In a complete category, intersection of every family of subobjects of a fixed object always exists?
$begingroup$
Proposition 4.2.4 of Borceux's Handbook of Categorical Algebra Vol. 1 states that in a complete category, the intersection of every family of subobjects of a fixed object always exists.
The proof begins by taking a nonempty family of monomorphisms $s_icolon S_ito A$, for $iin I$. By completeness, the limit $(L,(p_i)_{iin I})$ of the diagram constituted by the various morphisms $s_i$ exists. He then says all the composites $s_icirc p_icolon Lto A$ are equal by definition of a limit.
This is my one stumbling block in the proof. I don't see what property of the limit makes these all equal. Are they all solutions to the same universal mapping problem?
category-theory
asked Dec 14 '18 at 8:22
Denise GiDenise Gi
32417
$endgroup$
add a comment |
$begingroup$
Proposition 4.2.4 of Borceux's Handbook of Categorical Algebra Vol. 1 states that in a complete category, the intersection of every family of subobjects of a fixed object always exists.
The proof begins by taking a nonempty family of monomorphisms $s_icolon S_ito A$, for $iin I$. By completeness, the limit $(L,(p_i)_{iin I})$ of the diagram constituted by the various morphisms $s_i$ exists. He then says all the composites $s_icirc p_icolon Lto A$ are equal by definition of a limit.
This is my one stumbling block in the proof. I don't see what property of the limit makes these all equal. Are they all solutions to the same universal mapping problem?
category-theory
asked Dec 14 '18 at 8:22
Denise GiDenise Gi
32417
$endgroup$
add a comment |
$begingroup$
Proposition 4.2.4 of Borceux's Handbook of Categorical Algebra Vol. 1 states that in a complete category, the intersection of every family of subobjects of a fixed object always exists.
The proof begins by taking a nonempty family of monomorphisms $s_icolon S_ito A$, for $iin I$. By completeness, the limit $(L,(p_i)_{iin I})$ of the diagram constituted by the various morphisms $s_i$ exists. He then says all the composites $s_icirc p_icolon Lto A$ are equal by definition of a limit.
This is my one stumbling block in the proof. I don't see what property of the limit makes these all equal. Are they all solutions to the same universal mapping problem?
category-theory
asked Dec 14 '18 at 8:22
Denise GiDenise Gi
32417
$endgroup$
Proposition 4.2.4 of Borceux's Handbook of Categorical Algebra Vol. 1 states that in a complete category, the intersection of every family of subobjects of a fixed object always exists.
The proof begins by taking a nonempty family of monomorphisms $s_icolon S_ito A$, for $iin I$. By completeness, the limit $(L,(p_i)_{iin I})$ of the diagram constituted by the various morphisms $s_i$ exists. He then says all the composites $s_icirc p_icolon Lto A$ are equal by definition of a limit.
This is my one stumbling block in the proof. I don't see what property of the limit makes these all equal. Are they all solutions to the same universal mapping problem?
category-theory
category-theory
asked Dec 14 '18 at 8:22
Denise GiDenise Gi
32417
asked Dec 14 '18 at 8:22
Denise GiDenise Gi
32417
asked Dec 14 '18 at 8:22
Denise GiDenise Gi
32417
asked Dec 14 '18 at 8:22
Denise GiDenise Gi
32417
asked Dec 14 '18 at 8:22
Denise GiDenise Gi
32417
32417
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Recall that a limit of a diagram $D : mathcal I to mathcal C$ is in particular a cone over $D$, meaning an object $L$ together with a maps $(p_i)_{iinmathrm{Ob}(mathcal I)}$ such that for any map $k:i to j$ in $mathcal I$ it holds that $D(k)circ p_i = p_j$.
Here, starting from the family of monos $(s_i)_{iin I}$, the category $mathcal I$ is the category with objects:
- either $iin I$
- or a special object $star$
The non identity morphisms of $mathcal I$ are $k_i : i to star$ and there is exactly one such morphism for each $iin I$. In other words it is the category obtained from the discrete set $I$ by formally adding a terminal object.
The diagram $D$ you are looking at is $imapsto S_i$, $star mapsto A$ and $k_i mapsto s_i$. So given a cone $(L,(p_i)_{iin mathrm{Ob}(mathcal I)}$), one has for each $iin I$:
$$s_i circ p_i = D(k_i) circ p_i = p_star$$
So in particular all $s_icirc p_i$ are equal. This is by definition of the limit being a cone over $D$.
answered Dec 14 '18 at 8:45
PecePece
8,23511241
$endgroup$
$begingroup$
It's not the category (abstract diagram) $mathcal I$ you describe, but the 'abstract cone' over $mathcal I$.
$endgroup$
– Berci
Dec 14 '18 at 9:13
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3039108%2fin-a-complete-category-intersection-of-every-family-of-subobjects-of-a-fixed-ob%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Recall that a limit of a diagram $D : mathcal I to mathcal C$ is in particular a cone over $D$, meaning an object $L$ together with a maps $(p_i)_{iinmathrm{Ob}(mathcal I)}$ such that for any map $k:i to j$ in $mathcal I$ it holds that $D(k)circ p_i = p_j$.
Here, starting from the family of monos $(s_i)_{iin I}$, the category $mathcal I$ is the category with objects:
- either $iin I$
- or a special object $star$
The non identity morphisms of $mathcal I$ are $k_i : i to star$ and there is exactly one such morphism for each $iin I$. In other words it is the category obtained from the discrete set $I$ by formally adding a terminal object.
The diagram $D$ you are looking at is $imapsto S_i$, $star mapsto A$ and $k_i mapsto s_i$. So given a cone $(L,(p_i)_{iin mathrm{Ob}(mathcal I)}$), one has for each $iin I$:
$$s_i circ p_i = D(k_i) circ p_i = p_star$$
So in particular all $s_icirc p_i$ are equal. This is by definition of the limit being a cone over $D$.
answered Dec 14 '18 at 8:45
PecePece
8,23511241
$endgroup$
$begingroup$
It's not the category (abstract diagram) $mathcal I$ you describe, but the 'abstract cone' over $mathcal I$.
$endgroup$
– Berci
Dec 14 '18 at 9:13
add a comment |
$begingroup$
Recall that a limit of a diagram $D : mathcal I to mathcal C$ is in particular a cone over $D$, meaning an object $L$ together with a maps $(p_i)_{iinmathrm{Ob}(mathcal I)}$ such that for any map $k:i to j$ in $mathcal I$ it holds that $D(k)circ p_i = p_j$.
Here, starting from the family of monos $(s_i)_{iin I}$, the category $mathcal I$ is the category with objects:
- either $iin I$
- or a special object $star$
The non identity morphisms of $mathcal I$ are $k_i : i to star$ and there is exactly one such morphism for each $iin I$. In other words it is the category obtained from the discrete set $I$ by formally adding a terminal object.
The diagram $D$ you are looking at is $imapsto S_i$, $star mapsto A$ and $k_i mapsto s_i$. So given a cone $(L,(p_i)_{iin mathrm{Ob}(mathcal I)}$), one has for each $iin I$:
$$s_i circ p_i = D(k_i) circ p_i = p_star$$
So in particular all $s_icirc p_i$ are equal. This is by definition of the limit being a cone over $D$.
answered Dec 14 '18 at 8:45
PecePece
8,23511241
$endgroup$
$begingroup$
It's not the category (abstract diagram) $mathcal I$ you describe, but the 'abstract cone' over $mathcal I$.
$endgroup$
– Berci
Dec 14 '18 at 9:13
add a comment |
$begingroup$
Recall that a limit of a diagram $D : mathcal I to mathcal C$ is in particular a cone over $D$, meaning an object $L$ together with a maps $(p_i)_{iinmathrm{Ob}(mathcal I)}$ such that for any map $k:i to j$ in $mathcal I$ it holds that $D(k)circ p_i = p_j$.
Here, starting from the family of monos $(s_i)_{iin I}$, the category $mathcal I$ is the category with objects:
- either $iin I$
- or a special object $star$
The non identity morphisms of $mathcal I$ are $k_i : i to star$ and there is exactly one such morphism for each $iin I$. In other words it is the category obtained from the discrete set $I$ by formally adding a terminal object.
The diagram $D$ you are looking at is $imapsto S_i$, $star mapsto A$ and $k_i mapsto s_i$. So given a cone $(L,(p_i)_{iin mathrm{Ob}(mathcal I)}$), one has for each $iin I$:
$$s_i circ p_i = D(k_i) circ p_i = p_star$$
So in particular all $s_icirc p_i$ are equal. This is by definition of the limit being a cone over $D$.
answered Dec 14 '18 at 8:45
PecePece
8,23511241
$endgroup$
Recall that a limit of a diagram $D : mathcal I to mathcal C$ is in particular a cone over $D$, meaning an object $L$ together with a maps $(p_i)_{iinmathrm{Ob}(mathcal I)}$ such that for any map $k:i to j$ in $mathcal I$ it holds that $D(k)circ p_i = p_j$.
Here, starting from the family of monos $(s_i)_{iin I}$, the category $mathcal I$ is the category with objects:
- either $iin I$
- or a special object $star$
The non identity morphisms of $mathcal I$ are $k_i : i to star$ and there is exactly one such morphism for each $iin I$. In other words it is the category obtained from the discrete set $I$ by formally adding a terminal object.
The diagram $D$ you are looking at is $imapsto S_i$, $star mapsto A$ and $k_i mapsto s_i$. So given a cone $(L,(p_i)_{iin mathrm{Ob}(mathcal I)}$), one has for each $iin I$:
$$s_i circ p_i = D(k_i) circ p_i = p_star$$
So in particular all $s_icirc p_i$ are equal. This is by definition of the limit being a cone over $D$.
answered Dec 14 '18 at 8:45
PecePece
8,23511241
answered Dec 14 '18 at 8:45
PecePece
8,23511241
answered Dec 14 '18 at 8:45
PecePece
8,23511241
answered Dec 14 '18 at 8:45
PecePece
8,23511241
8,23511241
$begingroup$
It's not the category (abstract diagram) $mathcal I$ you describe, but the 'abstract cone' over $mathcal I$.
$endgroup$
– Berci
Dec 14 '18 at 9:13
add a comment |
$begingroup$
It's not the category (abstract diagram) $mathcal I$ you describe, but the 'abstract cone' over $mathcal I$.
$endgroup$
– Berci
Dec 14 '18 at 9:13
$begingroup$
It's not the category (abstract diagram) $mathcal I$ you describe, but the 'abstract cone' over $mathcal I$.
$endgroup$
– Berci
Dec 14 '18 at 9:13
$begingroup$
It's not the category (abstract diagram) $mathcal I$ you describe, but the 'abstract cone' over $mathcal I$.
$endgroup$
– Berci
Dec 14 '18 at 9:13
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3039108%2fin-a-complete-category-intersection-of-every-family-of-subobjects-of-a-fixed-ob%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
