SNAKE LEMMA IN INCOMPLETE RELATIVE HOMOLOGICAL ...

84 TAMAR JANELIDZE
vπ 1 = f ′ ϕ (see diagram (3.4) below). Using the fact that (Y × Z K w , π 1 , π 2 ) is the pullback
of g and k w and that k w = ker(w), an easy diagram chase proves that (Y × Z K w , π 1 , ϕ) is
the pullback of v and f ′ . Therefore, we obtain the commutative diagram
1 P
1 P
P
P
1 P 1 P 1 P
P
P
1 P
π 1 π 1 1 P
P
Y
P
P
1 P
π 2
X ′
π 1 1 Y
1 Y π 1 ϕ 1 X ′
1 X ′
K w
Y
1
Y
Kw
X ′
k w g
X ′
1 Y 1 Y
v f ′ 1 X ′ 1
X ′
q u
K w Z Y Y ′ X ′ Q u
1 P
P
ϕ
where P = Y × Z K w , and all the diamond parts are pullbacks. Since π 2 and q u are in E,
by condition 2.1(g) we have the factorization (unique up to an isomorphism)
1 Y ×Z Kw
Y × Z K w
Y × Z K w
r
ϕ
R X ′ (3.3)
π 2
q u
r 1
r 2
K w Q u
where r : Y × Z K w → R is a morphism in E and r 1 : R → K w and r 2 : R → Q u
are jointly monic morphisms in C. As follows from the definition of composition of
relations, (R, r 1 , r 2 ) is the composite relation q u f ′◦ vg ◦ k w from K w to Q u (Note, that since
the pullback (Y × Z K w , π 1 , π 2 ) of k w and g, and the pullback (Y × Z K w , π 1 , ϕ) of v and
f ′ exists in C, the composite relations g ◦ k w : K w → Y and f ′◦ v : Y → X ′ also exist.
Moreover, since π 2 and q u are in E, the composite q u (f ′◦ v)(g ◦ k w ) of the three relations
g ◦ k w , f ′◦ v, and q u also exists and we have q u (f ′◦ v)(g ◦ k w ) = q u f ′◦ vg ◦ k w ).
To prove that q u f ′◦ vg ◦ k w : K w → Q u is a morphism in C, consider the commutative