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## Main Question or Discussion Point

Hello,

One says that an odd derivation d is a differential modulo another differential d' if this two things hold:

[d,d']=0

d^2=-[d',d"] for some d"

- The first condition says that d defines a derivative on H*(d'), but how? To have a well-defined action of this derivation on H*(d') we must have:

dx belongs to H*(d') whenever x belongs to H*(d') i.e: d'(dx)=0 & dx#d'y for every y

- d'(dx)=-dd'x=0 OK

- How to show the second one?

- The second condition says that d is indeed a differential i.e: d^2=0 in H*(d') but for x belongs to H*(d') we have:

d^2x=-d'd"x-d"d'x=-d'd"x ????????, How does it vanish?

Thanks for every help....

One says that an odd derivation d is a differential modulo another differential d' if this two things hold:

[d,d']=0

d^2=-[d',d"] for some d"

- The first condition says that d defines a derivative on H*(d'), but how? To have a well-defined action of this derivation on H*(d') we must have:

dx belongs to H*(d') whenever x belongs to H*(d') i.e: d'(dx)=0 & dx#d'y for every y

- d'(dx)=-dd'x=0 OK

- How to show the second one?

- The second condition says that d is indeed a differential i.e: d^2=0 in H*(d') but for x belongs to H*(d') we have:

d^2x=-d'd"x-d"d'x=-d'd"x ????????, How does it vanish?

Thanks for every help....