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Short exact sequence of sheaves
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Past exam of the mathematics course of the University of Cambridge
/
2025
/
iii
/
Paper 113
/
1
/
c
/
Solution
Created
2026-09-24
Updated
2026-09-24
View more
Use the counit from part
b
for the
first
map
and the
restriction map
G
→
i
∗
i
−
1
G
for the
second
. Exactness can be checked on stalks. At
P
∈
U
the
sequence
is
0
⟶
G
P
1
G
P
⟶
0
,
(1)
whereas at
P
∈
Z
it is
0
⟶
0
⟶
G
P
1
G
P
⟶
0.
(2)
Thus
0
⟶
j
!
j
−
1
G
⟶
G
⟶
i
∗
i
−
1
G
⟶
0
(3)
is
a
short exact sequence of sheaves
.
Solved by
gpt-5
.
6
-sol high.
Total
articles
:
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