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In
mathematics, especially
homological algebra and other applications of
Abelian category theory, the '''five lemma''' is an important and widely used
lemma about
commutative diagrams.
The five lemma is valid ''not only'' for abelian categories but also works in
group theory, for example.
The five lemma can be thought of as a combination of two other theorems, the '''four lemmas''', which are
dual to each other.
==Statement==
Consider the following
commutative diagram in any
Abelian category (such as the category of
Abelian groups or the category of
vector spaces over a given
field), or in the category of
groups.
Image:FiveLemma.png
The five lemma states that, if the rows are
exact,
m and
p are
isomorphisms,
l is an
epimorphism, and
q is a
monomorphism, then
n is also an isomorphism.
The 2 four lemmas state:
# If m and p are epimorphisms and q is a monomorphism, then n is an epimorphism.
# If m and p are monomorphisms and l is an epimorphism, then n is a monomorphism.
==Proof==
The method of proof we shall use is commonly referred to as ''
diagram-chasing''. Although it may boggle the mind at first, once one has some practice at it, it is actually fairly routine. We shall prove the five lemma by individually proving each of the 2 four lemmas.
To perform diagram chasing, we assume that we are in a category of
modules over some
ring, so that we may speak of ''elements'' of the objects in the diagram and think of the morphisms of the diagram as ''
functions'' (in fact,
homomorphisms) acting on those elements.
Then a morphism is a monomorphism
if and only if it is
injective, and it is an epimorphism iff it is
surjective.
Similarly, to deal with exactness, we can think of
kernels and
images in a function-theoretic sense.
The proof will still apply to any Abelian category because of
Mitchell's embedding theorem, which states that any Abelian category can be represented as a category of modules over some ring.
For the category of groups, just turn all additive notation below into multiplicative notation, and note that commutativity is never used.
So, to prove (1), assume that
m and
p are surjective and
q is injective.
Image:FourLemma01.png
* Let
c' be an element of
C'.
* Let
d be an element of the
inverse image under
p of
t(
c');
d exists since
p is surjective.
* By commutativity of the diagram,
u(
p(
d)) =
q(
j(
d)).
* Since im
t = ker
u by exactness, 0 =
u(
t(
c')) =
u(
p(
d)) =
q(
j(
d)).
* Since
q is injective,
j(
d) = 0, so
d is in ker
j = im
h.
* Let
c in
C be such that
h(
c) =
d.
* Then
t(
n(
c)) =
p(
h(
c)) =
t(
c'), so
t(
c' −
n(
c)) = 0.
* By exactness,
c' −
n(
c) must be in the image of
s; let
b' be an element of the inverse image of
c'−
n(
c).
* Since
m is surjective, we can find
b in
B such that
b' =
m(
b).
* By commutativity,
n(
g(
b)) =
s(
m(
b)) =
c' −
n(
c).
* Since
n is a homomorphism,
n(
g(
b) +
c) =
n(
g(
b)) +
n(
c) =
c' −
n(
c) +
n(
c) =
c'.
* Therefore,
n is surjective.
Then, to prove (2), assume that
m and
p are injective and
l is surjective.
Image:FourLemma02.png
* Let
c in
C be such that
n(
c) = 0.
*
t(
n(
c)) is then 0.
* By commutativity,
p(
h(
c)) = 0.
* Since
p is injective,
h(
c) = 0.
* By exactness, there is an element
b of
B such that
g(
b) =
c.
* By commutativity,
s(
m(
b)) =
n(
g(
b)) =
n(
c) = 0.
* By exactness, there is then an element
a' of
A' such that
r(
a') =
m(
b).
* Since
l is surjective, there is
a in
A such that
l(
a) =
a'.
* By commutativity,
m(
f(
a)) =
r(
l(
a)) =
m(
b).
* Since
m is injective,
f(
a) =
b.
* So
c =
g(
f(
a)).
* By exactness, ker
g = im
f, so
c = 0.
* Therefore,
n is injective.
Combining the 2 four lemmas now proves the entire five lemma.
----
See also:
Short five lemma,
Snake lemma
