- #1
Bacle
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No, the boundary operator is not relative--sorry, Einstein . I mean,
the boundary operator in relative homology.
I have seen it defined in two different ways , which I do not
believe are equivalent to each other:
Given a pair (X,A), A<X, and Del is the Bdry. operator on X, and (c_n+C_n(A))
is a relative n-chain. The relative Del_XA has been defined like this :
i) Del_XA (c_n+ C_n(A)):= Del(c_n)+ Del(C_n(A))
ii) Del_XA (c_n+ C_n(A)):= Del(c_n)+ C_(n-1)(A)
Now, i makes sense , since Del is linear, and i agrees with the relative operator
induced by the map Del: C_n(X)--->C_(n-1)(X)
But both i , ii satisfy Del^2=0 . But the two are not equivalent, because
Del(C_n(A)) is not equal to C_(n-1)(A) , unless every chain in C_(n-1)(A)
is a boundary, which is not always the case --i.e., when H_n is not trivial, I
think ( Am I right.?)
Which brings me to another question: Is there more than one natural way
of defining a Del operator for a given homology theory.?
Got to go: I got to go visit my relatives . That's what the operator said.
Thanks in Advance ( and sorry for cheesy operator joke)
the boundary operator in relative homology.
I have seen it defined in two different ways , which I do not
believe are equivalent to each other:
Given a pair (X,A), A<X, and Del is the Bdry. operator on X, and (c_n+C_n(A))
is a relative n-chain. The relative Del_XA has been defined like this :
i) Del_XA (c_n+ C_n(A)):= Del(c_n)+ Del(C_n(A))
ii) Del_XA (c_n+ C_n(A)):= Del(c_n)+ C_(n-1)(A)
Now, i makes sense , since Del is linear, and i agrees with the relative operator
induced by the map Del: C_n(X)--->C_(n-1)(X)
But both i , ii satisfy Del^2=0 . But the two are not equivalent, because
Del(C_n(A)) is not equal to C_(n-1)(A) , unless every chain in C_(n-1)(A)
is a boundary, which is not always the case --i.e., when H_n is not trivial, I
think ( Am I right.?)
Which brings me to another question: Is there more than one natural way
of defining a Del operator for a given homology theory.?
Got to go: I got to go visit my relatives . That's what the operator said.
Thanks in Advance ( and sorry for cheesy operator joke)