PERVERSE COHERENT SHEAVES IN GOOD CHARACTERISTIC 7
Consider now the following commutative diagram:
Hom(E∗
M, M)
−◦ν
Ext1(P
, M)
E
id
E
We see that the natural map
Hom(E∗
⊗M, M))
Ext1(P
, M) is an isomorphism,
so from the long exact sequence associated to (2.3), we find that the map
(2.4) Hom(P , M) Hom(P, M)
is an isomorphism as well, and that
Ext1(P,
M)
Ext1(E∗
M, M)
is injective. But
Ext1(M,
M) = 0 because M is projective, so
(2.5)
Ext1(P,
M) = 0.
Before proceeding, we observe that R B. Otherwise, if R had a subquotient
isomorphic to L, there would be a nonzero map M R, since M is the projective
cover of L. But the composition M R M yields a nonscalar element of
Hom(M, M), a contradiction.
Next, for any object X B, we claim that
Hom(E∗
M, X) =
Ext1(E∗
M, X) = 0.
The former holds because L, the unique simple quotient of M, does not occur in
any composition series for X, and the latter holds because M is projective. Then,
from the long exact sequence obtained by applying
Exti(·,X)
to (2.3), we obtain
the following isomorphisms for any X B:
Hom(P , X)

Hom(P, X), (2.6)
Ext1(P
, X)


Ext1(P,
X).
Since P is a projective object of B, the latter isomorphism actually implies
(2.7)
Ext1(P
, X) =
Ext1(P,
X) = 0.
Form the following commutative diagram with exact rows:
Hom(P , R) Hom(P , M) Hom(P , L)
Ext1(P
, R)
Hom(P, R) Hom(P, M) Hom(P, L)
Ext1(P,
R)
The first vertical map is an isomorphism by (2.6), and the second by (2.4). Both
terms in the fourth column vanish by (2.7). Thus, the third vertical map is also an
isomorphism. But Hom(P , L) = 0, so
Hom(P, L) = 0
as well. Combining this with (2.6) and the fact that L is the unique simple quotient
of P , we see that it is the unique simple quotient of P as well.
7
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