Johnstone, Topos theory: families of arrows inducing the same sheaf condition
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Johnstone, Topos theory, 0.3, page 13, asserts that, given a Grothendieck pretopology $P$, if the equalizer condition on a presheaf $F$ is satisfied for a family of arrows $R={U_ito U}$, then it is satisfied for every family $S={W_jto U}supset R$.
By equalizer condition I mean that $$F(U)to prod_R F(U_i)rightrightarrows prod_{Rtimes R}F(U_itimes_U U_j),$$ with the natural restriction maps, is an equalizer.
I can't see why this is true. First of all, should I assume that $R$ is a covering in the given pretopology? Otherwise it would sound very strange to me. For example, take a discrete space of two points $X={p,q}$, the pretopology being given by the topological coverings, and take $R={pto X}, S={pto X,qto X}$; then the equalizer condition for $R$ means that every local section on $p$ comes from a global section on $X$. So take $F(p):={0}, F(q)=mathbb Z, F(varnothing)={0},F(X)={0}$. This is a presheaf (with the natural maps, identity or zero) and satisfies the equalizer condition for $U=X$ and the family $R$ just defined: in
$${0}to {0} rightrightarrows {0}$$ one can always take the global section $0$ on the left. However, the equalizer for $S$ is: $${0}to {0}times mathbb Zrightrightarrows {0}times {0}times {0}times mathbb Z$$ where both arrows on the right are equal to $(0,1)mapsto (0,0,0,1)$ (note that the "mixed" fiber products $ptimes_X q$ and $qtimes _X p$ are the empty set). So take $(0,1)in {0}times mathbb Z$ in the central node of the last diagram. The maps on the right are the same, hence $F$ satisfies the equalizer condition for $S$ if and only if there exists $sin F(X)=0$ s.t. $s|_{{p}}=0,s|_{{q}}=1$, but this is of course impossible.
Is my conterexample correct?
Thank you in advance.
sheaf-theory topos-theory
asked Dec 9 '18 at 14:53
W. RetherW. Rether
738417
$endgroup$
|
show 2 more comments
$begingroup$
Johnstone, Topos theory, 0.3, page 13, asserts that, given a Grothendieck pretopology $P$, if the equalizer condition on a presheaf $F$ is satisfied for a family of arrows $R={U_ito U}$, then it is satisfied for every family $S={W_jto U}supset R$.
By equalizer condition I mean that $$F(U)to prod_R F(U_i)rightrightarrows prod_{Rtimes R}F(U_itimes_U U_j),$$ with the natural restriction maps, is an equalizer.
I can't see why this is true. First of all, should I assume that $R$ is a covering in the given pretopology? Otherwise it would sound very strange to me. For example, take a discrete space of two points $X={p,q}$, the pretopology being given by the topological coverings, and take $R={pto X}, S={pto X,qto X}$; then the equalizer condition for $R$ means that every local section on $p$ comes from a global section on $X$. So take $F(p):={0}, F(q)=mathbb Z, F(varnothing)={0},F(X)={0}$. This is a presheaf (with the natural maps, identity or zero) and satisfies the equalizer condition for $U=X$ and the family $R$ just defined: in
$${0}to {0} rightrightarrows {0}$$ one can always take the global section $0$ on the left. However, the equalizer for $S$ is: $${0}to {0}times mathbb Zrightrightarrows {0}times {0}times {0}times mathbb Z$$ where both arrows on the right are equal to $(0,1)mapsto (0,0,0,1)$ (note that the "mixed" fiber products $ptimes_X q$ and $qtimes _X p$ are the empty set). So take $(0,1)in {0}times mathbb Z$ in the central node of the last diagram. The maps on the right are the same, hence $F$ satisfies the equalizer condition for $S$ if and only if there exists $sin F(X)=0$ s.t. $s|_{{p}}=0,s|_{{q}}=1$, but this is of course impossible.
Is my conterexample correct?
Thank you in advance.
sheaf-theory topos-theory
asked Dec 9 '18 at 14:53
W. RetherW. Rether
738417
$endgroup$
$begingroup$
In your $F$, what is the morphism $F(q)to F(varnothing)$?
$endgroup$
– Malice Vidrine
Dec 9 '18 at 18:59
$begingroup$
The zero morphism. Is this a problem?
$endgroup$
– W. Rether
Dec 9 '18 at 19:16
$begingroup$
I think I'm just being confused by you notation. Writing ${0}$ as $0$, when $0$ is also common notation for an initial object, makes it look like you're trying to reason from an impossible presheaf. I think I see what you're doing now.
$endgroup$
– Malice Vidrine
Dec 9 '18 at 19:26
$begingroup$
Oh, sorry! Fixed. $F$ should be a presheaf, isn't it?
$endgroup$
– W. Rether
Dec 9 '18 at 20:33
$begingroup$
Another possible counterexample in the first displayed answer to math.stackexchange.com/questions/856564/…
$endgroup$
– W. Rether
Dec 9 '18 at 20:38
|
show 2 more comments
$begingroup$
Johnstone, Topos theory, 0.3, page 13, asserts that, given a Grothendieck pretopology $P$, if the equalizer condition on a presheaf $F$ is satisfied for a family of arrows $R={U_ito U}$, then it is satisfied for every family $S={W_jto U}supset R$.
By equalizer condition I mean that $$F(U)to prod_R F(U_i)rightrightarrows prod_{Rtimes R}F(U_itimes_U U_j),$$ with the natural restriction maps, is an equalizer.
I can't see why this is true. First of all, should I assume that $R$ is a covering in the given pretopology? Otherwise it would sound very strange to me. For example, take a discrete space of two points $X={p,q}$, the pretopology being given by the topological coverings, and take $R={pto X}, S={pto X,qto X}$; then the equalizer condition for $R$ means that every local section on $p$ comes from a global section on $X$. So take $F(p):={0}, F(q)=mathbb Z, F(varnothing)={0},F(X)={0}$. This is a presheaf (with the natural maps, identity or zero) and satisfies the equalizer condition for $U=X$ and the family $R$ just defined: in
$${0}to {0} rightrightarrows {0}$$ one can always take the global section $0$ on the left. However, the equalizer for $S$ is: $${0}to {0}times mathbb Zrightrightarrows {0}times {0}times {0}times mathbb Z$$ where both arrows on the right are equal to $(0,1)mapsto (0,0,0,1)$ (note that the "mixed" fiber products $ptimes_X q$ and $qtimes _X p$ are the empty set). So take $(0,1)in {0}times mathbb Z$ in the central node of the last diagram. The maps on the right are the same, hence $F$ satisfies the equalizer condition for $S$ if and only if there exists $sin F(X)=0$ s.t. $s|_{{p}}=0,s|_{{q}}=1$, but this is of course impossible.
Is my conterexample correct?
Thank you in advance.
sheaf-theory topos-theory
asked Dec 9 '18 at 14:53
W. RetherW. Rether
738417
$endgroup$
Johnstone, Topos theory, 0.3, page 13, asserts that, given a Grothendieck pretopology $P$, if the equalizer condition on a presheaf $F$ is satisfied for a family of arrows $R={U_ito U}$, then it is satisfied for every family $S={W_jto U}supset R$.
By equalizer condition I mean that $$F(U)to prod_R F(U_i)rightrightarrows prod_{Rtimes R}F(U_itimes_U U_j),$$ with the natural restriction maps, is an equalizer.
I can't see why this is true. First of all, should I assume that $R$ is a covering in the given pretopology? Otherwise it would sound very strange to me. For example, take a discrete space of two points $X={p,q}$, the pretopology being given by the topological coverings, and take $R={pto X}, S={pto X,qto X}$; then the equalizer condition for $R$ means that every local section on $p$ comes from a global section on $X$. So take $F(p):={0}, F(q)=mathbb Z, F(varnothing)={0},F(X)={0}$. This is a presheaf (with the natural maps, identity or zero) and satisfies the equalizer condition for $U=X$ and the family $R$ just defined: in
$${0}to {0} rightrightarrows {0}$$ one can always take the global section $0$ on the left. However, the equalizer for $S$ is: $${0}to {0}times mathbb Zrightrightarrows {0}times {0}times {0}times mathbb Z$$ where both arrows on the right are equal to $(0,1)mapsto (0,0,0,1)$ (note that the "mixed" fiber products $ptimes_X q$ and $qtimes _X p$ are the empty set). So take $(0,1)in {0}times mathbb Z$ in the central node of the last diagram. The maps on the right are the same, hence $F$ satisfies the equalizer condition for $S$ if and only if there exists $sin F(X)=0$ s.t. $s|_{{p}}=0,s|_{{q}}=1$, but this is of course impossible.
Is my conterexample correct?
Thank you in advance.
sheaf-theory topos-theory
sheaf-theory topos-theory
asked Dec 9 '18 at 14:53
W. RetherW. Rether
738417
asked Dec 9 '18 at 14:53
W. RetherW. Rether
738417
edited Dec 9 '18 at 20:33
W. Rether
asked Dec 9 '18 at 14:53
W. RetherW. Rether
738417
asked Dec 9 '18 at 14:53
W. RetherW. Rether
738417
asked Dec 9 '18 at 14:53
W. RetherW. Rether
738417
738417
$begingroup$
In your $F$, what is the morphism $F(q)to F(varnothing)$?
$endgroup$
– Malice Vidrine
Dec 9 '18 at 18:59
$begingroup$
The zero morphism. Is this a problem?
$endgroup$
– W. Rether
Dec 9 '18 at 19:16
$begingroup$
I think I'm just being confused by you notation. Writing ${0}$ as $0$, when $0$ is also common notation for an initial object, makes it look like you're trying to reason from an impossible presheaf. I think I see what you're doing now.
$endgroup$
– Malice Vidrine
Dec 9 '18 at 19:26
$begingroup$
Oh, sorry! Fixed. $F$ should be a presheaf, isn't it?
$endgroup$
– W. Rether
Dec 9 '18 at 20:33
$begingroup$
Another possible counterexample in the first displayed answer to math.stackexchange.com/questions/856564/…
$endgroup$
– W. Rether
Dec 9 '18 at 20:38
|
show 2 more comments
$begingroup$
In your $F$, what is the morphism $F(q)to F(varnothing)$?
$endgroup$
– Malice Vidrine
Dec 9 '18 at 18:59
$begingroup$
The zero morphism. Is this a problem?
$endgroup$
– W. Rether
Dec 9 '18 at 19:16
$begingroup$
I think I'm just being confused by you notation. Writing ${0}$ as $0$, when $0$ is also common notation for an initial object, makes it look like you're trying to reason from an impossible presheaf. I think I see what you're doing now.
$endgroup$
– Malice Vidrine
Dec 9 '18 at 19:26
$begingroup$
Oh, sorry! Fixed. $F$ should be a presheaf, isn't it?
$endgroup$
– W. Rether
Dec 9 '18 at 20:33
$begingroup$
Another possible counterexample in the first displayed answer to math.stackexchange.com/questions/856564/…
$endgroup$
– W. Rether
Dec 9 '18 at 20:38
$begingroup$
In your $F$, what is the morphism $F(q)to F(varnothing)$?
$endgroup$
– Malice Vidrine
Dec 9 '18 at 18:59
$begingroup$
In your $F$, what is the morphism $F(q)to F(varnothing)$?
$endgroup$
– Malice Vidrine
Dec 9 '18 at 18:59
$begingroup$
The zero morphism. Is this a problem?
$endgroup$
– W. Rether
Dec 9 '18 at 19:16
$begingroup$
The zero morphism. Is this a problem?
$endgroup$
– W. Rether
Dec 9 '18 at 19:16
$begingroup$
I think I'm just being confused by you notation. Writing ${0}$ as $0$, when $0$ is also common notation for an initial object, makes it look like you're trying to reason from an impossible presheaf. I think I see what you're doing now.
$endgroup$
– Malice Vidrine
Dec 9 '18 at 19:26
$begingroup$
I think I'm just being confused by you notation. Writing ${0}$ as $0$, when $0$ is also common notation for an initial object, makes it look like you're trying to reason from an impossible presheaf. I think I see what you're doing now.
$endgroup$
– Malice Vidrine
Dec 9 '18 at 19:26
$begingroup$
Oh, sorry! Fixed. $F$ should be a presheaf, isn't it?
$endgroup$
– W. Rether
Dec 9 '18 at 20:33
$begingroup$
Oh, sorry! Fixed. $F$ should be a presheaf, isn't it?
$endgroup$
– W. Rether
Dec 9 '18 at 20:33
$begingroup$
Another possible counterexample in the first displayed answer to math.stackexchange.com/questions/856564/…
$endgroup$
– W. Rether
Dec 9 '18 at 20:38
$begingroup$
Another possible counterexample in the first displayed answer to math.stackexchange.com/questions/856564/…
$endgroup$
– W. Rether
Dec 9 '18 at 20:38
|
show 2 more comments
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$begingroup$
In your $F$, what is the morphism $F(q)to F(varnothing)$?
$endgroup$
– Malice Vidrine
Dec 9 '18 at 18:59
$begingroup$
The zero morphism. Is this a problem?
$endgroup$
– W. Rether
Dec 9 '18 at 19:16
$begingroup$
I think I'm just being confused by you notation. Writing ${0}$ as $0$, when $0$ is also common notation for an initial object, makes it look like you're trying to reason from an impossible presheaf. I think I see what you're doing now.
$endgroup$
– Malice Vidrine
Dec 9 '18 at 19:26
$begingroup$
Oh, sorry! Fixed. $F$ should be a presheaf, isn't it?
$endgroup$
– W. Rether
Dec 9 '18 at 20:33
$begingroup$
Another possible counterexample in the first displayed answer to math.stackexchange.com/questions/856564/…
$endgroup$
– W. Rether
Dec 9 '18 at 20:38