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Barry Mitchell - Theory of Categories

Barry Mitchell - Theory of Categories

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1.
EXT~
163
with
E
exact, by
I,
20.3,
we can imbed this in
a
commutative diagram
8:0
-A
-B
vC-0
1/
ij
ly
E:O
-A-
B
---+
C-0
with Eexact. In fact, it suffices to take the right-hand square as
a
pullback, anddefine
A
+B
as the morphism induced by the given morphism
A
+
B
and thezero morphism from
A
to
C'.
Lemma
1.1.
Given a morphism
(a,
8,
y)
:
E'+E
of
short exact sequences, we canjind a commutative diagram
E
:O
-+
-
-
-
where
I?
is
exact and
p/3'
=
3.
Proof.
We form
i?
as in
(
1).
Then
8'
can be defined from the pullback propertyof
(1)
so
as to make the northeast corner of
(2)
commutative, and
so
that
b/3'
=
8.
That the northwest corner is commutative follows from the fact thatboth compositions yield the same thing when composed with both
B+C'
and
B+B.
I
Corollary
1.2.
The sequence
E
satisfying
(
1
)
is unique.
Proof.
This follows immediately by replacing
a
by 1, in
1.1.
I
In view
of
1.2
we shall denote the sequence
l?
of
(1)
by
Ey.
Dually, given
a
morphism
a
:
A
+A',
the sequence
aE
is defined by the commutative diagram
i
I
I/
aE:O -A'-
B
------+
C
-
Thus a morphism
(a,
8,
1)
:
E
+E'
expresses the fact that
E'
=
aE.
It will notusually be necessary to give a name to the middle morphism (which is not
 
164
VII.
EXTENSIONS
unique in any case). Therefore we shall write
(a,
1)
:
E+aE
and
(
1
,
y)
:
yE
-+
E.
Lemma
1.1
then says that given
a
morphism
(a,
y)
:
E'+E
we can find a factorizationHence
aE'
=
E
=
Ey.
Lemma
1.3.
The ollowing are true whenever either side is defined
(i)
1E
=
E
(i*)
El
=
E
(ii)
(a'a)E
=
a'(aE)
(iii)
(aE)y
=
a(Ey).
(ii*)
E(yy')
=
(Ey)y'.
Proof.
(i) and (ii) are obvious.
To
prove (iii) we consider the composition
(1,
,Y)
(a,
,I)
Ey+E----tuE.
Applying 1.1, this can be rewritten aswhich shows that
a(Ey)
=
?
=
(aE)y.
1
Lemma
1.3
enables us to write
a'aE,
Eyy',
and
aEy
without ambiguity.Let Ext$(C,
A)
denote the class of equivalence classes of short exactsequences of the type
O+A+B+C+O.
When no confusion can arise we shall write Ext'(C,
A).
A
logical difficulty(apart from the commonplace one that the members of Ext'(C,
A)
may notbe sets) arises from the fact that Ext'(C,
A)
may not be a set,
Of
course ifd ssmall, then Ext'(C,
A)
will be
a
set. Likewise it can be shown that Ext'(C,
A)
is a set if& has projectives or injectives (see the appendix to this chapter), orifdhas a generator or a cogenerator (exercise
1).
However, in order not torestrict ourselves to any particular class
of
abelian categories, we introduce atthis point the notion of a
big
abelian group.
This is defined in the same wayas an ordinary abelian group, except that the underlying class need not be a set.We are prevented from talking about "the category of big abelian groups"because the class of morphisms between a given pair
of
big groups need not be
a
set. Nevertheless this will not keep
us
from talking about kernels, cokernels,images, exact sequences, etc., for big abelian groups. These are defined
in
thesame set theoretic terms in which the corresponding notions for ordinaryabelian groups can be described. Nor will we be very inhibited in speaking
of
abig group valued functor from
a
category, and a natural transformation
of
twosuch functors. In fact, it is precisely the aim of this section to show that Ext'
 
1.
EXT~
165
is
a big group valued bifunctor, Henceforth in the chapter the term group willhe understood to mean big group.We define an addition in Ext'(C,
A)
by the rule
E
+E'
=
V(E
@E')d
and proceed to show that
+
makes Ext'(C,
A)
an abelian group. In thefollowing lemma we make use of the easy relation
aV
=
V(a
@a)
a
+
a'
=
V(a
@
a')d
and the relationof
I,
18.3.
(3)
(4)
Lemma
1.4.
Thefollowing are true whenever either side is dejined.
(i)
(a
@
a')(E
@
E')
=
aE
@
a'E'
(ii)
(a
+
a')E
=
aE
+
a'E
(iii)
a(E
+
E')
=
aE
+
aE'
(i*)
(E
0
')(y
0
')
=
Ey
0
'y'
(ii*)
E
(y
+
y')
=
Ey
+
Ey'
(iii*)
(E
+
E')y
=
Ey
+
E'y.
Proof.
The proof
of
(i) is trivial.
To
prove (ii) we observe the morphism
(d,
d,
A)
:
E+E
@
E
which shows that
dE
=
(E
@
E)d.
Then using (i)and
Eq.
(4)
we have
(a
+
a')E
=
V(a
@
a')dE
=
V(a
@
a')(E
@
E)d
=
V(aE
@
a'E)d
=
aE
+
a'E.
For
(iii) we use (i) and
Eq.
(3) to obtain
a(E
+
E')
=
aV(E
@
E')d
=
V(a
@
a)(E
@
E')d
=
V(aE
@
aE')d
=
aE
+uE'.
Theorem
1.5.
The
ofieration
+
gives
Ext'(C,
A)
the structure ofan abelian group.
Proof.
We first prove associativity. We have
I
E
+
(E'
+
E")
=
E
+
V(E'
@
E")d
=
V(E
@
V(E'
@
E")d)d
=
V((1
@
V)(E
@
(E'
@
E"))(l
0
d))d
=
V(l
@
V)(E
@
(E'
@
E"))(l
@
d)d.
(5)
(6)
Similarly, we have
(E
+
E')
+En
=
V(V
@
l)((E
@
E')
@
E")(d
@
1)d.
If we identify
E
@
(E'
@
E")
and
(E
@
E')
@
E"
in the obvious way,
it
i-
 
166
VlI. EXTENSIONS
easy
to
show that
V(l
@
V)
=
V(V
@
1)
and
(1
@
d)d
=
(d
@
1)d.
Henceassociativity follows from
(5)
and
(6).
Next we prove commutativity. The morphismshows that
r(E
@
E)
=
(E
@
E)T.
Also it
is
clear that
Vr
=
V
and
TA
=
A.
Hence
E
+
E
=
V(E
@
E’)d
=
VT(E
@
E’)A
=
V(E’
@
E)TA
=
V(E‘@E)A
=
E+E.
Now we show that the split exact sequence
E,:
O+A+A
@C-+C+O
acts as a zero element for Ext’(C,
A).
Consider an arbitrary sequence
Irh
E:O+A+B+C+O
and form the diagram
E:,@E:O
A@A
--+
(A@C)@B
-
@C
--+
0
(7)The morphism
a
is defined by the matrix
,
6
by means of the matrix
(3
(A
:),
y
by means of the matrix
(
:
)
,
and by means of the matrix
.
Then commutativity
of
the diagram comes from
a
few simple matrixmultiplications, and the middle row, being the coproduct of
E
with thesequence
0
+O
+C
=
C+O,
is exact. This shows that
E
=
E,
+
E
as required.Finally we show the existence of an additive inverse
for
each sequenc e(7).
 
2.
THE
EXACT
SEQUENCE
(SPECIAL
CASE)
167First we prove that
OE
=
Eo.
This follows from the commutative diagram
E:O
-
---+
B
4
-
0
where is represented by the matrix
.
Then we have, using 1.4, (ii),
(3
Eo
=
OE
=
(1
+
(-
1))E
=
1E+
(-
l)E
=
E+
(-
l)E.
Therefore
(-
l)E
acts as an additive inverse for
E.
I
When there can be no danger of confusion we shall write
0
in place of
E,.
Given a morphism
a
:
A +A',
we define a function
&
=
Ext'(C,
a)
Ext'(C,
A)
+-Ext'(C,
A')
by the relation
&(E)
aE.
imilarly, if
y
:
C' +-C
we define
9
=
Extl(y,
A)
:
Ext'(C,
A)
+Extl(C',
A)
by
9(E)
=
Ey.
Then by 1.3, 1.4, and 1.5 we see that Ext'
is
an additive groupvalued bifunctor, contravariant in the first variable and covariant in thesecond.
2.
The
Exact
Sequence (Special Case)
Consider any exact sequence
E:O+A+-B+C+O
and any object
X.
Then we have
a
function
6'
:
[X,
C]
+Ext'(X,
A)
defined by
O(y)
=
Ey.
It follows from 1.4 that
6'
is a group morphism. If
Y
+-
X
is any morphism, then by 1.3 we get a commutative diagram
 
168
VII.
EXTENSIONS
In other words,
8
s a natural transformation from
Hc
to Extl(
,
A).
Further-more, using
1.1,
a commutative diagram
0-A-B-C-0
with exact rows induces
a
commutative diagram
[X,
c]
-
xt'(X,
A)[X,
@I
-
xtl(X,
A')
The morphism
8
is called the
covariantconnecting
morphism
at Xrelativeto the short exact sequence
E.
Dually, one obtains a contravariant connectingmorphism
8
:
[A,
XI
+Extl(C, X)defined by
B(a)
=
aE.
Diagrams dual to
(1)
and
(3)
relative to morphisms
X-tY
and commutative diagrams
(2)
are obtained in this case.The proof of the following lemma is left to the reader.
Lemma
2.1.
Given a diagram
of
the
form
X
E:O
-
-B
- -
with
E
exact, then
y
can be factored through
B
+C
ifand
onb
ifEy
=
0.
I
Proposition
2.2.
Relative to an exact sequence
Lb.4
O+A-+B+C+O
the sequence
-
x
8
i
k
O+[&
A]
L[X,
B] +[X,
C]
+Ext'(X,
A)
+Extl(X,
B)
+Extl(X;C)
is
exact. Here we have denoted
[X,
p]
and
[X,
A]
by
and
A,
respectively.
 
2.
THE
EXACT
SEQUENCE
(SPECIAL
ASE)
169
Proof.
The exactness at places involving only the morphism functors isalready known. There are six things
to
show, namely, image
t
kernel andkernel
c
mage at each of the three remaining places in the sequence.
1.
It follows from
2.1
that
Ox(p)
=
0
for any
2.
An easy application of 2.1* shows that
$O(y)
=
0
for
any
y
E
[X,
].
3.
Since
hp
=
0,
we know
$2
=
0
by a property of the additive functor
4.
Suppose that
O(y)
=
0.
Then by
2.1,
y
=
x(p)
for some
5.
Suppose that
$(E)
=
0.
Consider the following commutative diagram:
0-A
'
+B-C-O,
E
[X,
B].
Ext'(X,
).
E
[X,
B].
A
E:
0
-
-
'
__*
X
-
O-B~B@X-+X-O
P
Now
pp
:
B'+B
makes the northwest corner of the diagram commutative.Defining
y
:
X
+Gas
the morphism making the northeast cornercommutative,we have
E
=
O(y).
6.
Suppose that
$(E)
=O.
Then we have the following commutativediagram?
0
E:O
-
- B'-
X
___+
0
o-c~c@x--+xdo
IP
~
t
The inelegant argument
on
page
170
can be avoided
if
one replaces the bottom row
of
thediagram
by
0
+
C=C
-+
0
+
0
and
uses
Lemma
2.1
*
in conjunction with the
9
emma.
 
170
VII.
EXTENSIONS
where
K
-+
B'
is defined as the kernel pp, the morphism
A
+
is defined since
A
+B -+B' +C
@
X+C
=
A
4
B +C
=
0,
and
K
+
Xis
just
K
+B' +X.
Wemust show that
E'
is exact. In the first place
A
+
is a monomorphism since
A
+
B
and
B
-+
B'
are monomorphisms.
Also
A+K+X
=
A+K+B'+X
=
A+B+B'+X
=
0.
Suppose that
Y+K+X=O.
Then
Y+K+B'+X=O
and
so
we obtain amorphism
Y+B
such that
Y4B+B'
=
Y+K+B'.
But thenand
so
we have
a
morphism
Y
+A
such that
Y+A+B
=
Y-+B.
It thenfollows from the fact that
K
+
B'
is
a
monomorphism that
Y
-+
=
Y
+
-+
K.
Therefore
A
+
is the kernel of
K
+
.
It remains to be shown that
K
+
isan epimorphism. Observe first that
8,
being opposite an epimorphism in apushout diagram, is an epimorphism. Also sincepp is an epimorphism, it is ;hecokernel of
K+B'.
Now suppose that
K+X+Z
=
0.
Then
Y+B+C= Y+B-+B'+C@X+C= Y+K+B'+C@X+C=O,K+B'-+C @X+X+Z
=
0,
and
so
there is a morphism
C4Z
uch that
B'+C@X+C+Z= B'+C@X+X+Z.
But since
B'
+C
@
X
is an epimorphism, we must have
C@X+X+Z=
C@X+C+Z,
and
so
it follows that
X+Z
=
0.
Therefore
E'
is exact, and
so
we have
E
=
@(El).
I
3.
Extn
Consider an exact sequence
A
E
0
+A
+B,l+
Bn-2
-+
. .
+
Bl+
Bo
+C+O.
We call
n
the
length
of
E,
and we call
A
and
C
the
left
and
right
ends
res-pectively of
E.
A morphism
E -+E'
f sequences of length
n
is a commutativediagram
E:O
+
A
+
B,-I
+
Bn-2
--+
.
.
.
+
B,
--+
Bo
+
C
+
0
"4
3.
4
4
3.
4y
E':O
-+
A'+
Bk-,
+
Bh-2
-+
.
. .
+
B;
-+
BA
+
C'+
0
If
E
nd
E'
have the same left end and the same right end and if
a
and
y
areidentity morphisms, then we shall say that the above morphism of sequenceshas
fixed ends.
Again we shall write
E
=
E'
f there is a morphism from
E
o
E'
with fixed endssuch that themorphisms
Bi
+B;
are all isomorphisms. However,

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