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Diagram Chasing: The Snake and the Five Lemma

Learn the craft of the subject: chasing elements around a commutative diagram. We prove the five lemma cleanly and run the snake lemma by hand to produce the connecting homomorphism out of thin air.

How to chase an element

A diagram commutes when every path between two nodes gives the same composite map. A short exact sequence 0 → A → B → C → 0 packs three facts: A → B is injective, B → C is surjective, and the image of A → B equals the kernel of B → C. Diagram chasing is the technique of starting with an element somewhere, pushing it along arrows, and using exactness and commutativity to pin down where it must go or come from.

The five lemma, chased in full

The five lemma says: in a commutative diagram with two exact rows of five terms, if the outer four vertical maps are isomorphisms, the middle one is too. In practice you only need a weaker hypothesis, and chasing it is the rite of passage. Here is the surjectivity half, written so you can follow each pull.

Two exact rows, vertical maps f1..f5:

  A1 --a--> A2 --b--> A3 --c--> A4 --d--> A5
  |f1       |f2       |f3       |f4       |f5
  v         v         v         v         v
  B1 --p--> B2 --q--> B3 --r--> B4 --s--> B5

Claim (surjectivity of f3): assume f2, f4 surjective, f5 injective.
Goal: every y in B3 has a preimage in A3.

1.  Push y down-stream: let z = r(y) in B4.
2.  f4 onto => pick x4 in A4 with f4(x4) = z.
3.  Check x4 dies under d: f5(d x4) = s(f4 x4) = s(r y) = 0
    (rows exact: s . r = 0). f5 injective => d(x4) = 0.
4.  Row exact at A4 => x4 = c(x3) for some x3 in A3.
5.  Compare f3(x3) with y: r(f3 x3) = f4(c x3) = f4(x4) = r(y),
    so r(y - f3 x3) = 0.  Exact at B3 => y - f3(x3) = q(w), w in B2.
6.  f2 onto => w = f2(x2), x2 in A2.  Then
    f3( x3 + b(x2) ) = f3(x3) + q(f2 x2) = f3(x3) + (y - f3 x3) = y.

Done: x3 + b(x2) is the preimage.  f3 is surjective.
Every step uses only exactness or commutativity of a single square. That is the whole game.

The injectivity half is the mirror image and uses f2, f4 injective and f1 surjective. Together they give: if f1, f2, f4, f5 are isomorphisms then f3 is. Memorize the moves, not the conclusion — every chase in the subject is assembled from exactly these pulls.

The snake lemma and the connecting map

Now the jewel. The snake lemma takes a commutative diagram with exact rows where two short exact sequences are stacked through vertical maps f, g, h. It produces a single long exact sequence weaving the kernels and cokernels — and crucially a brand-new map δ, the connecting homomorphism, that wasn't in the original data. δ is born from a diagram chase, and it is the engine behind every long exact sequence in the subject.

Rows exact, columns f, g, h:

     0 -> A --i--> B --p--> C -> 0
          |f       |g       |h
     0 -> A'--j--> B'--q--> C'-> 0

Snake lemma output (one long exact sequence):

  0 -> ker f -> ker g -> ker h --d--> coker f -> coker g -> coker h -> 0

Building the connecting map d: ker h -> coker f, by chasing c in ker h:
  - p is onto: pick b in B with p(b) = c.
  - g(b) maps to 0 in C':  q(g b) = h(p b) = h(c) = 0.
  - exact at B': g(b) = j(a') for a UNIQUE a' in A'  (j injective).
  - define d(c) = a' + im f  in coker f.

Well-defined: a different lift b -> b + i(a) changes g(b) by g(i a)
  = j(f a), shifting a' by f(a) -- which is 0 in coker f.  So d is honest.
The connecting homomorphism δ: ker h → coker f, conjured by lift-then-descend. This is the snake's bite.