Inverse limits of relative homology in Euclidean space
up vote
2
down vote
favorite
Let $C$ be a compact set in Euclidean space and $i$ an integer. Is the inverse limit
$$underset{Csubset U}{lim_leftarrow}H_i(U,C)$$
over all open sets $U$ containing $C$ of relative homology groups always zero?
algebraic-topology
migrated from mathoverflow.net Nov 28 at 10:08
This question came from our site for professional mathematicians.
|
show 4 more comments
up vote
2
down vote
favorite
Let $C$ be a compact set in Euclidean space and $i$ an integer. Is the inverse limit
$$underset{Csubset U}{lim_leftarrow}H_i(U,C)$$
over all open sets $U$ containing $C$ of relative homology groups always zero?
algebraic-topology
migrated from mathoverflow.net Nov 28 at 10:08
This question came from our site for professional mathematicians.
Hi Alexander, are you sure this is an inverse limit and not a direct limit? Anyway, your question is more suitable for MathStackExchange, so I've voted to move it there.
– Mark Grant
Nov 27 at 14:24
U containing C, so should be inverse limit.
– user43326
Nov 27 at 20:17
This is confusing. I assume that open subsets form a directed poset via standard inclusion ordering. I also assume that bonding maps are induced from inclusions. But $H_i$ is a covariant functor. How is that an inverse system?
– freakish
Nov 28 at 10:26
@freakish The system $mathfrak{U}(C)$ of open neigborhoods of $C$ is usually ordered by $U le V$ if $U supset V$ (reverse inclusion). $mathfrak{U}(C)$ is directed and may be understood as an inverse system of topological spaces: The bonding map $f^V_U : V to U$ is the inclusion.
– Paul Frost
Nov 28 at 11:15
@freakish The main reason why one doesn't consider the order $U le V$ if $U subset V$ is that $mathfrak{U}(C)$ would have $X = mathbb{R}^n$ as a maximum. That is, the direct system $mathfrak{U}(C)$ would be isomorphic to the trivial system ${ X }$.
– Paul Frost
Nov 28 at 14:26
|
show 4 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $C$ be a compact set in Euclidean space and $i$ an integer. Is the inverse limit
$$underset{Csubset U}{lim_leftarrow}H_i(U,C)$$
over all open sets $U$ containing $C$ of relative homology groups always zero?
algebraic-topology
Let $C$ be a compact set in Euclidean space and $i$ an integer. Is the inverse limit
$$underset{Csubset U}{lim_leftarrow}H_i(U,C)$$
over all open sets $U$ containing $C$ of relative homology groups always zero?
algebraic-topology
algebraic-topology
asked Nov 27 at 13:41
Alexander
migrated from mathoverflow.net Nov 28 at 10:08
This question came from our site for professional mathematicians.
migrated from mathoverflow.net Nov 28 at 10:08
This question came from our site for professional mathematicians.
Hi Alexander, are you sure this is an inverse limit and not a direct limit? Anyway, your question is more suitable for MathStackExchange, so I've voted to move it there.
– Mark Grant
Nov 27 at 14:24
U containing C, so should be inverse limit.
– user43326
Nov 27 at 20:17
This is confusing. I assume that open subsets form a directed poset via standard inclusion ordering. I also assume that bonding maps are induced from inclusions. But $H_i$ is a covariant functor. How is that an inverse system?
– freakish
Nov 28 at 10:26
@freakish The system $mathfrak{U}(C)$ of open neigborhoods of $C$ is usually ordered by $U le V$ if $U supset V$ (reverse inclusion). $mathfrak{U}(C)$ is directed and may be understood as an inverse system of topological spaces: The bonding map $f^V_U : V to U$ is the inclusion.
– Paul Frost
Nov 28 at 11:15
@freakish The main reason why one doesn't consider the order $U le V$ if $U subset V$ is that $mathfrak{U}(C)$ would have $X = mathbb{R}^n$ as a maximum. That is, the direct system $mathfrak{U}(C)$ would be isomorphic to the trivial system ${ X }$.
– Paul Frost
Nov 28 at 14:26
|
show 4 more comments
Hi Alexander, are you sure this is an inverse limit and not a direct limit? Anyway, your question is more suitable for MathStackExchange, so I've voted to move it there.
– Mark Grant
Nov 27 at 14:24
U containing C, so should be inverse limit.
– user43326
Nov 27 at 20:17
This is confusing. I assume that open subsets form a directed poset via standard inclusion ordering. I also assume that bonding maps are induced from inclusions. But $H_i$ is a covariant functor. How is that an inverse system?
– freakish
Nov 28 at 10:26
@freakish The system $mathfrak{U}(C)$ of open neigborhoods of $C$ is usually ordered by $U le V$ if $U supset V$ (reverse inclusion). $mathfrak{U}(C)$ is directed and may be understood as an inverse system of topological spaces: The bonding map $f^V_U : V to U$ is the inclusion.
– Paul Frost
Nov 28 at 11:15
@freakish The main reason why one doesn't consider the order $U le V$ if $U subset V$ is that $mathfrak{U}(C)$ would have $X = mathbb{R}^n$ as a maximum. That is, the direct system $mathfrak{U}(C)$ would be isomorphic to the trivial system ${ X }$.
– Paul Frost
Nov 28 at 14:26
Hi Alexander, are you sure this is an inverse limit and not a direct limit? Anyway, your question is more suitable for MathStackExchange, so I've voted to move it there.
– Mark Grant
Nov 27 at 14:24
Hi Alexander, are you sure this is an inverse limit and not a direct limit? Anyway, your question is more suitable for MathStackExchange, so I've voted to move it there.
– Mark Grant
Nov 27 at 14:24
U containing C, so should be inverse limit.
– user43326
Nov 27 at 20:17
U containing C, so should be inverse limit.
– user43326
Nov 27 at 20:17
This is confusing. I assume that open subsets form a directed poset via standard inclusion ordering. I also assume that bonding maps are induced from inclusions. But $H_i$ is a covariant functor. How is that an inverse system?
– freakish
Nov 28 at 10:26
This is confusing. I assume that open subsets form a directed poset via standard inclusion ordering. I also assume that bonding maps are induced from inclusions. But $H_i$ is a covariant functor. How is that an inverse system?
– freakish
Nov 28 at 10:26
@freakish The system $mathfrak{U}(C)$ of open neigborhoods of $C$ is usually ordered by $U le V$ if $U supset V$ (reverse inclusion). $mathfrak{U}(C)$ is directed and may be understood as an inverse system of topological spaces: The bonding map $f^V_U : V to U$ is the inclusion.
– Paul Frost
Nov 28 at 11:15
@freakish The system $mathfrak{U}(C)$ of open neigborhoods of $C$ is usually ordered by $U le V$ if $U supset V$ (reverse inclusion). $mathfrak{U}(C)$ is directed and may be understood as an inverse system of topological spaces: The bonding map $f^V_U : V to U$ is the inclusion.
– Paul Frost
Nov 28 at 11:15
@freakish The main reason why one doesn't consider the order $U le V$ if $U subset V$ is that $mathfrak{U}(C)$ would have $X = mathbb{R}^n$ as a maximum. That is, the direct system $mathfrak{U}(C)$ would be isomorphic to the trivial system ${ X }$.
– Paul Frost
Nov 28 at 14:26
@freakish The main reason why one doesn't consider the order $U le V$ if $U subset V$ is that $mathfrak{U}(C)$ would have $X = mathbb{R}^n$ as a maximum. That is, the direct system $mathfrak{U}(C)$ would be isomorphic to the trivial system ${ X }$.
– Paul Frost
Nov 28 at 14:26
|
show 4 more comments
1 Answer
1
active
oldest
votes
up vote
1
down vote
If you work with singular homology, then the answer is "no".
Let $C$ be the topologist's sine curve in $mathbb{R}^2$. There is a cofinal inverse sequence $(U_n)$ in $mathfrak{U}(C)$ such that each $U_n$ is homeomorphic to an open disk. Let $i_n : U_{n+1} to U_n$ denote inclusion. Consider the long exact reduced homology sequence of the pair $(U_n,C)$. We have $H_1(U_n) = tilde{H}_0(U_n) = 0$, hence we get an isomorphism
$$partial(n) =partial : H_1(U_n,C) to tilde{H}_0(C) = mathbb{Z} .$$
Concerning $tilde{H}_0(C)$ recall that $C$ has two path components. By naturality we get $partial(n) (i_n)_* = partial(n+1)$ which shows
$$underset{Csubset U}{lim_leftarrow}H_1(U,C) approx underset{nin mathbb{N}}{lim_leftarrow}H_1(U_n,C) approx mathbb{Z} .$$
answered Nov 28 at 11:40
Paul Frost
8,6371528
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',
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%2f3016984%2finverse-limits-of-relative-homology-in-euclidean-space%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
up vote
1
down vote
If you work with singular homology, then the answer is "no".
Let $C$ be the topologist's sine curve in $mathbb{R}^2$. There is a cofinal inverse sequence $(U_n)$ in $mathfrak{U}(C)$ such that each $U_n$ is homeomorphic to an open disk. Let $i_n : U_{n+1} to U_n$ denote inclusion. Consider the long exact reduced homology sequence of the pair $(U_n,C)$. We have $H_1(U_n) = tilde{H}_0(U_n) = 0$, hence we get an isomorphism
$$partial(n) =partial : H_1(U_n,C) to tilde{H}_0(C) = mathbb{Z} .$$
Concerning $tilde{H}_0(C)$ recall that $C$ has two path components. By naturality we get $partial(n) (i_n)_* = partial(n+1)$ which shows
$$underset{Csubset U}{lim_leftarrow}H_1(U,C) approx underset{nin mathbb{N}}{lim_leftarrow}H_1(U_n,C) approx mathbb{Z} .$$
answered Nov 28 at 11:40
Paul Frost
8,6371528
add a comment |
up vote
1
down vote
If you work with singular homology, then the answer is "no".
Let $C$ be the topologist's sine curve in $mathbb{R}^2$. There is a cofinal inverse sequence $(U_n)$ in $mathfrak{U}(C)$ such that each $U_n$ is homeomorphic to an open disk. Let $i_n : U_{n+1} to U_n$ denote inclusion. Consider the long exact reduced homology sequence of the pair $(U_n,C)$. We have $H_1(U_n) = tilde{H}_0(U_n) = 0$, hence we get an isomorphism
$$partial(n) =partial : H_1(U_n,C) to tilde{H}_0(C) = mathbb{Z} .$$
Concerning $tilde{H}_0(C)$ recall that $C$ has two path components. By naturality we get $partial(n) (i_n)_* = partial(n+1)$ which shows
$$underset{Csubset U}{lim_leftarrow}H_1(U,C) approx underset{nin mathbb{N}}{lim_leftarrow}H_1(U_n,C) approx mathbb{Z} .$$
answered Nov 28 at 11:40
Paul Frost
8,6371528
add a comment |
up vote
1
down vote
up vote
1
down vote
If you work with singular homology, then the answer is "no".
Let $C$ be the topologist's sine curve in $mathbb{R}^2$. There is a cofinal inverse sequence $(U_n)$ in $mathfrak{U}(C)$ such that each $U_n$ is homeomorphic to an open disk. Let $i_n : U_{n+1} to U_n$ denote inclusion. Consider the long exact reduced homology sequence of the pair $(U_n,C)$. We have $H_1(U_n) = tilde{H}_0(U_n) = 0$, hence we get an isomorphism
$$partial(n) =partial : H_1(U_n,C) to tilde{H}_0(C) = mathbb{Z} .$$
Concerning $tilde{H}_0(C)$ recall that $C$ has two path components. By naturality we get $partial(n) (i_n)_* = partial(n+1)$ which shows
$$underset{Csubset U}{lim_leftarrow}H_1(U,C) approx underset{nin mathbb{N}}{lim_leftarrow}H_1(U_n,C) approx mathbb{Z} .$$
answered Nov 28 at 11:40
Paul Frost
8,6371528
If you work with singular homology, then the answer is "no".
Let $C$ be the topologist's sine curve in $mathbb{R}^2$. There is a cofinal inverse sequence $(U_n)$ in $mathfrak{U}(C)$ such that each $U_n$ is homeomorphic to an open disk. Let $i_n : U_{n+1} to U_n$ denote inclusion. Consider the long exact reduced homology sequence of the pair $(U_n,C)$. We have $H_1(U_n) = tilde{H}_0(U_n) = 0$, hence we get an isomorphism
$$partial(n) =partial : H_1(U_n,C) to tilde{H}_0(C) = mathbb{Z} .$$
Concerning $tilde{H}_0(C)$ recall that $C$ has two path components. By naturality we get $partial(n) (i_n)_* = partial(n+1)$ which shows
$$underset{Csubset U}{lim_leftarrow}H_1(U,C) approx underset{nin mathbb{N}}{lim_leftarrow}H_1(U_n,C) approx mathbb{Z} .$$
answered Nov 28 at 11:40
Paul Frost
8,6371528
answered Nov 28 at 11:40
Paul Frost
8,6371528
answered Nov 28 at 11:40
Paul Frost
8,6371528
answered Nov 28 at 11:40
Paul Frost
8,6371528
8,6371528
add a comment |
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3016984%2finverse-limits-of-relative-homology-in-euclidean-space%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
Hi Alexander, are you sure this is an inverse limit and not a direct limit? Anyway, your question is more suitable for MathStackExchange, so I've voted to move it there.
– Mark Grant
Nov 27 at 14:24
U containing C, so should be inverse limit.
– user43326
Nov 27 at 20:17
This is confusing. I assume that open subsets form a directed poset via standard inclusion ordering. I also assume that bonding maps are induced from inclusions. But $H_i$ is a covariant functor. How is that an inverse system?
– freakish
Nov 28 at 10:26
@freakish The system $mathfrak{U}(C)$ of open neigborhoods of $C$ is usually ordered by $U le V$ if $U supset V$ (reverse inclusion). $mathfrak{U}(C)$ is directed and may be understood as an inverse system of topological spaces: The bonding map $f^V_U : V to U$ is the inclusion.
– Paul Frost
Nov 28 at 11:15
@freakish The main reason why one doesn't consider the order $U le V$ if $U subset V$ is that $mathfrak{U}(C)$ would have $X = mathbb{R}^n$ as a maximum. That is, the direct system $mathfrak{U}(C)$ would be isomorphic to the trivial system ${ X }$.
– Paul Frost
Nov 28 at 14:26