Questions: Long Exact Sequences and the Connecting Morphism

Question 1

A left-exact functor F is applied to a short exact sequence 0 → A → B → C → 0. What is the correct description of the result?

• A. A short exact sequence 0 → FA → FB → FC → 0, since functors preserve exactness
• B. A long exact sequence 0 → FA → FB → FC → R¹FA → R¹FB → R¹FC → ··· extending through derived functors
• C. Nothing useful — left-exact functors cannot be applied to short exact sequences
• D. A long exact sequence, but only if FA = 0

Answer: (B)

A left-exact functor preserves exactness on the left: 0 → FA → FB is exact, and FA → FB → FC is exact at FB. But left-exactness does not guarantee that FB → FC is surjective, so the sequence need not remain short exact on the right. The failure of surjectivity at FC is measured by R¹FA. The right-derived functors R^n F (e.g., Ext^n for Hom) extend the sequence to the right, with connecting morphisms linking degree n of one term to degree n+1 of the preceding term. This long exact sequence captures the complete homological information.

Question 2

The connecting morphism δ: R^n FC → R^(n+1) FA in a long exact sequence of derived functors is generated by which algebraic mechanism?

• A. It arises directly from the functoriality of F applied to the original short exact sequence
• B. It is constructed by the snake lemma applied to short exact sequences of resolutions at each homological degree
• C. It is defined by choosing a splitting of the original short exact sequence
• D. It comes from a natural transformation between derived functors of different orders

Answer: (B)

The connecting morphism is NOT produced by applying F to maps in the original sequence — those produce the maps FA → FB → FC but nothing that shifts homological degree. The connecting morphism arises from applying the snake lemma to the short exact sequence of projective (or injective) resolutions used to compute the derived functors. At each degree n, the snake lemma's diagram-chase produces a map from the cokernel at degree n (related to R^n FC) to the kernel at degree n+1 (related to R^(n+1) FA). This is why the snake lemma is the foundational engine of all connecting morphisms.

Question 3

Most functor applied to a short exact sequence produces a long exact sequence through its derived functors.

• T. True
• F. False

Answer: False

Only left-exact and right-exact functors (and their derived functors) produce long exact sequences in the standard sense. An exact functor — one that preserves all short exact sequences — yields nothing new: the result is still short exact, and all higher derived functors vanish on every object. Functors that are neither left- nor right-exact require more sophisticated machinery (like spectral sequences) and do not directly yield a simple long exact sequence. The long exact sequence is a consequence of left- or right-exactness failing at precisely one end.

Question 4

The snake lemma is the algebraic mechanism that generates the connecting morphisms appearing in long exact sequences of derived functors.

• T. True
• F. False

Answer: True

This is literally true, not merely metaphorical. To compute derived functors, one constructs projective or injective resolutions of A, B, and C and assembles them into a short exact sequence of chain complexes. Applying F levelwise produces a short exact sequence of cochain complexes where exactness may fail. At each degree n, the situation is exactly the setup of the snake lemma: a commutative diagram with exact rows. The snake lemma's connecting homomorphism, applied at each n, yields the connecting morphism δ: R^n FC → R^(n+1) FA in the long exact sequence.

Question 5

Suppose B is projective, so R^n F(B) = 0 for all n ≥ 1. What does the long exact sequence of derived functors then tell you about the relationship between R^n F(C) and R^(n+1) F(A)?

Model answer: The long exact sequence includes the segment ··· → R^n F(B) → R^n F(C) → R^(n+1) F(A) → R^(n+1) F(B) → ··· for each n ≥ 1. When B is projective, R^n F(B) = 0 for all n ≥ 1. Each segment becomes 0 → R^n F(C) → R^(n+1) F(A) → 0, which by exactness means R^n F(C) ≅ R^(n+1) F(A). This is the dimension shift formula: the derived invariants of C and A are isomorphic with a degree offset of 1, allowing computations at degree n to be bootstrapped to degree n+1.

Dimension shifting is one of the most powerful computational tools in homological algebra. By choosing a short exact sequence 0 → A → B → C → 0 with B projective (or free, or injective depending on context), the long exact sequence degenerates into a chain of isomorphisms between derived invariants of A and C at consecutive degrees. This is how one computes Ext^n(A, M) for modules of finite projective dimension, and how inductive arguments in homological algebra typically proceed — exploit a projective middle term to shift the computational burden by one degree at a time.