SNAKE LEMMA in INCOMPLETE RELATIVE HOMOLOGICAL CATEGORIES Dedicated to Dominique Bourn on the Occasion of His 60Th Birthday

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 0

fK gK Ku / Kv / Kw

ku kv kw f X / Y g / Z / 0 v u w f 0 g0 0 / X0 / Y 0 / Z0 d qu qv qw Ó Qu 0 / Qv 0 / Qw f Q g Q

0 0 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 in- cluded in the formulation of the lemma, but there is a problem: while the constructions of 0 0 fK , gK , fQ, and gQ as induced morphisms are straightforward, the several-step construc- 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 0◦ ◦ ◦ composite quf vg kw, where g is the relation opposite to g, etc. We do not know if the 0◦ ◦ non-abelian version of d = quf 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;

(d) If f ∈ E then ker(f) exists in C;

(e) A diagram of the form A0

f 0 0 g / A f B

g . B has a limit in C provided f and g are in E, and either (i) f = g and f 0 = g0, or (ii) f 0 and g0 are in E, (f, g) and (f 0, g0) are reﬂexive pairs, and f and g are jointly monic;

(f) If π 0 2 0 A ×B A / A

π1 f 0 / 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 ~ @@ h1 ~~ @@ h2 ~~ h @@ ~~ @@ ~ ~ @ A X B ~ @@ α ~~ @@ β ~~ @@ ~~ x1 x2 @@ ~~ @ CDw ' 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 f K / A / B

w K / 0 / B k0 A f 0 with f and f 0 in E and with k = ker(f) and k0 = ker(f 0), the morphism w is an isomorphism; (i) If in a commutative diagram

k f K / A / B

u w 0 / 0 / B K k0 A f 0 f, f 0, and u are in E, k = ker(f) and k0 = 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

f 0 0 g / A f B

g . B has a limit in C provided f, g, f 0, and g0 are in E, and either (i) f = g and f 0 = g0, or 0 0 0 0 0 0 (ii) (f, g) and (f , g ) are reﬂexive pairs (that is, fh = 1B = gh and f h = 1B = g h 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 = m0e0 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

k f K / A / B

u w (2.1) 0 / 0 / B K k0 A f 0

with k = ker(f), k0 = 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), k0 = 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

m¯ ek k0

~ 0 M m / A

Since u is in E and m is a monomorphism, by condition 2.1(a) there exists a unique morphismm ¯ : K0 → M withmu ¯ = ek and mm¯ = k0;m ¯ is a monomorphism since so k0. Since f 0mm¯ = 0,m ¯ is a monomorphism, and k0 = ker(f 0), we conclude thatm ¯ = ker(f 0m). By condition 2.1(c), f 0m is in E, therefore we can apply the E-Short Five Lemma to the diagram m¯ f 0m K0 / M / B

0 / 0 / B K k0 A f 0 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), k0 = 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). Con- sider the commutative diagram: f A / B

α β (2.2) 0 / 0 A f 0 B

(i) If α : A → A0 and β : B → B0 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 → B0 also factors as f 0 = m0e0 in which e0 is in E and m0 is monomorphism.

(ii) If α : A → A0 and β : B → B0 are monomorphisms and if f 0 : A0 → B0 factors as f 0 = m0e0 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 → C0 in E and a monomorphism m0 : C0 → B0 such that βm = m0γ. Consider the commutative diagram: α A / A0

e0 γe f 0 ~ 0 / 0 C m0 B 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 → C0 with e0α = γe and m0e0 = 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 = m0e0 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 S H π1 vv HH π2 vv e HH vv HH vv HH R {v T # S (2.3) HH v ? r1 ~ HH vv ?? s2 ~~ HH vv ? ~~ HH vv ?? ~ r2 H v ? ~ t1 H v s1 t2 ABCv $ zv ( 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 RR0R00 might exist even if neither RR0 nor R0R00 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

fi−1 fi ... / Ai−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

f g 0 / A / B / C is E-exact if and only if f = ker(g); and, if the sequence

f g A / 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 f 0 K0 / A0 / B0

u v w (2.4) / / K k A f B in C, k = ker(f) and w is a monomorphism, then k0 = 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 0 0

fK gK Ku / Kv / Kw

ku kv kw

f g X / Y / Z / 0 v u w (3.1)

f 0 g0 0 / X0 / Y 0 / Z0

q q q u d v w Ó Qu 0 / Qv 0 / Qw f Q g Q

0 0 0

in which all columns, the second and the third rows are E-exact sequences. If the morphism 0 0 0 0 0 0 g factors as g = g2g1 in which g1 is in E and g2 is a monomorphism, then:

0◦ ◦ (i) The composite d = quf vg kw : Kw → Qu is a morphism in C. (ii) The sequence

d Ku / Kv / Kw / Qu / Qv / Qw (3.2)

0◦ ◦ where d = quf 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. 0 0 0 0 Since f = ker(g ) and g vπ1 = 0, there exists a unique morphism ϕ : Y ×Z Kw → X with 84 TAMAR JANELIDZE

0 vπ1 = f 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 ÐÐ ? 1P ÐÐ ?? ÐÐ ?? ÐÐ ? P P >> AA 1P ÐÐ > 1P 1P A 1P ÐÐ >> AA ÐÐ >> AA ÐÐ > A P P P >> ?? } AA 1P ÐÐ > π1 π1 ÐÐ ? 1P 1P } A ϕ ÐÐ >> ÐÐ ?? }} AA ÐÐ >> ÐÐ ?? }} AA ÐÐ > ÐÐ ? ~}} A P Y P X0 } >> Ð >> AA AA π2 } > π1 1Y Ð > 1Y π1 A ϕ 1X0 }} A 1X0 }} >> ÐÐ >> AA }} AA }} >> ÐÐ >> AA }} AA ~}} > Ð > A } A Ð 0 ~} 0 Kw Y Y X X A Ð >> Ð ?? 0 } AA } BB 1Kw {{ AA kw g Ð > 1Y 1Y Ð ? v f } A 1X0 1X0 } B qu {{ AA ÐÐ >> ÐÐ ?? }} AA }} BB {{ AA ÐÐ >> ÐÐ ?? }} AA }} BB }{{ A Ð > Ð ? } A } B Ð Ð 0 ~} 0 ~} Kw ZY Y X 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 Kw 1Y × K w GG Z w ww GG ϕ ww r GG ww GG {ww G # 0 Y ×Z Kw R X (3.3) GG π2 ww G qu ww GG ww r1 r2 GG ww GG {wu #) Kw 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 0◦ ◦ relations, (R, r1, r2) is the composite relation quf 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 0 ◦ 0◦ 0 f exists in C, the composite relations g kw : Kw → Y and f v : Y → X also exist. 0◦ ◦ Moreover, since π2 and qu are in E, the composite qu(f v)(g kw) of the three relations ◦ 0◦ 0◦ ◦ 0◦ ◦ g kw, f v, and qu also exists and we have qu(f v)(g kw) = quf vg kw). 0◦ ◦ To prove that quf vg kw : Kw → Qu is a morphism in C, consider the commutative SNAKE LEMMA IN INCOMPLETE RELATIVE HOMOLOGICAL CATEGORIES 85 diagram

0 0 0

fK gK Ku / Kv / Kw w; h π2 ww ww kv ww # ww ku / Y ×Z Kw kw θ w π1 ww ww ww f {ww g X / Y / Z / 0 G ; G ww GG ww GG ww ϕ f1 GG ww f2 G# ww u X00 w 0 d u v { 0 / 0 r / 0 / 0 X f 0 Y g0 Z

qu qw qv

v Qu 0 / Qv 0 / Qw fQ gQ

0 0 0 (3.4) in which:

- All the horizontal and vertical arrows are as in diagram (3.1).

0 - π1 : Y ×Z Kw → Y , π2 : Y ×Z Kw → Kw, and ϕ : Y ×Z Kw → X are deﬁned as 0 above, i.e. (Y ×Z Kw, π1, π2) is the pullback of g and kw, and ϕ : Y ×Z Kw → X is 0 the unique morphism with vπ1 = f ϕ.

00 00 - f = f2f1 where f1 : X → X is a morphism in E and f2 : X → 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 00 θ : X → Y ×Z Kw with π1θ = f2 and π2θ = 0. Since π2 is in E and f2 = ker(g), we conclude that π2 = coker(θ).

0 0 0 0 00 0 - Since f = ker(g ) and g vf2 = 0, there exists a unique morphism u : X → X with 0 0 0 0 0 f u = vf2. It easily follows that u f1 = u, qu = coker(u ), and ϕθ = u .

- 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)).

0 Since quϕθ = quu = 0 and π2 = coker(θ), there exists a unique morphism d : Kw → Qu with quϕ = dπ2. We obtain the following factorization:

Y ×Z Kw 1 x FF Y ×Z Kw xx FF ϕ xx π2 FF xx FF |xx F " 0 Y ×Z Kw Kw X (3.5) FF π xx F qu 2 xx FF xx FF xx 1Kw d FF |xu ") Kw 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 0◦ ◦ quf 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 X0, that the kernel of u0 exists in C. Indeed, consider the commutative diagram X00 |> CC 0 f1 || CCu || u0 CC || 1 CC | ! 0 X / X / Qu (3.6) B u = qu BB {{ BB {{ u1 BB {{u2 B {{ M where u = u2u1 is the factorization of u with u2 = ker(qu) and u1 ∈ E (which does 0 0 exists since the ﬁrst column of the diagram (3.4) is E exact at X ), and u1 is the induced 0 0 morphism. Since f1 and u1 are in E, u1 is also in E, and therefore the kernel of u1 exists 0 0 in C. Since u2 is a monomorphism we conclude that Ker(u ) ≈ Ker(u1). SNAKE LEMMA IN INCOMPLETE RELATIVE HOMOLOGICAL CATEGORIES 87 Consider the following part of the diagram (3.4)

0 0 0

fK gK Ku / Kv / Kw < e m " ku K 0 kv kw u (3.7) ku0 f g X / Y / Z / 0 FF x; FF xx FF xx f1 F# xx f2 u X00 v w xx xxu0 {xx 0 / 0 / 0 / 0 X f 0 Y g0 Z

in which:

0 - ku0 = ker(u );

0 0 - Since kv = ker(v) and vf2ku0 = f u ku0 = 0, there exists a unique morphism m : Ku0 → Kv with kvm = f2ku0 ;

0 0 - Since ku0 = ker(u ) and u f1ku = uku = 0, there exists a unique morphism e : Ku → Ku0 with ku0 e = f1ku; 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 f1ku = ku0 e and f2ku0 = kvm 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 = dπ2h = quϕh = 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 = d2d1 where d2 is a monomorphism and d1 is in E. Indeed: since the second column of the Diagram 0 (3.4) is E-exact at Y , there exists a factorization v = v2v1, where v2 = ker(qv) is a monomorphism and v1 is in E. Applying Lemma 2.4(ii) to the diagram

ϕ 0 Y ×Z Kw / X

π1 f 0

0 Y v / Y 88 TAMAR JANELIDZE we conclude that ϕ = ϕ2ϕ1 where ϕ2 is a monomorphism and ϕ1 is in E, and then applying Lemma 2.4(i) to the diagram

ϕ 0 Y ×Z Kw / X

π2 qu

K / Q w d u 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 kded = gK , it remains to prove that ed is in E. For, consider the commutative diagram

ed Kv / Kd

0 h h kd

π 00 θ 2 X / Y ×Z Kw / Kw 9 rrr rr 0 ϕ s rr π2 0% Õ r u0 X × K d 1 2 Qu w θ0 ss ss ss π0 yss 1 K / 0 / Q qu u2 X qu u in which:

0 0 0 0 - π1, π2 are the pullback projections, and s = hϕ, π2i, θ = hu2, 0i, and h = h0, kdi 0 0 0 are the induced morphisms (the pullback (X ×Qu Kw, π1, π2) of qu and d does exist 0 in C since qu is in E); since qu is in E so is π2.

0 00 0 - u1 : X → Kqu and u2 : Kqu → X are as in diagram (3.6).

0 0 0 0 - Since u2 = ker(qu) and kd = ker(d), we conclude that θ = ker(π2) and h = ker(π1).

0 0 0 0 Since θ = ker(π2), θ = ker(π2), and the morphisms π2, π2, and u1 are in E, by Lemma 0 0 0 2.3, s : Y ×Z Kw → X ×Qu Kw is also in E. Therefore, since h = ker(ϕ) and h = ker(π1), by Lemma 2.7 and condition 2.1(f) the morphism ed : Kv → Kd is also in E, as desired. 0 0 E-exactness at Qu: Consider the commutative diagram (3.4), we have: fQdπ2 = fQquϕ = 0 0 qvf ϕ = qvvπ1 = 0, and since π2 is an epimorphism we conclude that fQd = 0. To prove 0 that the sequence (3.2) is E-exact at Qu, it suﬃces to prove that the kernel of fQ exists 0 in C and that the induced morphism from Kw to the kernel of fQ is in E. 0 0 0 It easily follows from Lemma 2.4(i) that there exists a factorization fQ = fQ2 fQ1 0 0 0 where fQ2 is a monomorphism and fQ1 is in E. Indeed: since f is a monomorphism and SNAKE LEMMA IN INCOMPLETE RELATIVE HOMOLOGICAL CATEGORIES 89 E contains all isomorphisms, applying Lemma 2.4(i) to the diagram

f 0 X0 / Y 0

qu qv Qu 0 / Qv fQ

0 0 0 we obtain the desired factorization of fQ; since fQ1 is in E and fQ2 is a monomorphism, 0 0 we conclude that the kernel of f exists in C. Let k 0 : K 0 → Q be the kernel of f Q fQ fQ u Q and let e 0 : K → K 0 be the induced unique morphism with e 0 k 0 = d, it remains fQ w fQ fQ fQ 0 to prove that e 0 is in E. Since q is in E, the pullback (K × X , p , p ) of k 0 and q fQ u Qu 1 2 fQ u exists in C and p1 is in E; therefore, we have the commutative diagram

0 Y 00 p

p f 0 0 2 0 0 K ×Qu X / X / Y

p1 qu qv

K 0 fQ / Qu 0 / Qv kf0 f Q Q in which v2 = ker(qv) (recall, that since the second column of the diagram (3.4) is E- 0 0 00 exact at Y , we have v = v2v1 where v1 ∈ E and v2 = ker(qv)) and p : K ×Qu X → Y is the induced morphism. Using the fact that v2 and p2 are monomorphisms and that 0 0 k 0 = ker(f ), an easy diagram chase proves that the square f p = v p is the pullback fQ Q 2 2 0 of f and v2. Next, consider the commutative diagram

π1 π2 Y o Y ×Z Kw / Kw

ϕ ef0 v1 ψ Q 0 v 00 o K × X / Kf 0 d Y p Qu p1 Q

v2 p2 kf0 Q Ó 0 ! 0 o / Qu Y f 0 X qu were ψ = hϕ, v π i; since k 0 is a monomorphism and the equalities 1 1 fQ

k 0 p ψ = q p ψ = q ϕ = dπ = k 0 e 0 π fQ 1 u 2 u 2 fQ fQ 2 90 TAMAR JANELIDZE

0 hold, we conclude that p ψ = e 0 π . Recall, that the square f ϕ = vπ is a pullback (see 1 fQ 2 1 the proof of (i)), therefore, (Y ×Z Kw, π1, ψ) is the pullback of p and v1, yielding that ψ is in E. Since p ψ = e 0 π , and ψ, p , and π are in E, we conclude that e 0 is also in E, 1 fQ 2 1 2 fQ as desired.

E-exactness at Qv: Consider the commutative diagram (3.4). According to the assump- 0 0 0 0 0 tions of the theorem, we have g = g2g1 were g1 is a morphism in E and g2 is a monomor- 0 0 0 0 phism. Then, by Lemma 2.4(i) there exists a factorization gQ = gQ2 gQ1 were gQ1 is a 0 0 morphism in E and gQ2 is a monomorphism, hence, the kernel of gQ exists in C. Since 0 0 0 0 0 0 gQfQqu = qwg f = 0 and qu is an epimorphism, we conclude that gQfQ = 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 g X / Y / Z BB v1 w1 | BB || BB || ¯ ¯ }| Y y¯ / Z u v 3 w 33 33 v2 w2 3 33 z¯ 3 f 0 Õ g0 3 X0 / Y 0 / Z0 ll5 (3.8) lll lll e1 e2 ll 0 ! ll p2 ) Ö 0l Kg0 / Qv×Q Z Q f 00 w qu ÙD BB uu qw Ù BB u Ù B qv uu ÙÙ BB uu Ùeg0 B uu 0 Ù Q kg0 B u p1 Ù Q BB uu ÙÙ BB uu Ù zuu Qu 0 / Qv 0 / Qw fQ gQ

in which:

0 - k 0 = ker(g ) and e 0 : Q → K 0 is the induced unique morphism with k 0 e 0 = gQ Q gQ u gQ gQ gQ 0 fQ.

0 0 0 0 -(Qv×Qw Z , p1, p2) is the pullback of gQ and qw (this pullback does exist since qw 0 00 is in E), and e = hq , g i and f = hk 0 , 0i are the canonical morphisms; since 2 v gQ 0 00 0 k 0 = ker(g ) we conclude that f = ker(p ). gQ Q 2

00 0 0 0 00 - Since f = ker(p ) and p e f = 0 there exists a unique morphism e : X → K 0 2 2 2 1 gQ 00 0 with f e = e f ; it easily follows that e 0 q = e . 1 2 gQ u 1 - Since the second and the third columns of the diagram (3.8) are E-exact at Y 0 and 0 Z respectively, we have the factorizations v = v2v1 and w = w2w1, where v1, w1 ∈ E and v2 = ker(qv) and w2 = ker(qw). SNAKE LEMMA IN INCOMPLETE RELATIVE HOMOLOGICAL CATEGORIES 91

0 -z ¯ = h0, w2i, and since w2 = ker(qw) we conclude thatz ¯ = ker(p1). 0 0 ¯ ¯ - Sincez ¯ = ker(p1) and p1e2v2 = 0, there exists a unique morphismy ¯ : Y → Z with e2v2 =z ¯y¯.

Since w1, g, and v1 are in E, we conclude thaty ¯ is in E; therefore, by Lemma 2.3, e2 is also in E. Then, Lemma 2.7 implies that e is in E, and since e 0 q = e we conclude 1 gQ u 1 that e 0 is also in E, as desired. gQ

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 Struc- tures, available online from March 2009.

[5] S. Mac Lane, Categories for the working mathematician, Graduate Texts in Mathe- matics 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: [email protected] This article may be accessed at http://www.tac.mta.ca/tac/ or by anonymous ftp at ftp://ftp.tac.mta.ca/pub/tac/html/volumes/23/4/23-04.{dvi,ps,pdf} THEORY AND APPLICATIONS OF CATEGORIES (ISSN 1201-561X) will disseminate articles that significantly advance the study of categorical algebra or methods, or that make significant new contributions to mathematical science using categorical methods. The scope of the journal includes: all areas of pure category theory, including higher dimensional categories; applications of category theory to algebra, geometry and topology and other areas of mathematics; applications of category theory to computer science, physics and other mathematical sciences; contributions to scientific knowledge that make use of categorical methods. Articles appearing in the journal have been carefully and critically refereed under the responsibility of members of the Editorial Board. Only papers judged to be both significant and excellent are accepted for publication. Full text of the journal is freely available in .dvi, Postscript and PDF from the journal’s server at http://www.tac.mta.ca/tac/ and by ftp. It is archived electronically and in printed paper format. Subscription information. Individual subscribers receive abstracts of articles by e-mail as they are published. To subscribe, send e-mail to [email protected] including a full name and postal address. For in- stitutional subscription, send enquiries to the Managing Editor, Robert Rosebrugh, [email protected]. Information for authors. The typesetting language of the journal is TEX, and LATEX2e strongly encouraged. Articles should be submitted by e-mail directly to a Transmitting Editor. Please obtain detailed information on submission format and style files at http://www.tac.mta.ca/tac/. Managing editor. Robert Rosebrugh, Mount Allison University: [email protected] TEXnical editor. Michael Barr, McGill University: [email protected] Assistant TEX editor. Gavin Seal, Ecole Polytechnique Fédéralede Lausanne: gavin [email protected] Transmitting editors. Clemens Berger, Universitéde Nice-Sophia Antipolis, [email protected] Richard Blute, Universitéd’ Ottawa: [email protected] Lawrence Breen, Universitéde Paris 13: [email protected] Ronald Brown, University of North Wales: ronnie.profbrown (at) btinternet.com Aurelio Carboni, Universitàdell Insubria: [email protected] Valeria de Paiva, Cuill Inc.: [email protected] Ezra Getzler, Northwestern University: getzler(at)northwestern(dot)edu Martin Hyland, University of Cambridge: [email protected] P. T. Johnstone, University of Cambridge: [email protected] Anders Kock, University of Aarhus: [email protected] Stephen Lack, University of Western Sydney: [email protected] F. William Lawvere, State University of New York at Buffalo: [email protected] Tom Leinster, University of Glasgow, [email protected] Jean-Louis Loday, Universitéde Strasbourg: [email protected] Ieke Moerdijk, University of Utrecht: [email protected] Susan Niefield, Union College: [email protected] Robert Paré,Dalhousie University: [email protected] Jiri Rosicky, Masaryk University: [email protected] Brooke Shipley, University of Illinois at Chicago: [email protected] James Stasheff, University of North Carolina: [email protected] Ross Street, Macquarie University: [email protected] Walter Tholen, York University: [email protected] Myles Tierney, Rutgers University: [email protected] Robert F. C. Walters, University of Insubria: [email protected] R. J. Wood, Dalhousie University: [email protected]