Homology Math Worksheet

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qnynbjnat

Mathematics

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There are 4 problems in the problem sets, one of them is optional. Please label the answer clearly and provide details of each questions.

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I resend the final and completed answer for your assignment as shown below.

Problem 1:





Alternatively, since H o ( M ) = f  0 | df = 0 , it can be interpreted as H o ( M ) , and also

Z o ( M ) , is the space of C  - functions f : M →

whose 1st derivative is 0 , i.e., df = 0 .

Therefore, H o ( M ) , and also Z o ( M ) , only contains locally constant function because df = 0
if and only if f is constant on each connected component M1 , M 2 ,..., M N of M as we denote

M = M1  M 2 

 M N . Since there is no form of dimension  0 , it implies that

b0 ( M ) = 0 which equals to the number of connected components of M and
H o (M ) Zo (M ) =
words, H o ( M )

N

, where N refers to that number of connected components. In other

( a , a ,..., a ) | a   where f ( x ) = a for all x  M , 1  i  N . Since
1

2

N

i

i

M is given to be connected, we must have N = 1, and therefore H

H o (M )

1

, both of which imply H 0 ( M )

i

o

( M ) a1   or

, as desired result.

Alternatively, we can let  be a closed 1 -form on M satisfying the condition d = 0 . Hence,
there exists some function f on M satisfying df =  , which immediately implies the exactness
of

 . Because all closed 1 -forms are exact, it follows that H 1 ( M ) = 0 and thus

H 0 (M )

since M is connected, as desired result.

Problem 2:
For k = 0 , we immediately have H k
supported in M =

n

( )=H ( )
n

0

n

0 since constants are not compactly

.



For k = n = 1, if  = 1dx1 then we have ...


Anonymous
Awesome! Perfect study aid.

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