(

A

;

ℤ

2

)

. By the commutativity of the diagram,

∂

⁡

(

f

*

⁢

(

[

R

±

]

)

)

=

f

*

⁢

(

∂

⁡

(

[

R

±

]

)

)

=

f

*

⁢

(

[

A

]

)

=

[

A

]

. Thus

f

*

⁢

(

[

R

+

]

)

=

[

R

±

]

and

f

*

⁢

(

[

R

-

]

)

=

[

R

∓

]

since

f

commutes with the antipodal map. Thus

f

*

is an isomorphism on

H

n

⁢

(

S

n

,

A

;

ℤ

2

)

. Since

H

n

⁢

(

A

,

ℤ

2

)

=

0

, by the exactness of the

sequence

i

:

H

n

⁢

(

S

n

;

ℤ

2

)

→

H

n

⁢

(

S

n

,

A

;

ℤ

2

)

is

injective

, and so by the commutativity of the diagram (or equivalently
by the

5

-lemma)

f

*

:

H

n

⁢

(

S

n

;

ℤ

2

)

→

H

n

⁢

(

S

n

;

ℤ

2

)

is an isomorphism. Thus

f

has odd degree.

The other statement of the Borsuk-Ulam theorem is:

There is no odd map

S

n

→

S

n

-

1

.

Proof: If

f

where such a map, consider

f

restricted to the equator

A

of

S

n

. This is an odd
map from

S

n

-

1

to

S

n

-

1

and thus has odd degree. But the map

f

*

⁢

H

n

-

1

⁢

(

A

)

→

H

n

-

1

⁢

(

S

n

-

1

)

factors through

H

n

-

1

⁢

(

S

n

)

=

0

, and so must be zero. Thus

f

|

A

has degree 0, a

contradiction

.

Title

proof of Borsuk-Ulam theorem

Canonical name

ProofOfBorsukUlamTheorem

Date of creation

2013-03-22 13:10:33

Last modified on

2013-03-22 13:10:33

Owner

bwebste (988)

Last modified by

bwebste (988)

Numerical id

5

Author

bwebste (988)

Entry type

Proof

Classification

msc 54C99

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