Limits in a product category: proof verification












3












$begingroup$


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 $g = g_{a'} = f_{a'}$.











share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Seems OK for me.
    $endgroup$
    – Oskar
    Dec 28 '18 at 23:19
















3












$begingroup$


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 $g = g_{a'} = f_{a'}$.











share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Seems OK for me.
    $endgroup$
    – Oskar
    Dec 28 '18 at 23:19














3












3








3





$begingroup$


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 $g = g_{a'} = f_{a'}$.











share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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














  • 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










0






active

oldest

votes












Your Answer








StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3054716%2flimits-in-a-product-category-proof-verification%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























0






active

oldest

votes








0






active

oldest

votes









active

oldest

votes






active

oldest

votes
















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3054716%2flimits-in-a-product-category-proof-verification%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Bundesstraße 106

Verónica Boquete

Ida-Boy-Ed-Garten