SNAKE LEMMA IN INCOMPLETE RELATIVE HOMOLOGICAL ...

88 TAMAR JANELIDZE
we conclude that ϕ = ϕ 2 ϕ 1 where ϕ 2 is a monomorphism and ϕ 1 is in E, and then
applying Lemma 2.4(i) to the diagram
Y × Z K w
ϕ
X ′ q u
π 2
K w
d
Q u
we obtain the desired factorization of d. Since d 1 is in E and d 2 is a monomorphism, we
conclude that the kernel of d exists in C (precisely, Ker(d) ≈ Ker(d 1 )). Let k d : K d → K w
be the kernel of d and let e d : K v → K d be the induced unique morphism with k d e d = g K ,
it remains to prove that e d is in E. For, consider the commutative diagram
K v
e d
K d
h
h ′
k d
X ′′ θ Y × Z K w
ϕ
u ′ 1
θ ′
π 2
K w
s
X ′ × Qu K w d
π 2
′
π 1
′
K qu
u2 X ′
q u
Q u
in which:
- π ′ 1, π ′ 2 are the pullback projections, and s = 〈ϕ, π 2 〉, θ ′ = 〈u 2 , 0〉, and h ′ = 〈0, k d 〉
are the induced morphisms (the pullback (X ′ × Qu K w , π ′ 1, π ′ 2) of q u and d does exist
in C since q u is in E); since q u is in E so is π ′ 2.
- u ′ 1 : X ′′ → K qu and u 2 : K qu → X ′ are as in diagram (3.6).
- Since u 2 = ker(q u ) and k d = ker(d), we conclude that θ ′ = ker(π ′ 2) and h ′ = ker(π ′ 1).
Since θ = ker(π 2 ), θ ′ = ker(π ′ 2), and the morphisms π 2 , π ′ 2, and u ′ 1 are in E, by Lemma
2.3, s : Y × Z K w → X ′ × Qu K w is also in E. Therefore, since h = ker(ϕ) and h ′ = ker(π ′ 1),
by Lemma 2.7 and condition 2.1(f) the morphism e d : K v → K d is also in E, as desired.
E-exactness at Q u : Consider the commutative diagram (3.4), we have: f Q ′ dπ 2 = f Q ′ q uϕ =
q v f ′ ϕ = q v vπ 1 = 0, and since π 2 is an epimorphism we conclude that f Q ′ d = 0. To prove
that the sequence (3.2) is E-exact at Q u , it suffices to prove that the kernel of f Q ′ exists
in C and that the induced morphism from K w to the kernel of f Q ′ is in E.
It easily follows from Lemma 2.4(i) that there exists a factorization f Q ′ = f Q ′ 2
f Q ′ 1
where f Q ′ 2
is a monomorphism and f Q ′ 1
is in E. Indeed: since f ′ is a monomorphism and