SNAKE LEMMA IN INCOMPLETE RELATIVE HOMOLOGICAL ...

86 TAMAR JANELIDZE
- Since (Y × Z K w , π 1 , π 2 ) is the pullback of g and k w , there exists a unique morphism
θ : X ′′ → Y × Z K w with π 1 θ = f 2 and π 2 θ = 0. Since π 2 is in E and f 2 = ker(g),
we conclude that π 2 = coker(θ).
- Since f ′ = ker(g ′ ) and g ′ vf 2 = 0, there exists a unique morphism u ′ : X ′′ → X ′ with
f ′ u ′ = vf 2 . It easily follows that u ′ f 1 = u, q u = coker(u ′ ), and ϕθ = u ′ .
- h : K v → Y × Z K w is the canonical morphism and an easy diagram chase proves
that h = ker(ϕ) (we do not need this fact now, but we shall need it below when
proving (ii)).
Since q u ϕθ = q u u ′ = 0 and π 2 = coker(θ), there exists a unique morphism d : K w → Q u
with q u ϕ = dπ 2 . We obtain the following factorization:
1 Y ×Z Kw
Y × Z K w
π 2
Y × Z K w K w
ϕ
X ′ (3.5)
π 2
q u
1 Kw
d
K w Q u
Comparing diagrams (3.3) and (3.5), we conclude that the relation (K w , 1 Kw , d) can be
identified with the relation (R, r 1 , r 2 ). Therefore, r 1 is an isomorphism, proving that
q u f ′◦ vg ◦ k w is a morphism in C.
(ii): To prove that the sequence (3.2) is E-exact, we need to prove that it is E-exact
at K v , K w , Q u , and Q v .
E-exactness at K v : It follows from the fact that the first column of the diagram (3.4)
is E-exact at X ′ , that the kernel of u ′ exists in C. Indeed, consider the commutative
diagram
X
X ′′
f
1
u ′
u ′ 1
u
u 1
u 2
M
X ′ q u
Q u
(3.6)
where u = u 2 u 1 is the factorization of u with u 2 = ker(q u ) and u 1 ∈ E (which does
exists since the first column of the diagram (3.4) is E exact at X ′ ), and u ′ 1 is the induced
morphism. Since f 1 and u 1 are in E, u ′ 1 is also in E, and therefore the kernel of u ′ 1 exists
in C. Since u 2 is a monomorphism we conclude that Ker(u ′ ) ≈ Ker(u ′ 1).