In the snake lemma, the connecting homomorphism δ: ker(h) → coker(f) is the 'non-obvious' map in the long exact sequence. What makes it non-obvious compared to the other maps?
ABecause it requires inverting one of the vertical maps, which may not exist
BBecause it crosses the diagram diagonally by composing maps from different rows, and its well-definedness requires proof since it involves a non-unique lift
CBecause it only exists when f and h are isomorphisms
DBecause it goes from a cokernel to a kernel, reversing the usual direction of maps
The maps ker(f) → ker(g) → ker(h) are simply the top-row maps restricted to kernels — completely natural. Similarly coker(f) → coker(g) → coker(h) come from the bottom row. But δ travels diagonally: take c ∈ ker(h), lift to b in the top row (non-unique!), apply g vertically, then lift backwards through the bottom row to A', then project to coker(f). The proof that δ is well-defined — independent of which lift b you choose — is the substantive mathematical content of the lemma. No inverses are needed; exactness provides the necessary uniqueness.
Question 2 Multiple Choice
In the snake lemma setup, suppose f: A → A' is surjective and h: C → C' is injective. What can you conclude from the long exact sequence?
Ag must be an isomorphism
Bker(g) ≅ ker(f) and coker(g) ≅ coker(h), but g need not be an isomorphism
CThe connecting homomorphism δ vanishes, so ker(h) ≅ 0
Dg must be surjective but not necessarily injective
If f is surjective then coker(f) = 0; if h is injective then ker(h) = 0. The long exact sequence becomes 0 → ker(f) → ker(g) → 0 → 0 → coker(g) → coker(h) → 0. Exactness forces ker(g) ≅ ker(f) and coker(g) ≅ coker(h). But to conclude g is an isomorphism we would also need f injective (ker(f) = 0) and h surjective (coker(h) = 0). With only f surjective and h injective, g is not generally an isomorphism. The snake lemma's exact sequence precisely encodes what can and cannot be concluded.
Question 3 True / False
The connecting homomorphism δ is well-defined because any two different choices of lift b for c ∈ ker(h) yield elements of A' that differ by an element of im(f), hence the same class in coker(f).
TTrue
FFalse
Answer: True
This is the heart of the well-definedness proof. Given c ∈ ker(h), any two lifts b and b₁ to B satisfy b − b₁ ∈ ker(B → C) = im(A → B) by exactness of the top row. Applying g to b − b₁ and tracing through shows the corresponding pre-images a', a'₁ in A' differ by an element in im(f). Therefore [a'] = [a'₁] in coker(f) = A'/im(f). Without this, δ would be a multi-valued relation rather than a function — the entire lemma would fail.
Question 4 True / False
The snake lemma requires the vertical maps f, g, h to be injective or surjective for the connecting homomorphism to be constructible.
TTrue
FFalse
Answer: False
The vertical maps f, g, h can be arbitrary homomorphisms — no injectivity or surjectivity is assumed. What IS required is that the two horizontal rows are exact (as given by the short exact sequences 0 → A → B → C → 0 and 0 → A'→ B'→ C'→ 0). The construction of δ uses the surjectivity of B → C and injectivity of A'→ B', which are properties of the horizontal maps, not the vertical ones. Arbitrary vertical maps still yield the connecting homomorphism and the full exact sequence of kernels and cokernels.
Question 5 Short Answer
Why does the snake lemma's connecting homomorphism δ produce long exact sequences in homology when applied to short exact sequences of chain complexes?
Think about your answer, then reveal below.
Model answer: A short exact sequence of chain complexes 0 → A_• → B_• → C_• → 0 gives a commutative diagram at each degree n with boundary maps ∂_n as vertical arrows. The snake lemma applied at degree n produces a six-term exact sequence involving H_n(A), H_n(B), H_n(C) and their shifted versions. The connecting homomorphism δ: H_n(C) → H_{n-1}(A) stitches the degree-n piece to the degree-(n-1) piece. Splicing these across all degrees yields the long exact sequence ⋯ → H_n(A) → H_n(B) → H_n(C) → H_{n-1}(A) → ⋯, the foundational tool underlying Mayer-Vietoris and the long exact sequence of a pair.
The snake lemma is not just an abstract algebraic curiosity — it is the machine that generates the computational tools of algebraic topology. Every time homology 'jumps' from one space to another via a connecting map, the snake lemma is operating behind the scenes.