Limits in a product category: proof verification
$begingroup$
First, I will fix a relevant definition. Given a small category $mathsf{I}$, a diagram $mathcal{D}colonmathsf{I}tomathsf{C}$, a functor $Fcolonmathsf{C}tomathsf{D}$ and a cone $lambdacolon XRightarrow mathcal{D}$, there is a cone $F(lambda)colon F(X) Rightarrow Fmathcal{D}$ defines as follows: for any $i in mathsf{I}$, $F(lambda)_i = F(lambda_i)$. It is straightforward to make sure that this data indeed defines a cone. Such constructions are important in theory of functors which preserve, reflect or create limits or colimits.
Let $mathsf{I}$ be a small category, let $A$ be a set and let $mathsf{C}$ be a category. Let $prod_{A}mathsf{C}$ be a product category of the family $(mathsf{C})_A$ (the family $(mathsf{C}_a)_{a in A}$ so that for any $a in A$ we have $mathsf{C}_a = mathsf{C}$). Let $(Pi_acolonprod_{A}mathsf{C}tomathsf{C})_{a in A}$ be a family of projection functors:
for any $(X_a)_{a in A}$, $Pi_a((X_a)_{a in A}) = X_a$,
for any $(f_acolon X_a to Y_a)_{a in A}$, $Pi_a((f_a)_{a in A}) = f_a$.
Let $mathcal{D}colonmathsf{I}toprod_{A}mathsf{C}$ be a diagram and let $lambdacolon (l_a)_{a in A} Rightarrow mathcal{D}$ be a limit cone. Then, for any $a in A$, $Pi_a(lambda)colon l_a Rightarrow Pi_amathcal{D}$ is also a limit cone.
I want to make sure that this assertion and the following proof of it are right.
Let $a' in A$. Let $mucolon Y Rightarrow Pi_{a'}mathcal{D}$ be a cone. Consider a family $(Y_a)_{a in A}$ so that, for any $a in A$, $Y_a$ is equal to $Y$ if $a = a'$, and is equal to $l_a$ otherwise. Consider a hypothetical cone $nucolon (Y_a)_{a in A} Rightarrow mathcal{D}$ so that, for any $a in A$, $nu_a$ is equal to $mu$ if $a = a'$, and is equal to $Pi_alambda$ otherwise. We claim that $nu$ is indeed a cone. Let $fcolon ito j$ be a morphism of $mathsf{I}$. As $mu$ is a cone, the we have $Pi_{a'}(mathcal{D}(f)) circ mu_i = mu_j$. As each $Pi_a(lambda)$ is a cone, then, for any $a in Asetminus{a'}$, we have $Pi_a(mathcal{D}(f))circPi_a(lambda_i) = Pi_a(lambda_j)$. As $mathcal{D}(f) = (Pi_a(mathcal{D}(f)))_{a in A}$ and considering the definition of $nu$, we thus have $mathcal{D}(f)circnu_i = nu_j$. Hence, the data defining $nu$ in fact defines a cone.
Since $lambda$ is a limit cone, there is a unique morphism $(f_a)_{a in A}colon (Y_a)_{a in A} to (l_a)_{a in A}$ so that for any $i in mathsf{I}$ we have $lambda_icirc (f_a)_{a in A} = nu_i$. In particular, it means that $Pi_a(lambda_i)circ f_{a'} = mu_i$ for any $i in mathsf{I}$. Let $gcolon Yto l_{a'}$ be another morphism for which we have $Pi_a(lambda_i)circ g = mu_i$ for any $i in mathsf{I}$. Consider a family $(g_acolon Y_ato l_a)$ so that $g_a = g$ is $a = a'$, and $g_a = f_a$ otherwise. It is clear that we then have $lambda_icirc (g_a)_{a in A} = nu_i$ for any $i in mathsf{I}$, which, in turn, implies that $(f_a)_{a in A} = (g_a)_{a in A}$, hence $g = g_{a'} = f_{a'}$.
proof-verification category-theory limits-colimits
$endgroup$
add a comment |
$begingroup$
First, I will fix a relevant definition. Given a small category $mathsf{I}$, a diagram $mathcal{D}colonmathsf{I}tomathsf{C}$, a functor $Fcolonmathsf{C}tomathsf{D}$ and a cone $lambdacolon XRightarrow mathcal{D}$, there is a cone $F(lambda)colon F(X) Rightarrow Fmathcal{D}$ defines as follows: for any $i in mathsf{I}$, $F(lambda)_i = F(lambda_i)$. It is straightforward to make sure that this data indeed defines a cone. Such constructions are important in theory of functors which preserve, reflect or create limits or colimits.
Let $mathsf{I}$ be a small category, let $A$ be a set and let $mathsf{C}$ be a category. Let $prod_{A}mathsf{C}$ be a product category of the family $(mathsf{C})_A$ (the family $(mathsf{C}_a)_{a in A}$ so that for any $a in A$ we have $mathsf{C}_a = mathsf{C}$). Let $(Pi_acolonprod_{A}mathsf{C}tomathsf{C})_{a in A}$ be a family of projection functors:
for any $(X_a)_{a in A}$, $Pi_a((X_a)_{a in A}) = X_a$,
for any $(f_acolon X_a to Y_a)_{a in A}$, $Pi_a((f_a)_{a in A}) = f_a$.
Let $mathcal{D}colonmathsf{I}toprod_{A}mathsf{C}$ be a diagram and let $lambdacolon (l_a)_{a in A} Rightarrow mathcal{D}$ be a limit cone. Then, for any $a in A$, $Pi_a(lambda)colon l_a Rightarrow Pi_amathcal{D}$ is also a limit cone.
I want to make sure that this assertion and the following proof of it are right.
Let $a' in A$. Let $mucolon Y Rightarrow Pi_{a'}mathcal{D}$ be a cone. Consider a family $(Y_a)_{a in A}$ so that, for any $a in A$, $Y_a$ is equal to $Y$ if $a = a'$, and is equal to $l_a$ otherwise. Consider a hypothetical cone $nucolon (Y_a)_{a in A} Rightarrow mathcal{D}$ so that, for any $a in A$, $nu_a$ is equal to $mu$ if $a = a'$, and is equal to $Pi_alambda$ otherwise. We claim that $nu$ is indeed a cone. Let $fcolon ito j$ be a morphism of $mathsf{I}$. As $mu$ is a cone, the we have $Pi_{a'}(mathcal{D}(f)) circ mu_i = mu_j$. As each $Pi_a(lambda)$ is a cone, then, for any $a in Asetminus{a'}$, we have $Pi_a(mathcal{D}(f))circPi_a(lambda_i) = Pi_a(lambda_j)$. As $mathcal{D}(f) = (Pi_a(mathcal{D}(f)))_{a in A}$ and considering the definition of $nu$, we thus have $mathcal{D}(f)circnu_i = nu_j$. Hence, the data defining $nu$ in fact defines a cone.
Since $lambda$ is a limit cone, there is a unique morphism $(f_a)_{a in A}colon (Y_a)_{a in A} to (l_a)_{a in A}$ so that for any $i in mathsf{I}$ we have $lambda_icirc (f_a)_{a in A} = nu_i$. In particular, it means that $Pi_a(lambda_i)circ f_{a'} = mu_i$ for any $i in mathsf{I}$. Let $gcolon Yto l_{a'}$ be another morphism for which we have $Pi_a(lambda_i)circ g = mu_i$ for any $i in mathsf{I}$. Consider a family $(g_acolon Y_ato l_a)$ so that $g_a = g$ is $a = a'$, and $g_a = f_a$ otherwise. It is clear that we then have $lambda_icirc (g_a)_{a in A} = nu_i$ for any $i in mathsf{I}$, which, in turn, implies that $(f_a)_{a in A} = (g_a)_{a in A}$, hence $g = g_{a'} = f_{a'}$.
proof-verification category-theory limits-colimits
$endgroup$
2
$begingroup$
Seems OK for me.
$endgroup$
– Oskar
Dec 28 '18 at 23:19
add a comment |
$begingroup$
First, I will fix a relevant definition. Given a small category $mathsf{I}$, a diagram $mathcal{D}colonmathsf{I}tomathsf{C}$, a functor $Fcolonmathsf{C}tomathsf{D}$ and a cone $lambdacolon XRightarrow mathcal{D}$, there is a cone $F(lambda)colon F(X) Rightarrow Fmathcal{D}$ defines as follows: for any $i in mathsf{I}$, $F(lambda)_i = F(lambda_i)$. It is straightforward to make sure that this data indeed defines a cone. Such constructions are important in theory of functors which preserve, reflect or create limits or colimits.
Let $mathsf{I}$ be a small category, let $A$ be a set and let $mathsf{C}$ be a category. Let $prod_{A}mathsf{C}$ be a product category of the family $(mathsf{C})_A$ (the family $(mathsf{C}_a)_{a in A}$ so that for any $a in A$ we have $mathsf{C}_a = mathsf{C}$). Let $(Pi_acolonprod_{A}mathsf{C}tomathsf{C})_{a in A}$ be a family of projection functors:
for any $(X_a)_{a in A}$, $Pi_a((X_a)_{a in A}) = X_a$,
for any $(f_acolon X_a to Y_a)_{a in A}$, $Pi_a((f_a)_{a in A}) = f_a$.
Let $mathcal{D}colonmathsf{I}toprod_{A}mathsf{C}$ be a diagram and let $lambdacolon (l_a)_{a in A} Rightarrow mathcal{D}$ be a limit cone. Then, for any $a in A$, $Pi_a(lambda)colon l_a Rightarrow Pi_amathcal{D}$ is also a limit cone.
I want to make sure that this assertion and the following proof of it are right.
Let $a' in A$. Let $mucolon Y Rightarrow Pi_{a'}mathcal{D}$ be a cone. Consider a family $(Y_a)_{a in A}$ so that, for any $a in A$, $Y_a$ is equal to $Y$ if $a = a'$, and is equal to $l_a$ otherwise. Consider a hypothetical cone $nucolon (Y_a)_{a in A} Rightarrow mathcal{D}$ so that, for any $a in A$, $nu_a$ is equal to $mu$ if $a = a'$, and is equal to $Pi_alambda$ otherwise. We claim that $nu$ is indeed a cone. Let $fcolon ito j$ be a morphism of $mathsf{I}$. As $mu$ is a cone, the we have $Pi_{a'}(mathcal{D}(f)) circ mu_i = mu_j$. As each $Pi_a(lambda)$ is a cone, then, for any $a in Asetminus{a'}$, we have $Pi_a(mathcal{D}(f))circPi_a(lambda_i) = Pi_a(lambda_j)$. As $mathcal{D}(f) = (Pi_a(mathcal{D}(f)))_{a in A}$ and considering the definition of $nu$, we thus have $mathcal{D}(f)circnu_i = nu_j$. Hence, the data defining $nu$ in fact defines a cone.
Since $lambda$ is a limit cone, there is a unique morphism $(f_a)_{a in A}colon (Y_a)_{a in A} to (l_a)_{a in A}$ so that for any $i in mathsf{I}$ we have $lambda_icirc (f_a)_{a in A} = nu_i$. In particular, it means that $Pi_a(lambda_i)circ f_{a'} = mu_i$ for any $i in mathsf{I}$. Let $gcolon Yto l_{a'}$ be another morphism for which we have $Pi_a(lambda_i)circ g = mu_i$ for any $i in mathsf{I}$. Consider a family $(g_acolon Y_ato l_a)$ so that $g_a = g$ is $a = a'$, and $g_a = f_a$ otherwise. It is clear that we then have $lambda_icirc (g_a)_{a in A} = nu_i$ for any $i in mathsf{I}$, which, in turn, implies that $(f_a)_{a in A} = (g_a)_{a in A}$, hence $g = g_{a'} = f_{a'}$.
proof-verification category-theory limits-colimits
$endgroup$
First, I will fix a relevant definition. Given a small category $mathsf{I}$, a diagram $mathcal{D}colonmathsf{I}tomathsf{C}$, a functor $Fcolonmathsf{C}tomathsf{D}$ and a cone $lambdacolon XRightarrow mathcal{D}$, there is a cone $F(lambda)colon F(X) Rightarrow Fmathcal{D}$ defines as follows: for any $i in mathsf{I}$, $F(lambda)_i = F(lambda_i)$. It is straightforward to make sure that this data indeed defines a cone. Such constructions are important in theory of functors which preserve, reflect or create limits or colimits.
Let $mathsf{I}$ be a small category, let $A$ be a set and let $mathsf{C}$ be a category. Let $prod_{A}mathsf{C}$ be a product category of the family $(mathsf{C})_A$ (the family $(mathsf{C}_a)_{a in A}$ so that for any $a in A$ we have $mathsf{C}_a = mathsf{C}$). Let $(Pi_acolonprod_{A}mathsf{C}tomathsf{C})_{a in A}$ be a family of projection functors:
for any $(X_a)_{a in A}$, $Pi_a((X_a)_{a in A}) = X_a$,
for any $(f_acolon X_a to Y_a)_{a in A}$, $Pi_a((f_a)_{a in A}) = f_a$.
Let $mathcal{D}colonmathsf{I}toprod_{A}mathsf{C}$ be a diagram and let $lambdacolon (l_a)_{a in A} Rightarrow mathcal{D}$ be a limit cone. Then, for any $a in A$, $Pi_a(lambda)colon l_a Rightarrow Pi_amathcal{D}$ is also a limit cone.
I want to make sure that this assertion and the following proof of it are right.
Let $a' in A$. Let $mucolon Y Rightarrow Pi_{a'}mathcal{D}$ be a cone. Consider a family $(Y_a)_{a in A}$ so that, for any $a in A$, $Y_a$ is equal to $Y$ if $a = a'$, and is equal to $l_a$ otherwise. Consider a hypothetical cone $nucolon (Y_a)_{a in A} Rightarrow mathcal{D}$ so that, for any $a in A$, $nu_a$ is equal to $mu$ if $a = a'$, and is equal to $Pi_alambda$ otherwise. We claim that $nu$ is indeed a cone. Let $fcolon ito j$ be a morphism of $mathsf{I}$. As $mu$ is a cone, the we have $Pi_{a'}(mathcal{D}(f)) circ mu_i = mu_j$. As each $Pi_a(lambda)$ is a cone, then, for any $a in Asetminus{a'}$, we have $Pi_a(mathcal{D}(f))circPi_a(lambda_i) = Pi_a(lambda_j)$. As $mathcal{D}(f) = (Pi_a(mathcal{D}(f)))_{a in A}$ and considering the definition of $nu$, we thus have $mathcal{D}(f)circnu_i = nu_j$. Hence, the data defining $nu$ in fact defines a cone.
Since $lambda$ is a limit cone, there is a unique morphism $(f_a)_{a in A}colon (Y_a)_{a in A} to (l_a)_{a in A}$ so that for any $i in mathsf{I}$ we have $lambda_icirc (f_a)_{a in A} = nu_i$. In particular, it means that $Pi_a(lambda_i)circ f_{a'} = mu_i$ for any $i in mathsf{I}$. Let $gcolon Yto l_{a'}$ be another morphism for which we have $Pi_a(lambda_i)circ g = mu_i$ for any $i in mathsf{I}$. Consider a family $(g_acolon Y_ato l_a)$ so that $g_a = g$ is $a = a'$, and $g_a = f_a$ otherwise. It is clear that we then have $lambda_icirc (g_a)_{a in A} = nu_i$ for any $i in mathsf{I}$, which, in turn, implies that $(f_a)_{a in A} = (g_a)_{a in A}$, hence $g = g_{a'} = f_{a'}$.
proof-verification category-theory limits-colimits
proof-verification category-theory limits-colimits
edited Dec 28 '18 at 23:18
Oskar
3,2831819
3,2831819
asked Dec 28 '18 at 9:20
Jxt921Jxt921
1,038719
1,038719
2
$begingroup$
Seems OK for me.
$endgroup$
– Oskar
Dec 28 '18 at 23:19
add a comment |
2
$begingroup$
Seems OK for me.
$endgroup$
– Oskar
Dec 28 '18 at 23:19
2
2
$begingroup$
Seems OK for me.
$endgroup$
– Oskar
Dec 28 '18 at 23:19
$begingroup$
Seems OK for me.
$endgroup$
– Oskar
Dec 28 '18 at 23:19
add a comment |
0
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Seems OK for me.
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– Oskar
Dec 28 '18 at 23:19