Lifting properties in cochain complexes #
Let C be an abelian category. Consider a commutative diagram
in the category CochainComplex C ℤ.
t
A ⟶ X
i| |p
v v
B ⟶ Y
b
Assume that there exists a degreewise lifting B.X n ⟶ X.X n for any n : ℤ,
that Q is a cokernel of i, and K is a kernel of p. In this situation,
we construct a cocycle in Cocycle Q K 1 and show that there exists
a lifting B ⟶ X if this cocycle is a coboundary.
The 0-cochain from B to X given by the degreewise liftings.
Equations
- CochainComplex.Lifting.cochain₀ sq hsq = CochainComplex.HomComplex.Cochain.ofHoms fun (n : ℤ) => (hsq n).l
Instances For
A 1-cocycle from B to X obtained as the boundary of
the 0-cochain cochain₀ sq hsq consisting of the degreewise liftings.
This is refined below as a 1-cocycle from Q to K where Q is a
cokernel of i : A ⟶ B and K a kernel of p : X ⟶ Y (see cocycle₁).
Equations
- One or more equations did not get rendered due to their size.
Instances For
The 1-cochain from Q to K which refines cocycle₁'.
Equations
- CochainComplex.Lifting.cochain₁ sq hsq hQ hK = CochainComplex.HomComplex.Cochain.mk fun (n m : ℤ) (hnm : n + 1 = m) => ⋯.choose
Instances For
A 1-cocycle from a cokernel Q of i : A ⟶ B to a kernel K of
p : X ⟶ Y. If this is a coboundary, then the square in CochainCompplex C ℤ
has a lifting, see the lemma hasLift below.
Equations
- CochainComplex.Lifting.cocycle₁ sq hsq hQ hK = CochainComplex.HomComplex.Cocycle.mk (CochainComplex.Lifting.cochain₁ sq hsq hQ hK) 2 CochainComplex.Lifting.cocycle₁'._proof_1 ⋯
Instances For
Consider a commutative square in the category CochainComplex C ℤ
where C is an abelian category.
t
A ⟶ X
i| |p
v v
B ⟶ Y
b
Assume that there exists a degreewise lifting B.X n ⟶ X.X n for any n : ℤ,
that Q is a cokernel of i, and K is a kernel of p.
If the cocycle cocycle₁ sq hsq hQ hK : Cocycle Q K 1 is a coboundary,
we show that the square admits a lifting B ⟶ X.