Characterization of irreducible spaces in terms of sheaves
$begingroup$
The following exercise appears in the algebraic geometry textbook by Görtz and Wedhorn (Exercise 2.13 (b)):
Let $X$ be a connected topological space and assume that there exists
a sheaf $mathcal{F}$ such that $mathcal{F} (X) to mathcal{F} (U)$
is bijective for all non-empty open sets $U subseteq X$ and such that
$mathcal{F} (X)$ contains more than one element. Show that $X$ is
irreducible.
Is the assumption that $X$ be connected necessary here?
Here's my argument that does not use connectedness.
Note that the assumption that all the restrictions $mathcal{F} (X) to mathcal{F} (U)$ are bijective for $U ne emptyset$ implies that $mathcal{F} (U) to mathcal{F} (V)$ is bijective for any pair of nonempty open subsets $V subseteq U$. Indeed, in this case the restriction $operatorname{res}^U_Vcolon mathcal{F} (U) to mathcal{F} (V)$ is given by
$$mathcal{F} (U) xrightarrow{(operatorname{res}^X_U)^{-1}} mathcal{F} (X) xrightarrow{operatorname{res}^X_V} mathcal{F} (V)$$
To show that $X$ is irreducible, we may show that any two nonempty open subsets $U,V$ have a nonempty intersection. For the sake of contradiction, assume that $U, V ne emptyset$ and $Ucap V = emptyset$. Then the sheaf axioms for $mathcal{F}$ imply that
$$mathcal{F} (Ucup V) xrightarrow{cong} mathcal{F} (U) times mathcal{F} (V)$$
is a bijection, given by $s mapsto (s|_U, s|_V)$. However, the restrictions $rho_Ucolon s mapsto s|_U$ and $rho_Vcolon s mapsto s|_V$ are bijective by our asumption on $mathcal{F}$, and this would imply that the diagonal map
$$Deltacolon mathcal{F} (Ucup V) xrightarrow{(rho_U,rho_V)} mathcal{F} (U) times mathcal{F} (V) xrightarrow{rho_U^{-1}times rho_V^{-1}} mathcal{F} (Ucup V) times mathcal{F} (Ucup V)$$
is bijective. But as $mathcal{F} (Ucup V) cong mathcal{F} (X)$ has at least two different elements $s ne t$ by our assumption, the element $(s,t)$ is clearly not in the image of $Delta$.
This contradiction shows that $Ucap V ne emptyset$.
Does this look right? Note that the fact that $X$ is connected was not used. (But I don't see how a disconnected $X$ can admit such a sheaf $mathcal{F}$.)
general-topology sheaf-theory
asked Jan 15 at 6:00
Sr. TacuacínSr. Tacuacín
212
$endgroup$
add a comment |
$begingroup$
The following exercise appears in the algebraic geometry textbook by Görtz and Wedhorn (Exercise 2.13 (b)):
Let $X$ be a connected topological space and assume that there exists
a sheaf $mathcal{F}$ such that $mathcal{F} (X) to mathcal{F} (U)$
is bijective for all non-empty open sets $U subseteq X$ and such that
$mathcal{F} (X)$ contains more than one element. Show that $X$ is
irreducible.
Is the assumption that $X$ be connected necessary here?
Here's my argument that does not use connectedness.
Note that the assumption that all the restrictions $mathcal{F} (X) to mathcal{F} (U)$ are bijective for $U ne emptyset$ implies that $mathcal{F} (U) to mathcal{F} (V)$ is bijective for any pair of nonempty open subsets $V subseteq U$. Indeed, in this case the restriction $operatorname{res}^U_Vcolon mathcal{F} (U) to mathcal{F} (V)$ is given by
$$mathcal{F} (U) xrightarrow{(operatorname{res}^X_U)^{-1}} mathcal{F} (X) xrightarrow{operatorname{res}^X_V} mathcal{F} (V)$$
To show that $X$ is irreducible, we may show that any two nonempty open subsets $U,V$ have a nonempty intersection. For the sake of contradiction, assume that $U, V ne emptyset$ and $Ucap V = emptyset$. Then the sheaf axioms for $mathcal{F}$ imply that
$$mathcal{F} (Ucup V) xrightarrow{cong} mathcal{F} (U) times mathcal{F} (V)$$
is a bijection, given by $s mapsto (s|_U, s|_V)$. However, the restrictions $rho_Ucolon s mapsto s|_U$ and $rho_Vcolon s mapsto s|_V$ are bijective by our asumption on $mathcal{F}$, and this would imply that the diagonal map
$$Deltacolon mathcal{F} (Ucup V) xrightarrow{(rho_U,rho_V)} mathcal{F} (U) times mathcal{F} (V) xrightarrow{rho_U^{-1}times rho_V^{-1}} mathcal{F} (Ucup V) times mathcal{F} (Ucup V)$$
is bijective. But as $mathcal{F} (Ucup V) cong mathcal{F} (X)$ has at least two different elements $s ne t$ by our assumption, the element $(s,t)$ is clearly not in the image of $Delta$.
This contradiction shows that $Ucap V ne emptyset$.
Does this look right? Note that the fact that $X$ is connected was not used. (But I don't see how a disconnected $X$ can admit such a sheaf $mathcal{F}$.)
general-topology sheaf-theory
asked Jan 15 at 6:00
Sr. TacuacínSr. Tacuacín
212
$endgroup$
2
$begingroup$
Yes your argument is correct, so in particular you don't need $X$ connected.
$endgroup$
– Nicolas Hemelsoet
Jan 15 at 6:17
$begingroup$
$F(U)$ is a member of some category, according to the wikipedia definition. You're mapping into $textrm{Set}$? Are you following that definition? or some other?
$endgroup$
– Henno Brandsma
Jan 15 at 6:35
$begingroup$
Could you expand the first paragraph starting with "Note that the assumption.." I don't see it.
$endgroup$
– Henno Brandsma
Jan 15 at 6:36
$begingroup$
@HennoBrandsma The sheaves are sheaves of sets. The statement you mention follows from the fact that the restriction $mathcal{F} (X) to mathcal{F} (V)$ is the composition $mathcal{F} (X) to mathcal{F} (U) to mathcal{F} (V)$.
$endgroup$
– Sr. Tacuacín
Jan 15 at 6:45
add a comment |
$begingroup$
The following exercise appears in the algebraic geometry textbook by Görtz and Wedhorn (Exercise 2.13 (b)):
Let $X$ be a connected topological space and assume that there exists
a sheaf $mathcal{F}$ such that $mathcal{F} (X) to mathcal{F} (U)$
is bijective for all non-empty open sets $U subseteq X$ and such that
$mathcal{F} (X)$ contains more than one element. Show that $X$ is
irreducible.
Is the assumption that $X$ be connected necessary here?
Here's my argument that does not use connectedness.
Note that the assumption that all the restrictions $mathcal{F} (X) to mathcal{F} (U)$ are bijective for $U ne emptyset$ implies that $mathcal{F} (U) to mathcal{F} (V)$ is bijective for any pair of nonempty open subsets $V subseteq U$. Indeed, in this case the restriction $operatorname{res}^U_Vcolon mathcal{F} (U) to mathcal{F} (V)$ is given by
$$mathcal{F} (U) xrightarrow{(operatorname{res}^X_U)^{-1}} mathcal{F} (X) xrightarrow{operatorname{res}^X_V} mathcal{F} (V)$$
To show that $X$ is irreducible, we may show that any two nonempty open subsets $U,V$ have a nonempty intersection. For the sake of contradiction, assume that $U, V ne emptyset$ and $Ucap V = emptyset$. Then the sheaf axioms for $mathcal{F}$ imply that
$$mathcal{F} (Ucup V) xrightarrow{cong} mathcal{F} (U) times mathcal{F} (V)$$
is a bijection, given by $s mapsto (s|_U, s|_V)$. However, the restrictions $rho_Ucolon s mapsto s|_U$ and $rho_Vcolon s mapsto s|_V$ are bijective by our asumption on $mathcal{F}$, and this would imply that the diagonal map
$$Deltacolon mathcal{F} (Ucup V) xrightarrow{(rho_U,rho_V)} mathcal{F} (U) times mathcal{F} (V) xrightarrow{rho_U^{-1}times rho_V^{-1}} mathcal{F} (Ucup V) times mathcal{F} (Ucup V)$$
is bijective. But as $mathcal{F} (Ucup V) cong mathcal{F} (X)$ has at least two different elements $s ne t$ by our assumption, the element $(s,t)$ is clearly not in the image of $Delta$.
This contradiction shows that $Ucap V ne emptyset$.
Does this look right? Note that the fact that $X$ is connected was not used. (But I don't see how a disconnected $X$ can admit such a sheaf $mathcal{F}$.)
general-topology sheaf-theory
asked Jan 15 at 6:00
Sr. TacuacínSr. Tacuacín
212
$endgroup$
The following exercise appears in the algebraic geometry textbook by Görtz and Wedhorn (Exercise 2.13 (b)):
Let $X$ be a connected topological space and assume that there exists
a sheaf $mathcal{F}$ such that $mathcal{F} (X) to mathcal{F} (U)$
is bijective for all non-empty open sets $U subseteq X$ and such that
$mathcal{F} (X)$ contains more than one element. Show that $X$ is
irreducible.
Is the assumption that $X$ be connected necessary here?
Here's my argument that does not use connectedness.
Note that the assumption that all the restrictions $mathcal{F} (X) to mathcal{F} (U)$ are bijective for $U ne emptyset$ implies that $mathcal{F} (U) to mathcal{F} (V)$ is bijective for any pair of nonempty open subsets $V subseteq U$. Indeed, in this case the restriction $operatorname{res}^U_Vcolon mathcal{F} (U) to mathcal{F} (V)$ is given by
$$mathcal{F} (U) xrightarrow{(operatorname{res}^X_U)^{-1}} mathcal{F} (X) xrightarrow{operatorname{res}^X_V} mathcal{F} (V)$$
To show that $X$ is irreducible, we may show that any two nonempty open subsets $U,V$ have a nonempty intersection. For the sake of contradiction, assume that $U, V ne emptyset$ and $Ucap V = emptyset$. Then the sheaf axioms for $mathcal{F}$ imply that
$$mathcal{F} (Ucup V) xrightarrow{cong} mathcal{F} (U) times mathcal{F} (V)$$
is a bijection, given by $s mapsto (s|_U, s|_V)$. However, the restrictions $rho_Ucolon s mapsto s|_U$ and $rho_Vcolon s mapsto s|_V$ are bijective by our asumption on $mathcal{F}$, and this would imply that the diagonal map
$$Deltacolon mathcal{F} (Ucup V) xrightarrow{(rho_U,rho_V)} mathcal{F} (U) times mathcal{F} (V) xrightarrow{rho_U^{-1}times rho_V^{-1}} mathcal{F} (Ucup V) times mathcal{F} (Ucup V)$$
is bijective. But as $mathcal{F} (Ucup V) cong mathcal{F} (X)$ has at least two different elements $s ne t$ by our assumption, the element $(s,t)$ is clearly not in the image of $Delta$.
This contradiction shows that $Ucap V ne emptyset$.
Does this look right? Note that the fact that $X$ is connected was not used. (But I don't see how a disconnected $X$ can admit such a sheaf $mathcal{F}$.)
general-topology sheaf-theory
general-topology sheaf-theory
asked Jan 15 at 6:00
Sr. TacuacínSr. Tacuacín
212
asked Jan 15 at 6:00
Sr. TacuacínSr. Tacuacín
212
edited Jan 15 at 6:43
Sr. Tacuacín
asked Jan 15 at 6:00
Sr. TacuacínSr. Tacuacín
212
asked Jan 15 at 6:00
Sr. TacuacínSr. Tacuacín
212
asked Jan 15 at 6:00
Sr. TacuacínSr. Tacuacín
212
212
2
$begingroup$
Yes your argument is correct, so in particular you don't need $X$ connected.
$endgroup$
– Nicolas Hemelsoet
Jan 15 at 6:17
$begingroup$
$F(U)$ is a member of some category, according to the wikipedia definition. You're mapping into $textrm{Set}$? Are you following that definition? or some other?
$endgroup$
– Henno Brandsma
Jan 15 at 6:35
$begingroup$
Could you expand the first paragraph starting with "Note that the assumption.." I don't see it.
$endgroup$
– Henno Brandsma
Jan 15 at 6:36
$begingroup$
@HennoBrandsma The sheaves are sheaves of sets. The statement you mention follows from the fact that the restriction $mathcal{F} (X) to mathcal{F} (V)$ is the composition $mathcal{F} (X) to mathcal{F} (U) to mathcal{F} (V)$.
$endgroup$
– Sr. Tacuacín
Jan 15 at 6:45
add a comment |
2
$begingroup$
Yes your argument is correct, so in particular you don't need $X$ connected.
$endgroup$
– Nicolas Hemelsoet
Jan 15 at 6:17
$begingroup$
$F(U)$ is a member of some category, according to the wikipedia definition. You're mapping into $textrm{Set}$? Are you following that definition? or some other?
$endgroup$
– Henno Brandsma
Jan 15 at 6:35
$begingroup$
Could you expand the first paragraph starting with "Note that the assumption.." I don't see it.
$endgroup$
– Henno Brandsma
Jan 15 at 6:36
$begingroup$
@HennoBrandsma The sheaves are sheaves of sets. The statement you mention follows from the fact that the restriction $mathcal{F} (X) to mathcal{F} (V)$ is the composition $mathcal{F} (X) to mathcal{F} (U) to mathcal{F} (V)$.
$endgroup$
– Sr. Tacuacín
Jan 15 at 6:45
2
2
$begingroup$
Yes your argument is correct, so in particular you don't need $X$ connected.
$endgroup$
– Nicolas Hemelsoet
Jan 15 at 6:17
$begingroup$
Yes your argument is correct, so in particular you don't need $X$ connected.
$endgroup$
– Nicolas Hemelsoet
Jan 15 at 6:17
$begingroup$
$F(U)$ is a member of some category, according to the wikipedia definition. You're mapping into $textrm{Set}$? Are you following that definition? or some other?
$endgroup$
– Henno Brandsma
Jan 15 at 6:35
$begingroup$
$F(U)$ is a member of some category, according to the wikipedia definition. You're mapping into $textrm{Set}$? Are you following that definition? or some other?
$endgroup$
– Henno Brandsma
Jan 15 at 6:35
$begingroup$
Could you expand the first paragraph starting with "Note that the assumption.." I don't see it.
$endgroup$
– Henno Brandsma
Jan 15 at 6:36
$begingroup$
Could you expand the first paragraph starting with "Note that the assumption.." I don't see it.
$endgroup$
– Henno Brandsma
Jan 15 at 6:36
$begingroup$
@HennoBrandsma The sheaves are sheaves of sets. The statement you mention follows from the fact that the restriction $mathcal{F} (X) to mathcal{F} (V)$ is the composition $mathcal{F} (X) to mathcal{F} (U) to mathcal{F} (V)$.
$endgroup$
– Sr. Tacuacín
Jan 15 at 6:45
$begingroup$
@HennoBrandsma The sheaves are sheaves of sets. The statement you mention follows from the fact that the restriction $mathcal{F} (X) to mathcal{F} (V)$ is the composition $mathcal{F} (X) to mathcal{F} (U) to mathcal{F} (V)$.
$endgroup$
– Sr. Tacuacín
Jan 15 at 6:45
add a comment |
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$begingroup$
Yes your argument is correct, so in particular you don't need $X$ connected.
$endgroup$
– Nicolas Hemelsoet
Jan 15 at 6:17
$begingroup$
$F(U)$ is a member of some category, according to the wikipedia definition. You're mapping into $textrm{Set}$? Are you following that definition? or some other?
$endgroup$
– Henno Brandsma
Jan 15 at 6:35
$begingroup$
Could you expand the first paragraph starting with "Note that the assumption.." I don't see it.
$endgroup$
– Henno Brandsma
Jan 15 at 6:36
$begingroup$
@HennoBrandsma The sheaves are sheaves of sets. The statement you mention follows from the fact that the restriction $mathcal{F} (X) to mathcal{F} (V)$ is the composition $mathcal{F} (X) to mathcal{F} (U) to mathcal{F} (V)$.
$endgroup$
– Sr. Tacuacín
Jan 15 at 6:45