# Math 501 - Problem Set 5 (due in class on Tue, Mar 26) Problem 1

```Math 501 - Problem Set 5 (due in class on Tue, Mar 26)
Problem 1. Show directly (without using the bijection with Ext1 ) that the
extension classes of the abelian group Z/(p) are represented by the split exact
sequence and the short exact sequences of the form
p
i
0 → Z/(p) → Z/(p2 ) → Z/(p) → 0.
where 1 ≤ i ≤ p − 1.
Problem 2. Let A be an abelian category with enough injectives, and let
A, B be objects of A. Given n-extensions (exact sequences)
ξ : 0 → B → X 1 → &middot; &middot; &middot; → X n → A → 0,
and
ξ 0 : 0 → B → Y 1 → &middot; &middot; &middot; → Y n → A → 0,
of A by B, we say that ξ is pre-equivalent to ξ 0 if there exists a commutative
diagram
ξ:
/
0
B
idB
ξ0 :
/
0
B
/
/
X1
Y1
/ ...
/ ...
/
/
Xn
Yn
/
/
/
A
0
idA
A
/
0.
We define Yoneda equivalence to be the smallest equivalence relation on the
set of n-extensions of A by B which contains the pre-equivalences and write
ξ ∼ ξ 0 if ξ and ξ 0 are Yoneda equivalent.
(1) Show that there is a natural bijection
∼
n-extensions of A by B
=
Θ:
−→ ExtnA (A, B)
up to Yoneda equivalence
generalizing the statement for 1-extensions proved in class. The map
Θ, called the Yoneda bijection, is constructed as follows. Given an
n-extension ξ as above, we consider ξ as a right resolution
'
B −→ (X 1 → X 2 → &middot; &middot; &middot; → X n → A → 0) =: X •
'
and choose an injective resolution B → I • . The identity map B → B
extends to a map of complexes f : X • → I • which is unique up to homotopy. We apply the functor Hom(A, −) and pass to nth cohomology
to obtain a canonical map Hom(A, A) → ExtnA (A, B). Define Θ(ξ) to
be the image of idA under this map.
(2) We extend the Baer sum defined in class for n = 1: For n &gt; 1 we define
the Baer sum of n-extensions ξ and ξ 0 to be the n-extension
`
0 → B → X 1 B Y 1 → X 2 ⊕Y 2 → &middot; &middot; &middot; → X n−1 ⊕Y n−1 → X n &times;A Y n → A → 0.
where the leftmost and rightmost terms denote a fibered coproduct
and fibered product, respectively. Show that this operation provides
an abelian group structure on the set of n-extension classes such that
the map Θ becomes a group isomorphism.
Problem 3. Let A be an abelian category with enough projectives. Consider
the abelian category Aop and define the map Θ = ΘAop as in Problem 2(1).
(1) Given objects A, B of A, show that there is a natural identification
of Yoneda equivalence classes of n-extensions of B by A in Aop with
equivalence classes of n-extensions of A by B in A. Show that this
identification is a group isomorphism with respect to the Baer sum.
Show that there is a natural identification ExtnAop (B, A) ∼
= ExtnA (A, B)
where both groups are computed via a projective resolution of P• →
A. Deduce that we obtain a natural group isomorphism Θ0 which is
uniquely determined by the commutativity of the diagram
ΘAop
n-extension classes
/ Extn op (B, A)
op
A
∼
=
in A of B by A
∼
=
∼
=
n-extension classes
in A of A by B
Θ0
/
ExtnA (A, B)
(2) Assume that A has enough projectives and injectives. Let A,B be
'
'
objects of A, let P• → A be a projective resolution, and let B → I •
be an injective resolution of B. We introduce the mapping complex
Hom• (P• , I • ) to be the cochain complex given by
M
Homn (P• , I • ) =
Hom(Pi , I j )
i+j=n
2
with differential on a map f of degree n given by
d(f ) = dI ◦ f − (−1)n f ◦ dP .
Show that there are natural quasi-isomorphisms
'
'
Hom(P• , B) −→ Hom• (P• , I • ) ←− Hom(A, I • )
which can be used to balance Ext∗ in analogy to the balancing isomorphisms for Tor∗ constructed in class. Show that under the resulting
identification of ExtnA (A, B), computed via a projective resolution of
A, and ExtnA (A, B) computed via an injective resolution of B, the two
maps Θ and Θ0 agree.
Problem 4. Let A be an abelian category with enough injectives. Let A, B
and C be objects of A. Given a map f : A → B and an n-extension
ξ : C → X1 → &middot; &middot; &middot; → Xn → B → 0
of B by C, we define an n-extension
ξ ◦ f : 0 → C → X 1 → &middot; &middot; &middot; → X n−1 → X n &times;B A → A → 0.
Given a map g : B → C and an m-extension
ν : B → Y 1 → &middot; &middot; &middot; → Y m → A → 0,
of A by B, we define an m-extension
`
g ◦ ν : 0 → C → Y 1 B C → Y 2 &middot; &middot; &middot; → Y m → A → 0.
Finally, given an n-extension ξ of B by C and an m-extension ν of A by B
as above, we define an n + m-extension
ξ ◦ ν : C → X1 → &middot; &middot; &middot; → Xn → Y 1 → &middot; &middot; &middot; → Y m → A → 0
by composing X n → B and B → Y 1 .
(1) Use the Yoneda interpretation of Ext∗ to show that, for every n, m ≥ 0,
these constructions provide a well-defined bilinear map
n+m
ExtnA (A, B) &times; Extm
(A, C)
A (B, C) → ExtA
called Yoneda composition.
3
(2) Given objects A, B of A, we define the graded abelian group
M
Ext∗A (A, B) :=
ExtnA (A, B).
n≥0
Show that Yoneda composition defines a category Ext∗ (A) with objects
given by the objects of A and morphisms between objects A,B of A
given by
M
HomExt∗ (A) (A, B) = Ext∗A (A, B) :=
ExtnA (A, B).
n≥0
The statement to verify is the associativity of the composition law.
Note that all morphism sets in this category carry the structure of
a graded abelian group which is respected by the composition law.
We say that Ext∗ (A) is enriched over graded abelian groups. Further
note that the category A is obtained from Ext∗ (A) by discarding all
morphisms of positive degree.
(3) Deduce that, given an object A of A, the graded abelian group Ext∗A (A, A)
carries a canonical associative graded Z-algebra structure.
(4) Let k be a field. Consider the k[x, y]-module k ∼
= k[x, y]/(x, y). Explic∗
itly compute the graded algebra Extk[x,y] (k, k).
(5) Bonus: Prove a statement which, for any abelian category A with
enough projectives, allows you to directly compute the algebra Ext∗A (A, A)
in terms of a projective resolution of A without reference to the Yoneda
isomorphism.
4
```