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.

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