Commutative algebra - Department of Mathematical Sciences - old ...

44 3. EXACT SEQUENCES OF MODULES
and
0
0
Im f
N
where the rows are short exact sequences.
Cok f
v
v
v
Im f ′
N ′ Cok f ′
3.2.2. Lemma. Given a commutative diagram of homomorphisms
0
M
u
M ′
f
f ′
N
v
N ′
g
g ′
L
w
L ′
where the rows exact sequences. The snake homomorphism δ : Ker w → Cok u
is well defined by: For z ∈ Ker w choose y ∈ N such that g(y) = z. The
element v(y) ∈ Ker g ′ so there is x ′ ∈ M ′ such that f ′ (x ′ ) = v(y). Then δ(z) =
x ′ + Im u ∈ Cok u.
Proof. Assume g(y ′ ) = z and f ′ (x ′′ ) = v(y ′ ). There is x ∈ M with f(x) = y−y ′ .
Now f ′ (u(x)) = v(f(x)) = v(y − y ′ ) = f ′ (x ′ − x ′′ ) so u(x) = x ′ − x ′′ since f ′
is injective. Then x ′ + Im u = x ′′ + Im u as wanted. The choices made respect
addition and scalar multiplication showing that δ is a homomorphism.
3.2.3. Remark. The snake is
Ker u
f
Ker v
0
0
f
g
M N L 0
u
v
w
0
M ′
f ′
N ′
g ′
L ′
Cok u
The construction of δ is schematically
M ′
Cok u
f ′
N
v
N ′
g
f ′
Ker w
L
Cok v
g
g ′
x ′
δ(z)
Ker w
Cok w
0
v(y)
z
y
z