如何追逐一个元素
当任意两点之间的每条路径都给出相同的复合映射时，图表交换。短正合列 0 → A → B → C → 0 打包了三个事实：A → B 是单射，B → C 是满射，且 A → B 的像等于 B → C 的核。图表追逐就是从某处的一个元素出发，沿箭头推动它，利用正合与交换来确定它必然去往何处或来自何处的技术。
五引理的完整追逐
五引理说：在两行各五项、行均正合的交换图中，若外侧四个竖直映射是同构，则中间那个也是。实践中你只需更弱的假设，而追逐它是入门礼。这里是满射那一半，写成你能跟上每一次拉动的样子。
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.单射那一半是镜像，使用 f2、f4 单射与 f1 满射。两半合起来给出：若 f1、f2、f4、f5 是同构，则 f3 也是。记住招式，而非结论——这门学问里的每一次追逐都正由这些拉动拼装而成。
蛇形引理与连接映射
现在是明珠。蛇形引理接受一个行均正合的交换图，其中两条短正合列通过竖直映射 f、g、h 叠起。它产出一条把核与余核编织在一起的长正合列——关键是一个全新的映射 δ，即连接同态，原始资料中并不存在它。δ 诞生于一次图表追逐，它是这门学问中每条长正合列背后的引擎。
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.