In this section we prove two well-known and important results of homological algebra for abelian catetories. The 5-lemma and the snake lemma. These results allow us to associate long exact sequences to short exact sequences of
complexes over an abelian category, in a functorial way. Later we prove analogous results for triangulated categories and the homotopy categories, respectively. We mainly follow [Bor94b, 1.10].
We begin by proving the 5-lemma.
Proof. By duality if suffices to prove thatf3is a monomorphism. We use 2.4.3 (ii). LetxP˚C1be a pseudo-element such thatf3pxq “0. We have
This shows thatf3is a monomorphism.
To give a proof of the snake lemma, we need the following lemma.
Lemma 2.6.2. LetA be an abelian category. Consider a pullback (resp. pushout) diagram
A B
Proof. By duality it is enough to prove the claim about the pullback diagram. Consider the morphismsk:KÑC and 0 :K Ñ B. By the pullback property there exists a unique morphism k1 such that g1k1 “k and f1k1 “0.
To show thatk1 is the kernel of f1, let hbe a morphism such thatf1h“0. Commutativity of the diagram gives that f g1h“0, so there exists a unique morphismj such thatg1k1j “kj “g1h. Uniqueness of the pullback gives k1j“h. This shows thatk1 is the kernel off1.
We are ready to prove a restricted version of the snake lemma.
Theorem 2.6.3 (Restricted Snake lemma). Let Abe an abelian category and
a commutative diagram inA with exact rows. Then we have an exact sequence
0 kerf f˜1 kerg g˜1 kerh ˜δ cokerf f˜2 cokerg ˜g2 cokerh 0 . (2.9)
Similarly we obtain the unique morphisms ˜g1and ˜g2making the diagram commutative.
Existence of δ:˜ Consider the diagram
by commutativity, so there exists a unique morphismφ: kerhÑX2 such that
X1ÑkerhÑX2“X1ÑB1ÑB2ÑX2. (2.12)
By commutativity and the fact thate1 is an epimorphism
kerhÑX2ÑC2“0, so there exists a unique morphism ˜δ: kerhÑcokerf such that
kerhÑcokerf ÑX2“kerhÑX2. δ˜on pseudo-elements: Consider the following diagram of pseudo-elements
xP˚kerh y epyq “˚kpxq
z gpyq 0
upzq
(2.13)
where existence ofyfollows from 2.4.3 (iii), becauseeis an epimorphism, and existence ofzfollows from 2.4.3 (iv), becausegpyqis mapped to zero.
We show that ˜δpxq “˚upzq. Since X1 is the pullback ofk ande, by 2.4.3 (vi) there exists a pseudo-element x0P˚X1 such thatk1px0q “˚y ande1px0q “˚x. Now
p1δpxq “˜ ˚ φpxq “˚ u1gk1px0q “˚u1ppzq “˚p1upzq.
The morphismp1 is a monomorphism by lemma 2.6.2, so by 2.4.3 (ii) we have that ˜δpxq “˚upzq.
Exactness at kerf: We need to show that ˜f1is a monomorphism. But this follows from the equality k2f˜1“mk1,
because k2, m, andk1 are monomorphisms.
Exactness at kerg: We use 2.4.3 (iv). By commutativity and the fact thatk3is a monomorphism, ˜g1f˜1is the zero morphism. LetxP˚kergsuch that ˜g1pxq “0. Now,ek2pxq “0, so by exactness of the second row, there exists a pseudo-elementa1P˚ A1 such thatmpa1q “˚k2pxq. We havefpa1q “0 becauseppfpa1qq “˚ gpk2pxqq “0 andpis a monomorphism. Thus, there exists a pseudo-elementyP˚kerf such thatk1pyq “˚a1. Sincek2 is a monomorphism, ˜f1pyq “˚x.
Exactness at kerh: We use 2.4.3 (iv). Let w P˚ kerg. Now gk2pwq “˚ 0, and by 2.4.3 (ii) 0 P˚ A2 is the unique pseudo-element, up to pseudo-equality, with pseudo-image gk2pwq. From the diagram (2.13) we get δ˜˜g1pwq “˚c1p0q “0.
Next, let xP˚kerhsuch that ˜δpxq “0. By the diagram (2.13) and exactness of the sequence A1 f A2 c1 cokerf 0
there existsa1P˚A1such thatfpa1q “˚ z. By commutativitygmpa1q “˚gpyqand by exactness of the second rowempa1q “˚0. Thus, from 2.4.3 (v) there existsb1P˚ B1such thatgpb1q “0 andepb1q “˚epyq “˚k3pxq.
Since in the diagram (2.13) the choice of the pseudo-elementymapping tok3pxqwas arbitrary, we can choose y to beb1. Finally, from exactness of
0 kerg B1 B2
k2 g
by 2.4.3 (iv) we get a pseudo-elementwP˚kergsuch thatk2pwq “˚b1. By commutativityk3˜g2pwq “˚k3pxq and 2.4.3 (ii) applied to k3 we get ˜g2pwq “˚x. This shows that the sequence is exact at kerh.
Exactness at cokerf: We use 2.4.3 (iv). LetxP˚ kerh. From the diagram (2.13) and commutativity, we see that f˜2pδpxqq “˜ ˚c2pgpyqq “˚0, becausec2g“0.
Let x1 P˚ cokerf such that ˜f2px1q “˚ 0. By 2.4.3 (iii) there exists z P˚ A2 with c1pzq “˚ x1. From commutativity of the diagram (2.11) and exactness of
B1 g B2 c2 cokerg 0
we get a pseudo-elementyP˚B1such thatgpb1q “˚ppzq. Nowngpb1q “˚0 by exactness of the third row, so from the exact sequence
0 kerh C1 C2
k3 h
we get a pseudo-elementxP˚kerhwithk3pxq “˚epyq. The choices ofz,y, andxfit into the diagram (2.13), showing that ˜δpxq “˚x1. This shows that the sequence is exact at cokerf.
Exactness at cokerg: We use 2.4.3 (iv). By commutativity ˜g2f˜2“0. LetxP˚ cokerg such that ˜g2pxq “ 0. By 2.4.3 (iii) there exists b2P˚B2 such thatc2pb2q “˚x. By exactness of the following sequence
C1 h C2 c3 cokerh 0 ,
there exists c1 P˚ C1 with hpc1q “ npb2q. By 2.4.3 (iii) there exists b1 P˚ B1 such that epb1q “˚ c1. By commutativity,ngpb1q “˚npb2q “˚ c2. By 2.4.3 (v), there existsb12P˚B2such thatnpb12q “0 andc2pb12q “˚x.
By exactness of the third row, there is a2 P˚ A2 such that ppa1q “˚ b12. Therefore c1pa2q P˚ cokerf is the pseudo-element which is mapped tox. This shows exactness at cokerg.
Exactness at cokerh: It suffices to show that ˜g2is an epimorphism. By commutativity
˜
g2c2“c3n
where c2, c3, andnare epimorphisms. Hence ˜g2 is an epimorphism.
To see that the morphism ˜δ in (2.9) is not in general the zero morphism, consider the following morphism of short exact sequences overAb.
0 0 Z Z 0
0 Z Z Z{2Z 0
Id
Id 2
We see that in (2.9) the last morphisms between kernels 0 – kerpZ ÑId Zq Ñ kerpZ Ñ Z{2q – 2Z – Z is not surjective andZ–cokerp0ÑZq ÑcokerpZÑId Zq –0 is not injective. In this case the sequence (2.9) is exact when δ˜“IdZ, which is not the zero morphism. If one embeddes this diagram to complexes overAbin degree 0, one gets that also in this case the morphism ˜δ, now between complexes, is not the zero morphism.
We wish to apply the previous result to certain kernel and cokernel sequences. But these are not short exact sequences as the above argument shows. Therefore we need to generalize theorem 2.6.3 to the following
Corollary 2.6.4 (Snake lemma). Let Abe an abelian category. Given a diagram
A1 B1 C1 0
0 A2 B2 C2
m f
e
g h
p n
with exact rows, there exists a morphismδ: kerhÑcokerf such that the following sequence is an exact sequence kerf f˜1 kerg g˜1 kerh δ cokerf f˜2 cokerg ˜g2 cokerh .
Proof. Consider the following commutative diagram
kerf kerg kerh
A1 A11 B1 C1 0
0 A2 B2 B21 C2
cokerf cokerg cokerh
α β
f g h
τ φ
where the objects A11 and B21 are given by theorem 2.2.10 applied to the morphisms A1 ÑB1 and B2 ÑC2. By theorem 2.2.10 we get unique morphismsA11ÑA2andC1ÑB21 which keep the diagram commutative. LetX11 be the kernel ofC1ÑB12andX21 the cokernel ofA11ÑB2. It is not hard to see that we obtain factorizationsβ “β2β1
andτ “τ2τ1through the kernelX11 and the cokernelX21. Let us denote byX1the pullback of the morphismsηand eand byX2 the pushout of the morphismspandA2ÑX21. We have obtained the following commutative diagram
X1 X11 kerh
A1 A11 B1 C1 0
0 A2 B2 B21 C2
kerf X21 X2
β2
η
e γ h
p η2
(2.14)
The dashed morphisms exist by lemma 2.6.2 and are the kernel and the cokernel of X1 Ñ X11 and X21 Ñ X2, respectively. From this diagram we obtain the following commutative diagram
X1 X11
0 A11 B1 C1 0
0 A2 B2 B21 0
X21 X2
By applying theorem 2.6.3 we obtain the morphism ˜δ:X11 ÑX21. We show that the morphismβ2:X11 Ñkerhis an epimorphism, and by duality thatτ1: cokerf ÑX21 is a monomorphism, so that we can define the mapδto be the compositeτ1´1δβ˜ 2´1.
Let c P˚ kerh, pη2γqpcq “˚ phqpcq “˚ 0 so that pγqpcq “ 0 by 2.4.3 (ii) since η2 is a monomorphism.
Since η is the kernel of γ, by 2.4.3 (iv), there existsuP˚ X11 such that ηpuq “˚ pcq. Finally, by commutativity pβ2qpuq “˚ ηpuq “˚ pcq, thusβ2puq “˚ c by 2.4.3 (ii), because is a monomorphism. This shows that β2 is an epimorpism and thus an isomorphism. By duality, the morphismτ1is also an isomorphism.
Lemma 2.6.5. Consider a morphismf :X‚ÑY‚ of two complexes over an abelian category. Then the diagram cokerdi´1X‚ cokerdi´1Y‚
kerdi`1X‚ kerdi`1Y‚
f˜2
φ1 φ2
f˜1
(2.15)
is commutative for alliPZ, where the morphismsφ1 andφ2 are from the diagram (2.5)and the morphismsf˜1and f˜2 are from the diagram (2.10).
Proof. Recall that the morphismsφ1,φ2, ˜f1 and ˜f2 are the unique morphisms which make the following diagrams commutative
Xi Yi
cokerdi´1X‚ cokerdi´1Y‚
f˜1
kerdi`1X‚ kerdi`1Y‚
Xi`1 Yi`1
f˜2
Xi cokerdi´1X‚
kerdi`1X‚ Xi`1
φ1
Yi cokerdi´1Y‚
kerdi`1Y‚ Yi`1
φ2 ,
where the lower two diagrams are of the from (2.5). By the fact thatfis a morphism of complexes and commutativity the equation. Thus we obtain that the diagram (2.15) is commutative.
Next we prove the main theorem of this section which assings to every short exact sequence of complexes over an abelian categoryAa long exact sequence overAin a functorial way.
Theorem 2.6.6 (Functorial long exact sequence). Let A be an abelian category. For any short exact sequence of objects in CpAq
This long exact sequence is functorial in the sense that given a morphism of two short exact sequences
0 X1‚ Y1‚ Z1‚ 0
Proof. By theorem 2.6.3 and lemma 2.6.5 we have following commutative diagram with exact rows cokerdi´1X‚ cokerdi´1Y‚ cokerdi´1Z‚ 0
for alliPZ. By lemma 2.5.3 and corollary 2.6.4 we get long exact sequence (2.16).
Let us show that the long exact sequence is functorial. By definition of a functor and proposition 2.5.4, the following diagrams are commutative for alliPZ
HipX1‚q HipY1‚q
It remais to show that the following diagram is commutative for all i HipZ1‚q Hi`1pX1‚q
By construction ofδ1and δ2 in corollary 2.6.4, we need to show that the following diagram is commutative HipZ1‚q W1 W2 Hi`1pX1‚q
Recall from the proof of corollary 2.6.4, the morphisms ˜δ1and ˜δ2 are constructed from the following commutative diagrams
Sinceh1andh3 are morphisms of complexes and by theorem 2.2.10, the following diagrams are commutative kerdi`1Y1 C2 kerdi`1Z1
kerdi`1Y2 C˜2 kerdi`1Z2
cokerdi´1X1 A1 cokerdi´1Y1
cokerdi´1X2 A˜1 cokerdi´1Y2
By the proof of corollary 2.6.4 W1 Ñ cokerdi´1Z1 and ˜W1 Ñ cokerdi´1Z2 are the kernels of cokerdi´1Z1 Ñ C2 and cokerdi´1Z2 Ñ C˜2 and the morphisms kerdi`1X1 Ñ W˜1 and kerdi`1X2 Ñ W˜2 are cokernels of A1 Ñ kerdi`1X1 and A˜1Ñkerdi`1X2 . By commutativity and the kernel property, there exists unique morphismsW1ÑW˜1andW2ÑW˜2
such that the following diagrams are commutative
W1 W˜1
cokerdi´1Z1 cokerdi´1Z2
kerdi`1X1 kerdi`1X2
W2 W˜2
(2.23)
We need to show that the following cubes commute
X1 W1
X˜1 W˜1
cokerdi´1Y1 cokerdi´1Z1
cokerdi´1Y2 cokerdi´1Z2
kerdi`1X1 kerdi`1Y1
kerdi`1X2 kerdi`1Y2
W2 X2
W˜2 X˜2
(2.24)
We see that both of the commutative diagrams (2.23) are embedded in both of the above cubes. Commutativity of
the following squares
kerdi`1X1 kerdi`1Y1
kerdi`1X2 kerdi`1Y2
cokerdi´1Y1 cokerdi´1Z1
cokerdi´1Y2 cokerdi´1Z2 follow from the equations
kerdi`1X1 Ñkerdi`1Y1 Ñkerdi`1Y2 ÑY2i`1“kerdi`1X1 ÑX1i`1ÑY1i`1ÑY2i`1
“kerdi`1X1 ÑX1i`1ÑX2i`1ÑY2i`1
“kerdi`1X1 Ñkerdi`1X2 Ñkerdi`1Y2 ÑY2i`1 and
Y1iÑcokerdi´1Y1 Ñcokerdi´1Z1 Ñcokerdi´1Z2 “Y1iÑZ1i ÑZ2iÑcokerdi´1Z2
“Y1iÑY2iÑZ2iÑcokerdi´1Z2
“Y1iÑcokerdi´1Y1 Ñcokerdi´1Y2 Ñcokerdi´1Z2 because kerdi`1Y2 ÑY2i`1 is a monomorphism andY1iÑcokerdi´1Y1 is an epimorphism. The squares
X1 W1
cokerdi´1Y1 cokerdi´1Z1
X˜1 W˜1
cokerdi´1Y2 cokerdi´1Z2 are commutative because they are pullback squares. Similarly, the squares
kerdi`1X1 kerdi`1Y1
W2 X2
kerdi`1X2 kerdi`1Y2
W˜2 X˜2
are commutative since they are pushout squares. From the following identities X1Ñcokerdi´1Y1 Ñcokerdi´1Y2 Ñcokerdi´1Z2
“X1Ñcokerdi´1Y1 Ñcokerdi´1Z1 Ñcokerdi´1Z2
“X1ÑW1Ñcokerdi´1Z1 Ñcokerdi´1Z2
“X1ÑW1ÑW˜1Ñcokerdi´1Z2 and
kerdi`1X1 ÑW2ÑW˜2ÑX˜2
“kerdi`1X1 Ñkerdi`1X2 ÑW˜2ÑX˜2
“kerdi`1X1 Ñkerdi`1X2 Ñkerdi`1Y2 ÑX˜2
“kerdi`1X1 Ñkerdi`1Y1 Ñkerdi`1Y2 ÑX˜2
and the pullback and pushout universality of ˜X1 and ˜X2, we obtain unique morphisms X1 ÑX˜1 and X2 Ñ X˜2
which make the remaining squares commutative.
Putting these together the left and right squares of the following diagram are commutative.
X1 cokerdi´1Y1 kerdi`1Y1 X2
X˜1 cokerdi´1Y2 kerdi`1Y2 X˜2
(2.25)
The middle square commutes by lemma 2.6.5, and so the diagram is commutative. Finally, from bottom square of the cube (2.24), commutativity of the right square of (2.23), definition ofHi`1ph3q, and the fact thatW2–Hi`1pX1‚q andW1–Hi`1pX2‚q(as shown in corollary 2.6.4), we have the following commutative diagrams
X1 W1 HipZ1‚q
X˜1 W˜1 HipZ2‚q
–
–
Hi`1pX1‚q W2 X2
Hi`1pX2‚q W˜2 X˜2 –
–
(2.26)
We have the following equalities
X1ÑHipZ1‚q ÑW1ÑW2ÑHi`1pX1‚q ÑHi`1pX2‚q ÑX˜2
“X1ÑW1ÑW2ÑX2ÑX˜2 (2.26)
“X1Ñcokerdi´1Y1 Ñkerdi`1Y1 ÑX2ÑX˜2 (2.21)
“X1ÑX˜1Ñcokerdi´1Y2 Ñkerdi`1Y2 ÑX˜2 (2.25)
“X1ÑX˜1ÑW˜1ÑW˜2ÑX˜2 (2.22)
“X1ÑHipZ1‚q ÑHipZ2‚q ÑW˜1ÑW˜2ÑHi`1pX2‚q ÑX˜2. (2.26)
The morphismX1ÑHipZ1‚qis an epimorphism andHi`1pX2‚q ÑX2is a monomorphism, so these can be cancelled out. Therefore we obtain commutativity of the diagram (2.20). This finishes the proof.