Is the pull-back of the structure sheaf the structure sheaf?
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Maybe this is a stupid question, but I got irritated by it: Suppose $f: X rightarrow Y$ is a morphism of schemes. That comes with a map of sheaves $f^#: mathcal{O}_Y rightarrow f_* mathcal{O}_X$. Because $f_*$ and $f^*$ are adjoint to each other, this map corresponds to a homomorphism $f^*mathcal{O}_Y rightarrow mathcal{O}_X$ of $mathcal{O}_X$-modules. But as far as I understand, $f^*mathcal{O}_Y = f^{-1}mathcal{O}_Y otimes_{f^{-1}mathcal{O}_Y}mathcal{O}_X = mathcal{O}_X$. So the map $f^#$ really is the same as an $mathcal{O}_X$-module homomorphism $mathcal{O}_X rightarrow mathcal{O}_X$, which is the same as giving a global section $s in Gamma(X, mathcal{O}_X)$, because the map is fully determined by the value of the global section $1$.
Is this reasoning correct, or did I make a mistake?
algebraic-geometry schemes coherent-sheaves pullback
asked Jan 18 at 8:18
red_trumpetred_trumpet
1,063319
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add a comment |
$begingroup$
Maybe this is a stupid question, but I got irritated by it: Suppose $f: X rightarrow Y$ is a morphism of schemes. That comes with a map of sheaves $f^#: mathcal{O}_Y rightarrow f_* mathcal{O}_X$. Because $f_*$ and $f^*$ are adjoint to each other, this map corresponds to a homomorphism $f^*mathcal{O}_Y rightarrow mathcal{O}_X$ of $mathcal{O}_X$-modules. But as far as I understand, $f^*mathcal{O}_Y = f^{-1}mathcal{O}_Y otimes_{f^{-1}mathcal{O}_Y}mathcal{O}_X = mathcal{O}_X$. So the map $f^#$ really is the same as an $mathcal{O}_X$-module homomorphism $mathcal{O}_X rightarrow mathcal{O}_X$, which is the same as giving a global section $s in Gamma(X, mathcal{O}_X)$, because the map is fully determined by the value of the global section $1$.
Is this reasoning correct, or did I make a mistake?
algebraic-geometry schemes coherent-sheaves pullback
asked Jan 18 at 8:18
red_trumpetred_trumpet
1,063319
$endgroup$
1
$begingroup$
Yes, the canonical map of sheaves on $X$: $f^*mathcal{O}_Y = f^{-1}mathcal{O}_Y otimes_{f^{-1}mathcal{O}_Y}mathcal{O}_X stackrel {cong} {to} mathcal{O}_X$ is an isomorphism.
$endgroup$
– Georges Elencwajg
Jan 18 at 13:04
add a comment |
$begingroup$
Maybe this is a stupid question, but I got irritated by it: Suppose $f: X rightarrow Y$ is a morphism of schemes. That comes with a map of sheaves $f^#: mathcal{O}_Y rightarrow f_* mathcal{O}_X$. Because $f_*$ and $f^*$ are adjoint to each other, this map corresponds to a homomorphism $f^*mathcal{O}_Y rightarrow mathcal{O}_X$ of $mathcal{O}_X$-modules. But as far as I understand, $f^*mathcal{O}_Y = f^{-1}mathcal{O}_Y otimes_{f^{-1}mathcal{O}_Y}mathcal{O}_X = mathcal{O}_X$. So the map $f^#$ really is the same as an $mathcal{O}_X$-module homomorphism $mathcal{O}_X rightarrow mathcal{O}_X$, which is the same as giving a global section $s in Gamma(X, mathcal{O}_X)$, because the map is fully determined by the value of the global section $1$.
Is this reasoning correct, or did I make a mistake?
algebraic-geometry schemes coherent-sheaves pullback
asked Jan 18 at 8:18
red_trumpetred_trumpet
1,063319
$endgroup$
Maybe this is a stupid question, but I got irritated by it: Suppose $f: X rightarrow Y$ is a morphism of schemes. That comes with a map of sheaves $f^#: mathcal{O}_Y rightarrow f_* mathcal{O}_X$. Because $f_*$ and $f^*$ are adjoint to each other, this map corresponds to a homomorphism $f^*mathcal{O}_Y rightarrow mathcal{O}_X$ of $mathcal{O}_X$-modules. But as far as I understand, $f^*mathcal{O}_Y = f^{-1}mathcal{O}_Y otimes_{f^{-1}mathcal{O}_Y}mathcal{O}_X = mathcal{O}_X$. So the map $f^#$ really is the same as an $mathcal{O}_X$-module homomorphism $mathcal{O}_X rightarrow mathcal{O}_X$, which is the same as giving a global section $s in Gamma(X, mathcal{O}_X)$, because the map is fully determined by the value of the global section $1$.
Is this reasoning correct, or did I make a mistake?
algebraic-geometry schemes coherent-sheaves pullback
algebraic-geometry schemes coherent-sheaves pullback
asked Jan 18 at 8:18
red_trumpetred_trumpet
1,063319
asked Jan 18 at 8:18
red_trumpetred_trumpet
1,063319
asked Jan 18 at 8:18
red_trumpetred_trumpet
1,063319
asked Jan 18 at 8:18
red_trumpetred_trumpet
1,063319
asked Jan 18 at 8:18
red_trumpetred_trumpet
1,063319
1,063319
1
$begingroup$
Yes, the canonical map of sheaves on $X$: $f^*mathcal{O}_Y = f^{-1}mathcal{O}_Y otimes_{f^{-1}mathcal{O}_Y}mathcal{O}_X stackrel {cong} {to} mathcal{O}_X$ is an isomorphism.
$endgroup$
– Georges Elencwajg
Jan 18 at 13:04
add a comment |
1
$begingroup$
Yes, the canonical map of sheaves on $X$: $f^*mathcal{O}_Y = f^{-1}mathcal{O}_Y otimes_{f^{-1}mathcal{O}_Y}mathcal{O}_X stackrel {cong} {to} mathcal{O}_X$ is an isomorphism.
$endgroup$
– Georges Elencwajg
Jan 18 at 13:04
1
1
$begingroup$
Yes, the canonical map of sheaves on $X$: $f^*mathcal{O}_Y = f^{-1}mathcal{O}_Y otimes_{f^{-1}mathcal{O}_Y}mathcal{O}_X stackrel {cong} {to} mathcal{O}_X$ is an isomorphism.
$endgroup$
– Georges Elencwajg
Jan 18 at 13:04
$begingroup$
Yes, the canonical map of sheaves on $X$: $f^*mathcal{O}_Y = f^{-1}mathcal{O}_Y otimes_{f^{-1}mathcal{O}_Y}mathcal{O}_X stackrel {cong} {to} mathcal{O}_X$ is an isomorphism.
$endgroup$
– Georges Elencwajg
Jan 18 at 13:04
add a comment |
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$begingroup$
Yes, the canonical map of sheaves on $X$: $f^*mathcal{O}_Y = f^{-1}mathcal{O}_Y otimes_{f^{-1}mathcal{O}_Y}mathcal{O}_X stackrel {cong} {to} mathcal{O}_X$ is an isomorphism.
$endgroup$
– Georges Elencwajg
Jan 18 at 13:04