So, the two definitions are equivalent.

[Bacle]

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)

 

[quasar987]

But um... how does i) even makes sense? If Del_XA is supposed to take relative n-chains in X to relative (n-1)-chains, then the image of a relative n-chain c_n+C_n(A) should be a relative (n-1)-chain, that is, a set of the form c'_{n-1} + C_{n-1}(A).

And as you said, C_{n-1}(A) is not always the same as Del(C_n(A)).

 

[Bacle]

Yes, Quasar, I agree, but this is part of the confusion. The operator in i)
agrees with the operator induced by the global boundary operator Del, by
the map Del: C_n(X)-->C_(n-1)(X) , which preserves chains in A.

 

[quasar987]

I do not see what you mean by that.

What I said is that the operator in i) is ill-defined as a map C_n(X,A)-->C_{n-1}(X,A).

 

[Bacle]

What I mean is that the format of induced maps is this:

Given groups G,G' , with respective normal subgroups N,N'

and a homomorphism h:G-->G' , such that h(N)<N' (this last condition

is necessary for well-definedness of induced map)

This gives us the induced map (by passing to the quotient) h_*: G/N --->G'/N'

defined by : h_*[( g+N)] := [h(g+N)]+N'

In our case, we have Del:C_n(X) -->C_(n-1)(X)

which sends the normal subgroup C_n(A)<C_n(X) ,

to the normal subgroup C_(n-1)(A) <C_(n-1)(X), which gives us the

induced map:

Del_* [( c_n+C_n(A))]:= [Del(c_n+C_n(A)]= (by linearity of del)

Delc_n+Del(C_n(A)].

And I agree that the map is not well, defined, but

Del_XA =Del_* it is the map induced by Del:C_n(X)-->C_(n-1)(X)

 

[Bacle]

Quasar:
Please let me clarify my point, which I think I did not make too clearly:

AFAIK, the relative Del operator on (X,A) , is the operator induced by

the Del. operator on X. But this induced operator seems to make no sense.

Thanks for your comments, though.

 

[quasar987]

Del_* [( c_n+C_n(A))]:= [Del(c_n+C_n(A)]= (by linearity of del)

[Delc_n+Del(C_n(A)]

={ Delc_n+Del(C_n(A)) } + C_{n-1}(A) = Delc_n+C_{n-1}(A) since Del(C_n(A) is contained in C_{n-1}(A).