Pushforward of projection map exact?
0
$begingroup$
Let
$$0to F to G to Hto 0$$
be a short exact sequence of coherent sheaves on the product spaces $Xtimes Y$, and $p$ be the projection to $X$. Is it true that
$$0to p_*F to p_*G to p_*Hto 0$$
is still exact?
algebraic-geometry exact-sequence
asked Dec 16 '18 at 20:34
User XUser X
32911
$endgroup$
add a comment |
0
$begingroup$
Let
$$0to F to G to Hto 0$$
be a short exact sequence of coherent sheaves on the product spaces $Xtimes Y$, and $p$ be the projection to $X$. Is it true that
$$0to p_*F to p_*G to p_*Hto 0$$
is still exact?
algebraic-geometry exact-sequence
asked Dec 16 '18 at 20:34
User XUser X
32911
$endgroup$
1
$begingroup$
No (even when $X$ is the spectrum of your base field).
$endgroup$
– Sasha
Dec 16 '18 at 20:43
$begingroup$
@Sasha Is it true for some special case like $F,G,H$ are line bundles and $X,Y$ are projective spaces?
$endgroup$
– User X
Dec 16 '18 at 20:47
$begingroup$
One case this is true is when $Y$ is affine. If $Y$ is not affine, in general, this is false.
$endgroup$
– Mohan
Dec 16 '18 at 23:10
1
$begingroup$
@UserX: $F$, $G$, and $H$ cannot be line bundles at the same time. A sufficient condition for the exactness is $R^1p_*F = 0$. This holds true, for instance, when $F$ is a sum of tensor products of line bundles pulled back from $X$ and $Y$, and $Y$ is a projective space of dimension $ge 2$.
$endgroup$
– Sasha
Dec 17 '18 at 7:18
$begingroup$
@Sasha Yeah this is enough for my case. Thx! You can post it as an answer if you want.
$endgroup$
– User X
Dec 17 '18 at 8:24
add a comment |
0
0
0
$begingroup$
Let
$$0to F to G to Hto 0$$
be a short exact sequence of coherent sheaves on the product spaces $Xtimes Y$, and $p$ be the projection to $X$. Is it true that
$$0to p_*F to p_*G to p_*Hto 0$$
is still exact?
algebraic-geometry exact-sequence
asked Dec 16 '18 at 20:34
User XUser X
32911
$endgroup$
Let
$$0to F to G to Hto 0$$
be a short exact sequence of coherent sheaves on the product spaces $Xtimes Y$, and $p$ be the projection to $X$. Is it true that
$$0to p_*F to p_*G to p_*Hto 0$$
is still exact?
algebraic-geometry exact-sequence
algebraic-geometry exact-sequence
asked Dec 16 '18 at 20:34
User XUser X
32911
asked Dec 16 '18 at 20:34
User XUser X
32911
asked Dec 16 '18 at 20:34
User XUser X
32911
asked Dec 16 '18 at 20:34
User XUser X
32911
asked Dec 16 '18 at 20:34
User XUser X
32911
32911
1
$begingroup$
No (even when $X$ is the spectrum of your base field).
$endgroup$
– Sasha
Dec 16 '18 at 20:43
$begingroup$
@Sasha Is it true for some special case like $F,G,H$ are line bundles and $X,Y$ are projective spaces?
$endgroup$
– User X
Dec 16 '18 at 20:47
$begingroup$
One case this is true is when $Y$ is affine. If $Y$ is not affine, in general, this is false.
$endgroup$
– Mohan
Dec 16 '18 at 23:10
1
$begingroup$
@UserX: $F$, $G$, and $H$ cannot be line bundles at the same time. A sufficient condition for the exactness is $R^1p_*F = 0$. This holds true, for instance, when $F$ is a sum of tensor products of line bundles pulled back from $X$ and $Y$, and $Y$ is a projective space of dimension $ge 2$.
$endgroup$
– Sasha
Dec 17 '18 at 7:18
$begingroup$
@Sasha Yeah this is enough for my case. Thx! You can post it as an answer if you want.
$endgroup$
– User X
Dec 17 '18 at 8:24
add a comment |
1
$begingroup$
No (even when $X$ is the spectrum of your base field).
$endgroup$
– Sasha
Dec 16 '18 at 20:43
$begingroup$
@Sasha Is it true for some special case like $F,G,H$ are line bundles and $X,Y$ are projective spaces?
$endgroup$
– User X
Dec 16 '18 at 20:47
$begingroup$
One case this is true is when $Y$ is affine. If $Y$ is not affine, in general, this is false.
$endgroup$
– Mohan
Dec 16 '18 at 23:10
1
$begingroup$
@UserX: $F$, $G$, and $H$ cannot be line bundles at the same time. A sufficient condition for the exactness is $R^1p_*F = 0$. This holds true, for instance, when $F$ is a sum of tensor products of line bundles pulled back from $X$ and $Y$, and $Y$ is a projective space of dimension $ge 2$.
$endgroup$
– Sasha
Dec 17 '18 at 7:18
$begingroup$
@Sasha Yeah this is enough for my case. Thx! You can post it as an answer if you want.
$endgroup$
– User X
Dec 17 '18 at 8:24
1
1
$begingroup$
No (even when $X$ is the spectrum of your base field).
$endgroup$
– Sasha
Dec 16 '18 at 20:43
$begingroup$
No (even when $X$ is the spectrum of your base field).
$endgroup$
– Sasha
Dec 16 '18 at 20:43
$begingroup$
@Sasha Is it true for some special case like $F,G,H$ are line bundles and $X,Y$ are projective spaces?
$endgroup$
– User X
Dec 16 '18 at 20:47
$begingroup$
@Sasha Is it true for some special case like $F,G,H$ are line bundles and $X,Y$ are projective spaces?
$endgroup$
– User X
Dec 16 '18 at 20:47
$begingroup$
One case this is true is when $Y$ is affine. If $Y$ is not affine, in general, this is false.
$endgroup$
– Mohan
Dec 16 '18 at 23:10
$begingroup$
One case this is true is when $Y$ is affine. If $Y$ is not affine, in general, this is false.
$endgroup$
– Mohan
Dec 16 '18 at 23:10
1
1
$begingroup$
@UserX: $F$, $G$, and $H$ cannot be line bundles at the same time. A sufficient condition for the exactness is $R^1p_*F = 0$. This holds true, for instance, when $F$ is a sum of tensor products of line bundles pulled back from $X$ and $Y$, and $Y$ is a projective space of dimension $ge 2$.
$endgroup$
– Sasha
Dec 17 '18 at 7:18
$begingroup$
@UserX: $F$, $G$, and $H$ cannot be line bundles at the same time. A sufficient condition for the exactness is $R^1p_*F = 0$. This holds true, for instance, when $F$ is a sum of tensor products of line bundles pulled back from $X$ and $Y$, and $Y$ is a projective space of dimension $ge 2$.
$endgroup$
– Sasha
Dec 17 '18 at 7:18
$begingroup$
@Sasha Yeah this is enough for my case. Thx! You can post it as an answer if you want.
$endgroup$
– User X
Dec 17 '18 at 8:24
$begingroup$
@Sasha Yeah this is enough for my case. Thx! You can post it as an answer if you want.
$endgroup$
– User X
Dec 17 '18 at 8:24
add a comment |
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1
$begingroup$
No (even when $X$ is the spectrum of your base field).
$endgroup$
– Sasha
Dec 16 '18 at 20:43
$begingroup$
@Sasha Is it true for some special case like $F,G,H$ are line bundles and $X,Y$ are projective spaces?
$endgroup$
– User X
Dec 16 '18 at 20:47
$begingroup$
One case this is true is when $Y$ is affine. If $Y$ is not affine, in general, this is false.
$endgroup$
– Mohan
Dec 16 '18 at 23:10
1
$begingroup$
@UserX: $F$, $G$, and $H$ cannot be line bundles at the same time. A sufficient condition for the exactness is $R^1p_*F = 0$. This holds true, for instance, when $F$ is a sum of tensor products of line bundles pulled back from $X$ and $Y$, and $Y$ is a projective space of dimension $ge 2$.
$endgroup$
– Sasha
Dec 17 '18 at 7:18
$begingroup$
@Sasha Yeah this is enough for my case. Thx! You can post it as an answer if you want.
$endgroup$
– User X
Dec 17 '18 at 8:24