Theory and Applications of Categories, Vol. 23, No. 4, 2010, pp. 76–91.
SNAKE LEMMA IN INCOMPLETE RELATIVE HOMOLOGICAL
CATEGORIES
Dedicated to Dominique Bourn on the occasion of his 60th birthday
TAMAR JANELIDZE
Abstract. The purpose of this paper is to prove a new, incomplete-relative, version
of Non-abelian Snake Lemma, where “relative” refers to a distinguished class of normal
epimorphisms in the ground category, and “incomplete” refers to omitting all completeness/cocompleteness assumptions not involving that class.
1. Introduction
The classical Snake Lemma known for abelian categories (see e.g. [5]) has been extended
to homological categories by D. Bourn; see F. Borceux and D. Bourn [1], and references
there. In [3], we extended it to the context of relative homological categories: here “relative” refers to a distinguished class E of regular epimorphisms in a ground category C
satisfying certain conditions which, in particular, make (C, E) relative homological when
(a) C is a homological category and E is the class of all regular epimorphisms in C; (b) C is
a pointed finitely complete category satisfying certain cocompleteness conditions, and E is
the class of all isomorphisms in C. In this paper we drop the completeness/cocompleteness
assumption and extend Snake Lemma further to the context of what was called an incomplete relative homological category in [4] (see, however, the correction below Definition
2.1).
Let us recall the formulation of Snake Lemma from [1] (see Theorem 4.4.2 of [1]). It
Partially supported by the International Student Scholarship of the University of Cape Town, and
Georgian National Science Foundation (GNSF/ST06/3-004).
Received by the editors 2009-1-17 and, in revised form, 2009-09-29.
Published on 2010-01-29 in the Bourn Festschrift.
2000 Mathematics Subject Classification: 18G50, 18G25, 18A20, 18B10.
Key words and phrases: incomplete relative homological category, homological category, normal
epimorphism, snake lemma.
c Tamar Janelidze, 2010. Permission to copy for private use granted.
76
SNAKE LEMMA IN INCOMPLETE RELATIVE HOMOLOGICAL CATEGORIES
77
says: “Let C be a homological category. Consider the diagram
0
0
fK
Ku
ku
/
Kv
0
gK
/
Kw
kw
kv
f
X
/Y
/
g
Z
/
0
v
u
0
/
w
f0
X0
d
qu
0
Y0
g0
/
Z0
qv
Qu
/
f0
/
qw
Qv
Q
0
g0
/
Qw
Q
0
where all squares of plain arrows are commutative and all sequences of plain arrows are
exact. There exists an exact sequence of dotted arrows making all squares commutative”.
The proof given in [1] contains explicit constructions of the dotted arrows, which makes
the result far more precise. This seems to suggest that these constructions should be included in the formulation of the lemma, but there is a problem: while the constructions of
0
as induced morphisms are straightforward, the several-step construcfK , gK , fQ0 , and gQ
tion of the “connecting morphism” d : Kw → Qu is too long and technical. A natural
solution of this problem, well-known in the abelian case, involves partial composition of
internal relations in C and goes back at least to S. Mac Lane [6] (see also e.g. [2] for the
so-called calculus of relations in regular categories): one should simply define d as the
composite qu f 0 ◦ vg ◦ kw , where g ◦ is the relation opposite to g, etc. We do not know if the
non-abelian version of d = qu f 0 ◦ vg ◦ kw is mentioned anywhere in the literature, but we use
its relative version in our relative Snake Lemma (Theorem 3.1), which is the main result
of this paper.
For the reader’s convenience we are mostly using the same notation as in [1].
2. Incomplete relative homological categories
Throughout the paper we assume that C is a pointed category and E is a class of morphisms in C containing all isomorphisms. Let us recall from [4]:
2.1. Definition. A pair (C, E) is said to be an incomplete relative homological category
if it satisfies the following conditions:
(a) Every morphism in E is a normal epimorphism;
78
TAMAR JANELIDZE
(b) The class E is closed under composition;
(c) If f ∈ E and gf ∈ E then g ∈ E;
(d) If f ∈ E then ker(f ) exists in C;
(e) A diagram of the form
A0
f0
A
f
/
g0
B
.B
g
has a limit in C provided f and g are in E, and either (i) f = g and f 0 = g 0 , or
(ii) f 0 and g 0 are in E, (f, g) and (f 0 , g 0 ) are reflexive pairs, and f and g are jointly
monic;
(f ) If
π2
A ×B A0
/
A0
f0
π1
A
f
/
B
is a pullback and f is in E, then π2 is also in E;
(g) If h1 : H → A and h2 : H → B are jointly monic morphisms in C and if α : A → C
and β : B → D are morphisms in E, then there exists a morphism h : H → X in
E and jointly monic morphisms x1 : X → C and x2 : X → D in C making the
diagram
H@
@@
~~
@@h2
~~
@@
~
h
~
@@
~
~
~~
h1
A
~~
~~
~
x1
~~
~w ~
X
B@
α
C
x2
@@
@@β
@@
@@
'
D
commutative (it easily follows from the fact that every morphism in E is a normal
epimorphism, that such factorization is unique up to an isomorphism);
SNAKE LEMMA IN INCOMPLETE RELATIVE HOMOLOGICAL CATEGORIES
79
(h) The E-Short Five Lemma holds in C, i.e. in every commutative diagram of the
form
K
/
k
f
A
/
B
w
K
k0
/
A0
/
f0
B
with f and f 0 in E and with k = ker(f ) and k 0 = ker(f 0 ), the morphism w is an
isomorphism;
(i) If in a commutative diagram
K
/
k
A
u
f
/
B
/
B
w
K0
/
k0
A0
f0
f , f 0 , and u are in E, k = ker(f ) and k 0 = ker(f 0 ), then there exists a morphism
e : A → M in E and a monomorphism m : M → A0 in C such that w = me.
Let us use this opportunity to make a correction to conditions 2.1(c) and 3.1(c) of [4].
They should be replaced, respectively, with:
(a) A diagram of the form
A0
f0
A
f
/
g
g0
B
.B
has a limit in C provided f , g, f 0 , and g 0 are in E, and either (i) f = g and f 0 = g 0 , or
(ii) (f, g) and (f 0 , g 0 ) are reflexive pairs (that is, f h = 1B = gh and f 0 h0 = 1B = g 0 h0
for some h and h0 ), and f and g are jointly monic.
(b) Condition 2.1(e) of the present paper.
Note that this replacement will not affect any results/arguments of [4], except that without
it a pair (C, Iso(C)) would be an incomplete relative homological category only when C
(is pointed and) admits equalizers of isomorphisms.
2.2. Remark. As easily follows from condition 2.1(g), if a morphism f : A → B in
C factors as f = em in which e is in E and m is a monomorphism, then it also factors
(essentially uniquely) as f = m0 e0 in which m0 is a monomorphism and e0 is in E.
80
TAMAR JANELIDZE
2.3. Lemma. If a pair (C, E) satisfies conditions 2.1(a)-2.1(d), then (C, E) satisfies
conditions 2.1(h) and 2.1(i) if and only if in every commutative diagram of the form
f
/A
k
K
u
/B
(2.1)
w
K0
/
k0
A0
/B
f0
with k = ker(f ), k 0 = ker(f 0 ), and with f , f 0 , and u in E, the morphism w is also in E.
Proof. Suppose (C, E) satisfies conditions 2.1(a)-2.1(d), 2.1(h) and 2.1(i). Consider the
commutative diagram (2.1) with k = ker(f ), k 0 = ker(f 0 ), and with f , f 0 , and u in E. By
condition 2.1(i), w = me where e : A → M is in E and m : M → A0 is a monomorphism.
Consider the commutative diagram:
/
u
K
K0
m̄
ek
k0
~
M
/
m
A0
Since u is in E and m is a monomorphism, by condition 2.1(a) there exists a unique
morphism m̄ : K 0 → M with m̄u = ek and mm̄ = k 0 ; m̄ is a monomorphism since so k 0 .
Since f 0 mm̄ = 0, m̄ is a monomorphism, and k 0 = ker(f 0 ), we conclude that m̄ = ker(f 0 m).
By condition 2.1(c), f 0 m is in E, therefore we can apply the E-Short Five Lemma to the
diagram
K0
m̄
/
M
f 0m
/
B
m
K0
k0
/
A0
f0
/
B
and conclude that m is an isomorphism. Hence, by condition 2.1(b) w is in E, as desired.
Conversely, suppose for every commutative diagram (2.1) with k = ker(f ), k 0 =
ker(f 0 ), and with f and f 0 in E, if u is in E then w is also in E. It is a well know
fact that under the assumptions of condition 2.1(h), ker(w) = 0; moreover, since E contains all isomorphisms and f and f 0 are in E, w : A → A0 is also in E. Since every
morphism in E is a normal epimorphism, we conclude that w is an isomorphism, proving
condition 2.1(h). The proof of condition 2.1(i) is trivial.
SNAKE LEMMA IN INCOMPLETE RELATIVE HOMOLOGICAL CATEGORIES
81
2.4. Lemma. Let (C, E) be a pair satisfying conditions 2.1(a)-2.1(d) and 2.1(g). Consider the commutative diagram:
f
A
/
B
α
β
A0
f0
(2.2)
/
B0
(i) If α : A → A0 and β : B → B 0 are in E and if f : A → B factors as f = me
in which e is in E and m is a monomorphism, then f 0 : A0 → B 0 also factors as
f 0 = m0 e0 in which e0 is in E and m0 is monomorphism.
(ii) If α : A → A0 and β : B → B 0 are monomorphisms and if f 0 : A0 → B 0 factors
as f 0 = m0 e0 in which e0 is in E and m0 is a monomorphism, then f : A → B also
factors as f = me in which e is in E and m is a monomorphism.
Proof. (i): Consider the commutative diagram (2.2) and suppose α and β are in E and
f = me in which e : A → C is in E and m : C → B is a monomorphism. Since β is
in E and m is a monomorphism, by Remark 2.2 there exists a morphism γ : C → C 0 in
E and a monomorphism m0 : C 0 → B 0 such that βm = m0 γ. Consider the commutative
diagram:
α
/ A0
A
e0
γe
~
C0
m0
/
f0
B0
Since α is in E and m0 is a monomorphism, conditions 2.1(a) and 2.1(d) imply the existence
of a unique morphism e0 : A0 → C 0 with e0 α = γe and m0 e0 = f 0 . Since α, e, and γ are in
E, conditions 2.1(b) and 2.1(c) imply that e0 is also in E. Hence, f 0 = m0 e0 in which e0 is
in E and m0 is a monomorphism, as desired.
(ii) can be proved similarly.
Let (C, E) be an incomplete relative homological category. We will need to compose
certain relations in C:
Let R = (R, r1 , r2 ) : A → B be a relation from A to B, i.e. a pair of jointly monic
morphisms r1 : R → A and r2 : R → B with the same domain, and let S = (S, s1 , s2 ) :
B → C be a relation from B to C. If the pullback (R ×B S, π1 , π2 ) of r2 and s1 exists
in C, and if there exists a morphism e : R ×B S → T in E and a jointly monic pair of
82
TAMAR JANELIDZE
morphisms t1 : T → A and t2 : T → C in C making the diagram
R ×B HS
HH π
v
HH 2
e
v
HH
v
v
HH
v
{v
#
T
S?
HH
v
?? s
HH
vv
v
HH
??2
v
v
HH
??
vsv
v
r2 HH$
zv 1
t2
(
π1 vvv
R
A
~
r1 ~~
~
~
~v ~ t1
B
(2.3)
C
commutative, then we will say that (T, t1 , t2 ) : A → C is the composite of (R, r1 , r2 ) :
A → B and (S, s1 , s2 ) : B → C. One can similarly define partial composition for three or
more relations satisfying a suitable associativity condition. Omitting details, let us just
mention that, say, a composite RR0 R00 might exist even if neither RR0 nor R0 R00 does.
2.5. Convention. We will say that a relation R = (R, r1 , r2 ) : A → B is a morphism
in C if r1 is an isomorphism.
2.6. Definition. Let (C, E) be an incomplete relative homological category. A sequence
of morphisms
f
fi
/ ...
/ Ai−1 i−1 / Ai
/ Ai+1
...
is said to be:
(i) E-exact at Ai , if the morphism fi−1 admits a factorization fi−1 = me, in which
e ∈ E and m = ker(fi );
(ii) an E-exact sequence, if it is E-exact at Ai for each i (unless the sequence either
begins with Ai or ends with Ai ).
As easily follows from Definition 2.6, the sequence
/
0
f
A
g
/B
/
C
is E-exact if and only if f = ker(g); and, if the sequence
f
A
/
g
B
/
/
C
0
is E-exact then g = coker(f ) and g is in E.
In the next section we will often use the following simple fact:
2.7.
Lemma. (Lemma 4.2.4(1) of [1]) If in a commutative diagram
K0
k0
/
A0
u
f0
/
B0
v
K
k
/
w
A
f
/
(2.4)
B
in C, k = ker(f ) and w is a monomorphism, then k 0 = ker(f 0 ) if and only if the left hand
square of the diagram (2.4) is a pullback.
SNAKE LEMMA IN INCOMPLETE RELATIVE HOMOLOGICAL CATEGORIES
83
3. The Snake Lemma
This section is devoted to our main result which generalizes Theorem 4.4.2 of [1] and its
relative version mentioned in [3]. Formulating it, we use the same notation as in [1].
3.1. Theorem. [Snake Lemma] Let (C, E) be an incomplete relative homological category. Consider the commutative diagram
0
Ku
0
fK
/
ku
0
gK
Kv
/
Kw
kv
X
kw
/Y
f
g
/
/
Z
0
v
u
/
0
(3.1)
w
X0
qu
f0
g0
Y0
/
Z0
qw
qv
d
Qu
/
f 0Q
/
Qv
g0 Q
0
/
Qw
0
0
in which all columns, the second and the third rows are E-exact sequences. If the morphism
g 0 factors as g 0 = g20 g10 in which g10 is in E and g20 is a monomorphism, then:
(i) The composite d = qu f 0 ◦ vg ◦ kw : Kw → Qu is a morphism in C.
(ii) The sequence
Ku
/
Kv
/ Kw
d
/
Qu
/ Qv
/
Qw
(3.2)
where d = qu f 0 ◦ vg ◦ kw , is E-exact.
Proof. (i): Under the assumptions of the theorem, let (Y ×Z Kw , π1 , π2 ) be the pullback
of g and kw (by condition 2.1(e) this pullback does exist in C); since g is in E, by
condition 2.1(f) the morphism π2 is also in E, and since kw is a monomorphism so is π1 .
Since f 0 = ker(g 0 ) and g 0 vπ1 = 0, there exists a unique morphism ϕ : Y ×Z Kw → X 0 with
84
TAMAR JANELIDZE
vπ1 = f 0 ϕ (see diagram (3.4) below). Using the fact that (Y ×Z Kw , π1 , π2 ) is the pullback
of g and kw and that kw = ker(w), an easy diagram chase proves that (Y ×Z Kw , π1 , ϕ) is
the pullback of v and f 0 . Therefore, we obtain the commutative diagram
P?
?
?? 1P
??
??
1P

P>
>
>> 1P
>>
>>
1P

P>
>
>> π1
>>
>>
1P

P
} >>
>> π1
>>
>>
}}
}}
}
}
~}
π2
{{
{{
{
{
}{
{
1Kw
Kw
Kw A
AA
AAkw
AA
A
g
Z


Y >
>

Y >
>
>> 1Y
>>
>>
1Y
Y

PA
A
AA 1P
AA
AA
P?
?
?? 1P
??
??
π1
>> 1Y
>>
>>
1Y




 
1P
Y

π1 




??
?? v
??
??
}}
}}
}
}
}~ }
1P
PA
A
AA ϕ
AA
AA
X0 A
PA
A
Y0
}
AA ϕ
1X 0 }}
AA
}
}
AA
}
~}}
X0
} AAA 1 0
f 0 }}
AA X
}
AA
}}
A
}
}~
AA 1 0
AA X
AA
A
X0
X0 B
}}
}}
}
}
}~ }
1X 0
BB
BBqu
BB
B
Qu
where P = Y ×Z Kw , and all the diamond parts are pullbacks. Since π2 and qu are in E,
by condition 2.1(g) we have the factorization (unique up to an isomorphism)
Y ×Z K
Gw
GG
w
GGϕ
ww
ww
GG
r
w
w
GG
w{ w
#
1Y ×Z Kw
Y ×Z Kw
Kw
w
π2 www
w
w
w
{wu w
r1
R
X 0 GG
r2
GG qu
GG
GG
G#
)
(3.3)
Qu
where r : Y ×Z Kw → R is a morphism in E and r1 : R → Kw and r2 : R → Qu
are jointly monic morphisms in C. As follows from the definition of composition of
relations, (R, r1 , r2 ) is the composite relation qu f 0 ◦ vg ◦ kw from Kw to Qu (Note, that since
the pullback (Y ×Z Kw , π1 , π2 ) of kw and g, and the pullback (Y ×Z Kw , π1 , ϕ) of v and
f 0 exists in C, the composite relations g ◦ kw : Kw → Y and f 0◦ v : Y → X 0 also exist.
Moreover, since π2 and qu are in E, the composite qu (f 0 ◦ v)(g ◦ kw ) of the three relations
g ◦ kw , f 0◦ v, and qu also exists and we have qu (f 0 ◦ v)(g ◦ kw ) = qu f 0 ◦ vg ◦ kw ).
To prove that qu f 0 ◦ vg ◦ kw : Kw → Qu is a morphism in C, consider the commutative
85
SNAKE LEMMA IN INCOMPLETE RELATIVE HOMOLOGICAL CATEGORIES
diagram
0
0
fK
Ku
0
/
gK
Kv
h
kv
/
ku
f
u
X 00
/Y
;
ww
w
w
ww
ww
ww f2
/
{
r
X0
g
kw
/
Z
f0
/
d
/ Z0
Y0
g0
qw
qv
0
v
0
fQ
0
ϕ
v
qu
Qu
/
w
u0
0
Y ×Z Kw
w
π1 www
w
w
w{ ww
θ
X GG
GG
GG
G
f1 GG
#
#
/ Kw
w;
π2 www
w
ww
ww
/
Qv
0
0
gQ
/
Qw
0
(3.4)
in which:
- All the horizontal and vertical arrows are as in diagram (3.1).
- π1 : Y ×Z Kw → Y , π2 : Y ×Z Kw → Kw , and ϕ : Y ×Z Kw → X 0 are defined as
above, i.e. (Y ×Z Kw , π1 , π2 ) is the pullback of g and kw , and ϕ : Y ×Z Kw → X 0 is
the unique morphism with vπ1 = f 0 ϕ.
- f = f2 f1 where f1 : X → X 00 is a morphism in E and f2 : X 00 → Y is the kernel of
g (such factorization of f does exist in C since the second row of diagram (3.1) is
E-exact).
86
TAMAR JANELIDZE
- Since (Y ×Z Kw , π1 , π2 ) is the pullback of g and kw , there exists a unique morphism
θ : X 00 → Y ×Z Kw with π1 θ = f2 and π2 θ = 0. Since π2 is in E and f2 = ker(g),
we conclude that π2 = coker(θ).
- Since f 0 = ker(g 0 ) and g 0 vf2 = 0, there exists a unique morphism u0 : X 00 → X 0 with
f 0 u0 = vf2 . It easily follows that u0 f1 = u, qu = coker(u0 ), and ϕθ = u0 .
- h : Kv → Y ×Z Kw 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 qu ϕθ = qu u0 = 0 and π2 = coker(θ), there exists a unique morphism d : Kw → Qu
with qu ϕ = dπ2 . We obtain the following factorization:
Y ×Z K
Fw
FF
x
xx
FFϕ
π2
xx
FF
x
x
FF
x| x
"
1Y ×Z Kw
Y ×Z Kw
Kw
x
π2 xxx
x
xx
1Kw
|xu x
Kw
X 0 FF
d
FF qu
FF
FF
F"
)
(3.5)
Qu
Comparing diagrams (3.3) and (3.5), we conclude that the relation (Kw , 1Kw , d) can be
identified with the relation (R, r1 , r2 ). Therefore, r1 is an isomorphism, proving that
qu f 0 ◦ vg ◦ kw 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 Kv , Kw , Qu , and Qv .
E-exactness at Kv : It follows from the fact that the first column of the diagram (3.4)
is E-exact at X 0 , that the kernel of u0 exists in C. Indeed, consider the commutative
diagram
00
>X C
CC 0
|
CCu
||
|
CC
|| u01
C!
|
|
/ X0
u
BB
{=
BB
{
B
{{
{{ u2
u1 BBB
{
{
f1
X
qu
/
Qu
(3.6)
M
where u = u2 u1 is the factorization of u with u2 = ker(qu ) and u1 ∈ E (which does
exists since the first column of the diagram (3.4) is E exact at X 0 ), and u01 is the induced
morphism. Since f1 and u1 are in E, u01 is also in E, and therefore the kernel of u01 exists
in C. Since u2 is a monomorphism we conclude that Ker(u0 ) ≈ Ker(u01 ).
SNAKE LEMMA IN INCOMPLETE RELATIVE HOMOLOGICAL CATEGORIES
87
Consider the following part of the diagram (3.4)
0
0
fK
Ku
e
ku
"
/
0
gK
< Kv
/
kv
ku0
kw
f
xx
xx u0
x
{
x
X0
Kw
m
Ku0
g
/
X FF
;Y
FF
xx
x
FF
xx
f1 F# xx f2
v
u
X 00
0
/
/
f0
/
Z
/
(3.7)
0
w
Y0
/
g0
Z0
in which:
- ku0 = ker(u0 );
- Since kv = ker(v) and vf2 ku0 = f 0 u0 ku0 = 0, there exists a unique morphism m :
Ku0 → Kv with kv m = f2 ku0 ;
- Since ku0 = ker(u0 ) and u0 f1 ku = uku = 0, there exists a unique morphism e : Ku →
Ku0 with ku0 e = f1 ku ; since kv is a monomorphism, we conclude that me = fK .
The E-exactness at Kv will be proved if we show that e ∈ E and m = ker(gK ). The latter,
however, easily follows from Lemma 2.7. Indeed: by Lemma 2.7, the squares f1 ku = ku0 e
and f2 ku0 = kv m are pullbacks; therefore, since f1 is in E we conclude that e also is in E,
and since kw is a monomorphism, by the same lemma m = ker(gK ).
E-exactness at Kw : Consider the commutative diagram (3.4), we have: dgK = dπ2 h =
qu ϕh = 0. To prove that the sequence (3.2) is E-exact at Kw , it suffices to prove that the
kernel of d exists in C and that the induced morphism from Kv to the kernel of d is in E.
It easily follows from Lemma 2.4 that there exists a factorization d = d2 d1 where d2
is a monomorphism and d1 is in E. Indeed: since the second column of the Diagram
(3.4) is E-exact at Y 0 , there exists a factorization v = v2 v1 , where v2 = ker(qv ) is a
monomorphism and v1 is in E. Applying Lemma 2.4(ii) to the diagram
ϕ
Y ×Z Kw
/
X0
f0
π1
Y
v
/
Y0
88
TAMAR JANELIDZE
we conclude that ϕ = ϕ2 ϕ1 where ϕ2 is a monomorphism and ϕ1 is in E, and then
applying Lemma 2.4(i) to the diagram
ϕ
Y ×Z Kw
/
X0
qu
π2
Kw
/
d
Qu
we obtain the desired factorization of d. Since d1 is in E and d2 is a monomorphism, we
conclude that the kernel of d exists in C (precisely, Ker(d) ≈ Ker(d1 )). Let kd : Kd → Kw
be the kernel of d and let ed : Kv → Kd be the induced unique morphism with kd ed = gK ,
it remains to prove that ed is in E. For, consider the commutative diagram
ed
Kv
/ Kd
h0
h
X 00
/
θ
s
ϕ
u01
θ0
Kqu
u2
π2
Y ×Z K w
/
%
0
/ Kw
9
rrr
r
r
0
rrr π2
2 X ×Q u Kw
s
ss
ss 0
s
yss π1
X0
kd
qu
d
/ Qu
in which:
- π10 , π20 are the pullback projections, and s = hϕ, π2 i, θ0 = hu2 , 0i, and h0 = h0, kd i
are the induced morphisms (the pullback (X 0 ×Qu Kw , π10 , π20 ) of qu and d does exist
in C since qu is in E); since qu is in E so is π20 .
- u01 : X 00 → Kqu and u2 : Kqu → X 0 are as in diagram (3.6).
- Since u2 = ker(qu ) and kd = ker(d), we conclude that θ0 = ker(π20 ) and h0 = ker(π10 ).
Since θ = ker(π2 ), θ0 = ker(π20 ), and the morphisms π2 , π20 , and u01 are in E, by Lemma
2.3, s : Y ×Z Kw → X 0 ×Qu Kw is also in E. Therefore, since h = ker(ϕ) and h0 = ker(π10 ),
by Lemma 2.7 and condition 2.1(f) the morphism ed : Kv → Kd is also in E, as desired.
E-exactness at Qu : Consider the commutative diagram (3.4), we have: fQ0 dπ2 = fQ0 qu ϕ =
qv f 0 ϕ = qv vπ1 = 0, and since π2 is an epimorphism we conclude that fQ0 d = 0. To prove
that the sequence (3.2) is E-exact at Qu , it suffices to prove that the kernel of fQ0 exists
in C and that the induced morphism from Kw to the kernel of fQ0 is in E.
It easily follows from Lemma 2.4(i) that there exists a factorization fQ0 = fQ0 2 fQ0 1
where fQ0 2 is a monomorphism and fQ0 1 is in E. Indeed: since f 0 is a monomorphism and
SNAKE LEMMA IN INCOMPLETE RELATIVE HOMOLOGICAL CATEGORIES
89
E contains all isomorphisms, applying Lemma 2.4(i) to the diagram
f0
X0
/
Y0
qu
qv
Qu
/ Qv
0
fQ
we obtain the desired factorization of fQ0 ; since fQ0 1 is in E and fQ0 2 is a monomorphism,
we conclude that the kernel of fQ0 exists in C. Let kfQ0 : KfQ0 → Qu be the kernel of fQ0
and let efQ0 : Kw → KfQ0 be the induced unique morphism with efQ0 kfQ0 = d, it remains
to prove that efQ0 is in E. Since qu is in E, the pullback (K ×Qu X 0 , p1 , p2 ) of kfQ0 and qu
exists in C and p1 is in E; therefore, we have the commutative diagram
0 Y 00
p
v2
p2
K ×Q u X 0
/
p1
f0
X0
/
Y0
qu
KfQ0
/
kf 0
qv
Qu
0
fQ
Q
/ Qv
in which v2 = ker(qv ) (recall, that since the second column of the diagram (3.4) is Eexact at Y 0 , we have v = v2 v1 where v1 ∈ E and v2 = ker(qv )) and p : K ×Qu X 0 → Y 00
is the induced morphism. Using the fact that v2 and p2 are monomorphisms and that
kfQ0 = ker(fQ0 ), an easy diagram chase proves that the square f 0 p2 = v2 p is the pullback
of f 0 and v2 .
Next, consider the commutative diagram
π1
Y o
ϕ
v1
v
Y
o
00
Y
0
p1
o
! f0
X
/ Kf 0
Q
d
kf 0
p2
0
w
Q
K ×Qu X
p
/K
ef 0
ψ
v2
π2
Y ×Z Kw
Q
0
qu
/ Qu
were ψ = hϕ, v1 π1 i; since kfQ0 is a monomorphism and the equalities
kfQ0 p1 ψ = qu p2 ψ = qu ϕ = dπ2 = kfQ0 efQ0 π2
90
TAMAR JANELIDZE
hold, we conclude that p1 ψ = efQ0 π2 . Recall, that the square f 0 ϕ = vπ1 is a pullback (see
the proof of (i)), therefore, (Y ×Z Kw , π1 , ψ) is the pullback of p and v1 , yielding that ψ is
in E. Since p1 ψ = efQ0 π2 , and ψ, p1 , and π2 are in E, we conclude that efQ0 is also in E,
as desired.
E-exactness at Qv : Consider the commutative diagram (3.4). According to the assumptions of the theorem, we have g 0 = g20 g10 were g10 is a morphism in E and g20 is a monomor0
0
0
phism. Then, by Lemma 2.4(i) there exists a factorization gQ
= gQ
g 0 were gQ
is a
2 Q1
1
0
0
morphism in E and gQ2 is a monomorphism, hence, the kernel of gQ exists in C. Since
0 0
0 0
gQ
fQ qu = qw g 0 f 0 = 0 and qu is an epimorphism, we conclude that gQ
fQ = 0, therefore,
to prove that the sequence (3.2) is E-exact at Qv it suffices to prove that the induced
0
morphism from Qu to the kernel of gQ
is in E. For, consider the commutative diagram
f
X
/
g
Y B
B
BBv1
BB
Ȳ
u
f0
X0
e1
qu
!
K
0
gQ
Y0
/
)
Z̄ 3
33
w
33
3
w2 3
33
z̄
3 g0
/5 Z 0
l
lll
lll
l
l
l
lll p02
(3.8)
Qv ×Qw Z 0
f 00
BB
uu
BB
uu
BB
u
u
BB qv
uu
uu p01
kg0 BBB
u
uu
BB
Q
zuuu
/ Qv
0
fQ
/
ȳ
v
v2
e2
D
0
egQ
Qu
/
/Z
|
|
||
|} |
w1
qw
0
gQ
/ Qw
in which:
0
- kgQ0 = ker(gQ
) and egQ0 : Qu → KgQ0 is the induced unique morphism with kgQ0 egQ0 =
0
fQ .
0
- (Qv ×Qw Z 0 , p01 , p02 ) is the pullback of gQ
and qw (this pullback does exist since qw
0
00
is in E), and e2 = hqv , g i and f = hkgQ0 , 0i are the canonical morphisms; since
0
kgQ0 = ker(gQ
) we conclude that f 00 = ker(p02 ).
- Since f 00 = ker(p02 ) and p02 e2 f 0 = 0 there exists a unique morphism e1 : X 00 → KgQ0
with f 00 e1 = e2 f 0 ; it easily follows that egQ0 qu = e1 .
- Since the second and the third columns of the diagram (3.8) are E-exact at Y 0 and
Z 0 respectively, we have the factorizations v = v2 v1 and w = w2 w1 , where v1 , w1 ∈ E
and v2 = ker(qv ) and w2 = ker(qw ).
SNAKE LEMMA IN INCOMPLETE RELATIVE HOMOLOGICAL CATEGORIES
91
- z̄ = h0, w2 i, and since w2 = ker(qw ) we conclude that z̄ = ker(p01 ).
- Since z̄ = ker(p01 ) and p01 e2 v2 = 0, there exists a unique morphism ȳ : Ȳ → Z̄ with
e2 v2 = z̄ ȳ.
Since w1 , g, and v1 are in E, we conclude that ȳ is in E; therefore, by Lemma 2.3, e2 is
also in E. Then, Lemma 2.7 implies that e1 is in E, and since egQ0 qu = e1 we conclude
that egQ0 is also in E, as desired.
References
[1] F. Borceux, D. Bourn, Mal’cev, protomodular, homological and semi-abelian categories, Mathematics and its Applications, Kluwer, 2004.
[2] A. Carboni, G. M. Kelly, and M. C. Pedicchio, Some remarks on Mal’cev and Goursat
categories, Applied Categorical Structures 1, 1993, 385-421.
[3] T. Janelidze, Relative homological categories, Journal of Homotopy and Related
Structures 1, 2006, 185-194.
[4] T. Janelidze, Incomplete relative semi-abelian categories, Applied Categorical Structures, available online from March 2009.
[5] S. Mac Lane, Categories for the working mathematician, Graduate Texts in Mathematics 5, Springer-Verlag, 1971.
[6] S. Mac Lane, An algebra of additive relations, Proc. Nat. Acad. Sci. USA 47, 1961,
1043-1051.
Department of Mathematics and Applied Mathematics, University of Cape Town,
Rondebosch 7701, Cape Town, South Africa
Email: tamar.janelidze@uct.ac.za
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ftp://ftp.tac.mta.ca/pub/tac/html/volumes/23/4/23-04.{dvi,ps,pdf}
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