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This question is motivated by showing that the category $\mathbf{Sheaves} (X)$ from the open subset excluding the empty set category to the category of abelian group $\mathbf{Ab}$ has enough injective objects.

Prerequisition

 

Let $\mathscr{F} \in \mathbf{Sheaves} (X)$ and let $F = \underset{\rightarrow U}{\mathrm{Lim}} \mathscr{F}(U)$ where $U$ is taken over all open sets of $X$. Now let me point out what $F$ is. Let $\mathfrak{S}$ be the set of open sets of $X$. Then by the knowledge of category theory, we can take coproduct then coequalizer to get colimits. See This. So $F = \bigoplus_\limits{S \in \mathfrak{S}} \mathscr{F}(S) / A$ and $\mathscr{F}(S) \rightarrow F$ is the composition of caronical injection and quotient, where $A$ is generated by $$(a)_S - (f(a))_\mathrm{Codom\{f\}}$$ The subscript means which the summand coordinate the element is in. $a \in \mathscr{F}(S), S \in \mathfrak{S}$ and $f$ is a morphism induced by an inclusion in $\mathfrak{S}$.

Then my question is

How to show that $a \in \mathscr{F}(S) \mapsto 0 \in F$ through the map in the universal cone of the colimit iff there are $a'\in \mathscr{F}(S')$ and $f_1 $, $f_2$ induced by inclusions such that $$f_1(a') = a$$ and $$f_2(a') = 0$$

The if part is easy since $(a)_S = [(a')_{S'} - (f_2(a'))] - [ (a')_{S'} - (f_1(a'))_S] \in A$. To show the only if part, maybe we should show that if $(a)_S$ can be decomposed to sum of $(a')_{S'} - (f(a'))_\mathrm{Codom\{f\}}$ then it can be decomposed to

$$[(a')_{S'} - (f_2(a'))] - [ (a')_{S'} - (f_1(a'))_S]$$

Or more specifically, let $\mathfrak{S}_x$ be the set of all subsets containing $x \in X$. Show that $a \in \mathscr{F}(S_x)$ maps to zero in $\underset{\rightarrow}{\mathrm{Lim}} \mathscr{F}(\mathfrak{S}_x)$ iff $f(a) = 0$ for some $f$ induced by an inclusion. This proposition is more closed to my original purpose.

This question is motivated by showing that the category $\mathbf{Sheaves} (X)$ from the open subset excluding the empty set category to the category of abelian group $\mathbf{Ab}$ has enough injective objects.

Prerequisition

 

Let $\mathscr{F} \in \mathbf{Sheaves} (X)$ and let $F = \underset{\rightarrow U}{\mathrm{Lim}} \mathscr{F}(U)$ where $U$ is taken over all open sets of $X$. Now let me point out what $F$ is. Let $\mathfrak{S}$ be the set of open sets of $X$. Then by the knowledge of category theory, we can take coproduct then coequalizer to get colimits. See This. So $F = \bigoplus_\limits{S \in \mathfrak{S}} \mathscr{F}(S) / A$ and $\mathscr{F}(S) \rightarrow F$ is the composition of caronical injection and quotient, where $A$ is generated by $$(a)_S - (f(a))_\mathrm{Codom\{f\}}$$ The subscript means which the summand coordinate the element is in. $a \in \mathscr{F}(S), S \in \mathfrak{S}$ and $f$ is a morphism induced by an inclusion in $\mathfrak{S}$.

Then my question is

How to show that $a \in \mathscr{F}(S) \mapsto 0 \in F$ through the map in the universal cone of the colimit iff there are $a'\in \mathscr{F}(S')$ and $f_1 $, $f_2$ induced by inclusions such that $$f_1(a') = a$$ and $$f_2(a') = 0$$

The if part is easy since $(a)_S = [(a')_{S'} - (f_2(a'))] - [ (a')_{S'} - (f_1(a'))_S] \in A$. To show the only if part, maybe we should show that if $(a)_S$ can be decomposed to sum of $(a')_{S'} - (f(a'))_\mathrm{Codom\{f\}}$ then it can be decomposed to

$$[(a')_{S'} - (f_2(a'))] - [ (a')_{S'} - (f_1(a'))_S]$$

Or more specifically, let $\mathfrak{S}_x$ be the set of all subsets containing $x \in X$. Show that $a \in \mathscr{F}(S_x)$ maps to zero in $\underset{\rightarrow}{\mathrm{Lim}} \mathscr{F}(\mathfrak{S}_x)$ iff $f(a) = 0$ for some $f$ induced by an inclusion. This proposition is more closed to my original purpose.

This question is motivated by showing that the category $\mathbf{Sheaves} (X)$ from the open subset excluding the empty set category to the category of abelian group $\mathbf{Ab}$ has enough injective objects.

Prerequisition

Let $\mathscr{F} \in \mathbf{Sheaves} (X)$ and let $F = \underset{\rightarrow U}{\mathrm{Lim}} \mathscr{F}(U)$ where $U$ is taken over all open sets of $X$. Now let me point out what $F$ is. Let $\mathfrak{S}$ be the set of open sets of $X$. Then by the knowledge of category theory, we can take coproduct then coequalizer to get colimits. See This. So $F = \bigoplus_\limits{S \in \mathfrak{S}} \mathscr{F}(S) / A$ and $\mathscr{F}(S) \rightarrow F$ is the composition of caronical injection and quotient, where $A$ is generated by $$(a)_S - (f(a))_\mathrm{Codom\{f\}}$$ The subscript means which the summand coordinate the element is in. $a \in \mathscr{F}(S), S \in \mathfrak{S}$ and $f$ is a morphism induced by an inclusion in $\mathfrak{S}$.

Then my question is

How to show that $a \in \mathscr{F}(S) \mapsto 0 \in F$ through the map in the universal cone of the colimit iff there are $a'\in \mathscr{F}(S')$ and $f_1 $, $f_2$ induced by inclusions such that $$f_1(a') = a$$ and $$f_2(a') = 0$$

The if part is easy since $(a)_S = [(a')_{S'} - (f_2(a'))] - [ (a')_{S'} - (f_1(a'))_S] \in A$. To show the only if part, maybe we should show that if $(a)_S$ can be decomposed to sum of $(a')_{S'} - (f(a'))_\mathrm{Codom\{f\}}$ then it can be decomposed to

$$[(a')_{S'} - (f_2(a'))] - [ (a')_{S'} - (f_1(a'))_S]$$

Or more specifically, let $\mathfrak{S}_x$ be the set of all subsets containing $x \in X$. Show that $a \in \mathscr{F}(S_x)$ maps to zero in $\underset{\rightarrow}{\mathrm{Lim}} \mathscr{F}(\mathfrak{S}_x)$ iff $f(a) = 0$ for some $f$ induced by an inclusion. This proposition is more closed to my original purpose.

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XT Chen
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This question is motivated by showing that the category $\mathbf{Sheaves} (X)$ from the open subset excluding the empty set category to the category of abelian group $\mathbf{Ab}$ has enough injective objects.

Prerequisition

Let $\mathscr{F} \in \mathbf{Sheaves} (X)$ and let $F = \underset{\rightarrow U}{\mathrm{Lim}} \mathscr{F}(U)$ where $U$ is taken over all open sets of $X$. Now let me point out what $F$ is. Let $\mathfrak{S}$ be the set of open sets of $X$. Then by the knowledge of category theory, we can take coproduct then coequalizer to get colimits. See This. So $F = \bigoplus_\limits{S \in \mathfrak{S}} \mathscr{F}(S) / A$ and $\mathscr{F}(S) \rightarrow F$ is the composition of caronical injection and quotient, where $A$ is generated by $$(a)_S - (f(a))_\mathrm{Codom\{f\}}$$ The subscript means which the summand coordinate the element is in. $a \in \mathscr{F}(S), S \in \mathfrak{S}$ and $f$ is a morphism induced by an inclusion in $\mathfrak{S}$.

Then my question is

How to show that $a \in \mathscr{F}(S) \mapsto 0 \in F$ through the map in the universal cone of the colimit iff there are $a'\in \mathscr{F}(S')$ and $f_1 $, $f_2$ induced by inclusions such that $$f_1(a') = a$$ and $$f_2(a') = 0$$

The if part is easy since $(a)_S = [(a')_{S'} - (f_2(a'))] - [ (a')_{S'} - (f_1(a'))_S] \in A$. To show the only if part, maybe we should show that if $(a)_S$ can be decomposed to sum of $(a')_{S'} - (f(a'))_\mathrm{Codom\{f\}}$ then it can be decomposed to

$$[(a')_{S'} - (f_2(a'))] - [ (a')_{S'} - (f_1(a'))_S]$$

Or more specifically, let $\mathfrak{S}_x$ be the set of all subsets containing $x \in X$. Show that $a \in \mathscr{F}(S_x)$ maps to zero in $\underset{\rightarrow}{\mathrm{Lim}} \mathscr{F}(\mathfrak{S}_x)$ iff $f(a) = 0$ for some $f$ induced by an inclusion. This proposition is more closed to my original purpose.

This question is motivated by showing that the category $\mathbf{Sheaves} (X)$ from the open subset category to the category of abelian group $\mathbf{Ab}$ has enough injective objects.

Prerequisition

Let $\mathscr{F} \in \mathbf{Sheaves} (X)$ and let $F = \underset{\rightarrow U}{\mathrm{Lim}} \mathscr{F}(U)$ where $U$ is taken over all open sets of $X$. Now let me point out what $F$ is. Let $\mathfrak{S}$ be the set of open sets of $X$. Then by the knowledge of category theory, we can take coproduct then coequalizer to get colimits. See This. So $F = \bigoplus_\limits{S \in \mathfrak{S}} \mathscr{F}(S) / A$ and $\mathscr{F}(S) \rightarrow F$ is the composition of caronical injection and quotient, where $A$ is generated by $$(a)_S - (f(a))_\mathrm{Codom\{f\}}$$ The subscript means which the summand coordinate the element is in. $a \in \mathscr{F}(S), S \in \mathfrak{S}$ and $f$ is a morphism induced by an inclusion in $\mathfrak{S}$.

Then my question is

How to show that $a \in \mathscr{F}(S) \mapsto 0 \in F$ through the map in the universal cone of the colimit iff there are $a'\in \mathscr{F}(S')$ and $f_1 $, $f_2$ induced by inclusions such that $$f_1(a') = a$$ and $$f_2(a') = 0$$

The if part is easy since $(a)_S = [(a')_{S'} - (f_2(a'))] - [ (a')_{S'} - (f_1(a'))_S] \in A$. To show the only if part, maybe we should show that if $(a)_S$ can be decomposed to sum of $(a')_{S'} - (f(a'))_\mathrm{Codom\{f\}}$ then it can be decomposed to

$$[(a')_{S'} - (f_2(a'))] - [ (a')_{S'} - (f_1(a'))_S]$$

Or more specifically, let $\mathfrak{S}_x$ be the set of all subsets containing $x \in X$. Show that $a \in \mathscr{F}(S_x)$ maps to zero in $\underset{\rightarrow}{\mathrm{Lim}} \mathscr{F}(\mathfrak{S}_x)$ iff $f(a) = 0$ for some $f$ induced by an inclusion. This proposition is more closed to my original purpose.

This question is motivated by showing that the category $\mathbf{Sheaves} (X)$ from the open subset excluding the empty set category to the category of abelian group $\mathbf{Ab}$ has enough injective objects.

Prerequisition

Let $\mathscr{F} \in \mathbf{Sheaves} (X)$ and let $F = \underset{\rightarrow U}{\mathrm{Lim}} \mathscr{F}(U)$ where $U$ is taken over all open sets of $X$. Now let me point out what $F$ is. Let $\mathfrak{S}$ be the set of open sets of $X$. Then by the knowledge of category theory, we can take coproduct then coequalizer to get colimits. See This. So $F = \bigoplus_\limits{S \in \mathfrak{S}} \mathscr{F}(S) / A$ and $\mathscr{F}(S) \rightarrow F$ is the composition of caronical injection and quotient, where $A$ is generated by $$(a)_S - (f(a))_\mathrm{Codom\{f\}}$$ The subscript means which the summand coordinate the element is in. $a \in \mathscr{F}(S), S \in \mathfrak{S}$ and $f$ is a morphism induced by an inclusion in $\mathfrak{S}$.

Then my question is

How to show that $a \in \mathscr{F}(S) \mapsto 0 \in F$ through the map in the universal cone of the colimit iff there are $a'\in \mathscr{F}(S')$ and $f_1 $, $f_2$ induced by inclusions such that $$f_1(a') = a$$ and $$f_2(a') = 0$$

The if part is easy since $(a)_S = [(a')_{S'} - (f_2(a'))] - [ (a')_{S'} - (f_1(a'))_S] \in A$. To show the only if part, maybe we should show that if $(a)_S$ can be decomposed to sum of $(a')_{S'} - (f(a'))_\mathrm{Codom\{f\}}$ then it can be decomposed to

$$[(a')_{S'} - (f_2(a'))] - [ (a')_{S'} - (f_1(a'))_S]$$

Or more specifically, let $\mathfrak{S}_x$ be the set of all subsets containing $x \in X$. Show that $a \in \mathscr{F}(S_x)$ maps to zero in $\underset{\rightarrow}{\mathrm{Lim}} \mathscr{F}(\mathfrak{S}_x)$ iff $f(a) = 0$ for some $f$ induced by an inclusion. This proposition is more closed to my original purpose.

Notice added Canonical answer required by XT Chen
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Source Link
XT Chen
  • 1.2k
  • 7
  • 15

This question is motivated by showing that the category $\mathbf{Sheaves} (X)$ from the open subset category to the category of abelian group $\mathbf{Ab}$ has enough injective objects.

Prerequisition

Let $\mathscr{F} \in \mathbf{Sheaves} (X)$ and let $F = \underset{\rightarrow}{\mathrm{Lim}} \mathscr{F}$$F = \underset{\rightarrow U}{\mathrm{Lim}} \mathscr{F}(U)$ where $U$ is taken over all open sets of $X$. Now let me point out what $F$ is. Let $\mathfrak{S}$ be the set of open sets of $X$. Then by the knowledge of category theory, we can take coproduct then coequalizer to get colimits. See This. So $F = \bigoplus_\limits{S \in \mathfrak{S}} \mathscr{F}(S) / A$ and $\mathscr{F}(S) \rightarrow F$ is the composition of caronical injection and quotient, where $A$ is generated by $$(a)_S - (f(a))_\mathrm{Codom\{f\}}$$ The subscript means which the summand coordinate the element is in. $a \in \mathscr{F}(S), S \in \mathfrak{S}$ and $f$ is a morphism induced by an inclusion in $\mathfrak{S}$.

Then my question is

How to show that $a \in \mathscr{F}(S) \mapsto 0 \in F$ through the map in the universal cone of the colimit iff there are $a'\in \mathscr{F}(S')$ and $f_1 $, $f_2$ induced by inclusions such that $$f_1(a') = a$$ and $$f_2(a') = 0$$

The if part is easy since $(a)_S = [(a')_{S'} - (f_2(a'))] - [ (a')_{S'} - (f_1(a'))_S] \in A$. To show the only if part, maybe we should show that if $(a)_S$ can be decomposed to sum of $(a')_{S'} - (f(a'))_\mathrm{Codom\{f\}}$ then it can be decomposed to

$$[(a')_{S'} - (f_2(a'))] - [ (a')_{S'} - (f_1(a'))_S]$$

Or more specifically, let $\mathfrak{S}_x$ be the set of all subsets containing $x \in X$. Show that $a \in \mathscr{F}(S_x)$ maps to zero in $\underset{\rightarrow}{\mathrm{Lim}} \mathscr{F}(\mathfrak{S}_x)$ iff $f(a) = 0$ for some $f$ induced by an inclusion. This proposition is more closed to my original purpose.

This question is motivated by showing that the category $\mathbf{Sheaves} (X)$ from the open subset category to the category of abelian group $\mathbf{Ab}$ has enough injective objects.

Prerequisition

Let $\mathscr{F} \in \mathbf{Sheaves} (X)$ and let $F = \underset{\rightarrow}{\mathrm{Lim}} \mathscr{F}$. Now let me point out what $F$ is. Let $\mathfrak{S}$ be the set of open sets of $X$. Then by the knowledge of category theory, we can take coproduct then coequalizer to get colimits. See This. So $F = \bigoplus_\limits{S \in \mathfrak{S}} \mathscr{F}(S) / A$ and $\mathscr{F}(S) \rightarrow F$ is the composition of caronical injection and quotient, where $A$ is generated by $$(a)_S - (f(a))_\mathrm{Codom\{f\}}$$ The subscript means which the summand coordinate the element is in. $a \in \mathscr{F}(S), S \in \mathfrak{S}$ and $f$ is a morphism induced by an inclusion in $\mathfrak{S}$.

Then my question is

How to show that $a \in \mathscr{F}(S) \mapsto 0 \in F$ through the map in the universal cone of the colimit iff there are $a'\in \mathscr{F}(S')$ and $f_1 $, $f_2$ induced by inclusions such that $$f_1(a') = a$$ and $$f_2(a') = 0$$

The if part is easy since $(a)_S = [(a')_{S'} - (f_2(a'))] - [ (a')_{S'} - (f_1(a'))_S] \in A$. To show the only if part, maybe we should show that if $(a)_S$ can be decomposed to sum of $(a')_{S'} - (f(a'))_\mathrm{Codom\{f\}}$ then it can be decomposed to

$$[(a')_{S'} - (f_2(a'))] - [ (a')_{S'} - (f_1(a'))_S]$$

Or more specifically, let $\mathfrak{S}_x$ be the set of all subsets containing $x \in X$. Show that $a \in \mathscr{F}(S_x)$ maps to zero in $\underset{\rightarrow}{\mathrm{Lim}} \mathscr{F}(\mathfrak{S}_x)$ iff $f(a) = 0$ for some $f$ induced by an inclusion. This proposition is more closed to my original purpose.

This question is motivated by showing that the category $\mathbf{Sheaves} (X)$ from the open subset category to the category of abelian group $\mathbf{Ab}$ has enough injective objects.

Prerequisition

Let $\mathscr{F} \in \mathbf{Sheaves} (X)$ and let $F = \underset{\rightarrow U}{\mathrm{Lim}} \mathscr{F}(U)$ where $U$ is taken over all open sets of $X$. Now let me point out what $F$ is. Let $\mathfrak{S}$ be the set of open sets of $X$. Then by the knowledge of category theory, we can take coproduct then coequalizer to get colimits. See This. So $F = \bigoplus_\limits{S \in \mathfrak{S}} \mathscr{F}(S) / A$ and $\mathscr{F}(S) \rightarrow F$ is the composition of caronical injection and quotient, where $A$ is generated by $$(a)_S - (f(a))_\mathrm{Codom\{f\}}$$ The subscript means which the summand coordinate the element is in. $a \in \mathscr{F}(S), S \in \mathfrak{S}$ and $f$ is a morphism induced by an inclusion in $\mathfrak{S}$.

Then my question is

How to show that $a \in \mathscr{F}(S) \mapsto 0 \in F$ through the map in the universal cone of the colimit iff there are $a'\in \mathscr{F}(S')$ and $f_1 $, $f_2$ induced by inclusions such that $$f_1(a') = a$$ and $$f_2(a') = 0$$

The if part is easy since $(a)_S = [(a')_{S'} - (f_2(a'))] - [ (a')_{S'} - (f_1(a'))_S] \in A$. To show the only if part, maybe we should show that if $(a)_S$ can be decomposed to sum of $(a')_{S'} - (f(a'))_\mathrm{Codom\{f\}}$ then it can be decomposed to

$$[(a')_{S'} - (f_2(a'))] - [ (a')_{S'} - (f_1(a'))_S]$$

Or more specifically, let $\mathfrak{S}_x$ be the set of all subsets containing $x \in X$. Show that $a \in \mathscr{F}(S_x)$ maps to zero in $\underset{\rightarrow}{\mathrm{Lim}} \mathscr{F}(\mathfrak{S}_x)$ iff $f(a) = 0$ for some $f$ induced by an inclusion. This proposition is more closed to my original purpose.

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XT Chen
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