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The effacement extension is independent of the effacing morphism

Statement

In either case of A partial morphism of delta functors extends through one dimension shift, the new component defined from a chosen effacement of $A$ is independent of which effacing morphism is used.

Facts & Assumptions

Given: Two chosen effacements of the same object $A$ .

[L1]

Item 19 defines the next-degree component from any chosen effacement and proves naturality for morphisms covered by morphisms between chosen effacement sequences (A partial morphism of delta functors extends through one dimension shift).

[L2]

Finite coproducts of projectives are projective and finite products of injectives are injective (A coproduct of projectives is projective and a product of injectives is injective).

Proof

technique · direct

1.1L1L2givenconstruct

In the homological case, let $p_{i} : P_{i} ↠ A$ be two effacements of the relevant target value. Form the epimorphism $p = (p_{1}, p_{2}) : P_{1} \oplus P_{2} ↠ A .$ By [L2], $P_{1} \oplus P_{2}$ is projective. Since $T_{n}$ is additive, the canonical biproduct identification gives $T_{n} (P_{1} \oplus P_{2}) ≅ T_{n} (P_{1}) \oplus T_{n} (P_{2}),$ and under this identification the map $T_{n} (p)$ has components $T_{n} (p_{1})$ and $T_{n} (p_{2})$ , both zero. Thus $T_{n} (p) = 0$ , so $p$ is again an admissible effacement.

2.1L1step 1.1algebra

The inclusions $ι_{i} : P_{i} \to P_{1} \oplus P_{2}$ satisfy $p ι_{i} = p_{i}$ , so they give morphisms from each original effacement to the dominating effacement of step 1.1 over the identity of $A$ . By the naturality part of [L1], the component defined from $p_{i}$ agrees with the one defined from $p$ for each $i$ . Hence the components defined from $p_{1}$ and $p_{2}$ are equal.

3.1L1L2givenalgebra∎

The cohomological case is dual: if $e_{i} : A ↣ I_{i}$ are two effacements, then $e = (e_{1}, e_{2}) : A ↣ I_{1} \times I_{2}$ is again an admissible effacement by [L2], and the projections $I_{1} \times I_{2} \to I_{i}$ compare it with each original choice. Applying [L1] as in step 2.1 shows that the resulting degree- $(n + 1)$ component is independent of the chosen injective effacement.

The effacement extension is independent of the effacing morphism

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources