About a Kan fibration (Postnikov towers for simplicial sets)
I am trying to understand a specific construction of Postnikov towers for simplicial sets, as explained for instance here (under "absolute Postnikov tower")
So you start with a simplicial set $X$ (I don't know if it works for all simplicial sets, I don't mind assuming it's a Kan complex if it's necessary; or even if it makes the argument substantially easier) and define, for all $ngeq 0$ the following relation on simplices of $X$: if $alpha,beta in X_q$, seen as $alpha,beta :Delta^q to X$ , then $alphasim_n beta$ if and only if $sk_n(alpha) = sk_n(beta) : sk_nDelta^q to X$.
It is quite clear that this relation is compatible with boundaries and degeneracies, because if you look at $alpha in X_q$ as $alpha: Delta^qto X$, then $d_i(alpha) = alphacirc delta_i$, and $sk_n$ is a functor so everything works well here; and we may define $X(n) = X/sim_n$ as a simplicial set; and we have canonical projection maps $Xto X(n), X(n+1)to X(n)$ for all $ngeq 0$.
It's easy to see that the $X(n)$ satisfy the homotopy groups part of the requirements for a Postnikov tower; but I'm struggling to see why $X(n+1)to X(n)$ should be a Kan fibration. My question is :
Why is it a Kan fibration ?
Here's what I worked out : Let $p:X(n+1)to X(n)$ the wannabe-fibration; and start with $(m-1)$-simplices $x_0,...,widehat{x_k},...,x_min X(n+1)_{m-1}$ with compatible boundaries, and $yin X(n)_m$ such that $d_i(y) = p(x_i)$ for $ineq k$.
Then as $p$ is surjective, find $tilde{x} in X(n+1)_m$ such that for all $ineq k$, $d_i(tilde{x}) sim_n x_i$.
If $m-1 leq n$, then this implies $d_i(tilde{x}) = x_i$ for $ineq k$, so $tilde{x}$ is an appropriate filling. But we must now deal with the case $m-1> n$, i.e. $m-1geq n+1$.
Now this $n+1$ sort of makes me happy because in $X(n+1)$ we're only dealing with things up to their $(n+1)$-skeleton; but I don't really know what to do with it...
If we try to take $Delta^{n+1}overset{alpha}{to} Delta^{m-1} overset{x_i}{underset{d_i(tilde{x})}{rightrightarrows}} X(n+1)$ and show that they are equal, then if $alpha$ in its normal form has some codegeneracies, then we are done because then we pass to the $n$-skeleton, but if $alpha$ is only made up of cofaces, I don't know what to do.. I don't even know whether the same $tilde{x}$ will work, or if I somehow have to cook up another $z$ (and if I have to, this makes me think that perhaps I actually need $X$ to be a Kan complex, because otherwise I don't know how to "make simplices appear"); so that's why I need help.
EDIT : I noticed that I didn't mention that I knew how to prove that $Xto X(n)$ was a Kan fibration if $X$ is a Kan complex. Maybe that can help ?
EDIT 2 : Update, thanks to the first edit, I know it suffices to prove that if $qcirc p$ and $p$ are surjective fibrations, then $q$ is a fibration. For that I reduced it to the following statement (which I saw online and so think is true) : if $p: Xto Y$ is a surjective fibration, then any map $Lambda^n_k to Y$ lifts to $Lambda^n_k to X$; I'll try to prove this; but unless you have a better proof for the old question, this should be the new question : Why is this lifting property true ?
algebraic-topology homotopy-theory simplicial-stuff fibration
asked Dec 27 '18 at 12:29
Max
12.9k11040
add a comment |
I am trying to understand a specific construction of Postnikov towers for simplicial sets, as explained for instance here (under "absolute Postnikov tower")
So you start with a simplicial set $X$ (I don't know if it works for all simplicial sets, I don't mind assuming it's a Kan complex if it's necessary; or even if it makes the argument substantially easier) and define, for all $ngeq 0$ the following relation on simplices of $X$: if $alpha,beta in X_q$, seen as $alpha,beta :Delta^q to X$ , then $alphasim_n beta$ if and only if $sk_n(alpha) = sk_n(beta) : sk_nDelta^q to X$.
It is quite clear that this relation is compatible with boundaries and degeneracies, because if you look at $alpha in X_q$ as $alpha: Delta^qto X$, then $d_i(alpha) = alphacirc delta_i$, and $sk_n$ is a functor so everything works well here; and we may define $X(n) = X/sim_n$ as a simplicial set; and we have canonical projection maps $Xto X(n), X(n+1)to X(n)$ for all $ngeq 0$.
It's easy to see that the $X(n)$ satisfy the homotopy groups part of the requirements for a Postnikov tower; but I'm struggling to see why $X(n+1)to X(n)$ should be a Kan fibration. My question is :
Why is it a Kan fibration ?
Here's what I worked out : Let $p:X(n+1)to X(n)$ the wannabe-fibration; and start with $(m-1)$-simplices $x_0,...,widehat{x_k},...,x_min X(n+1)_{m-1}$ with compatible boundaries, and $yin X(n)_m$ such that $d_i(y) = p(x_i)$ for $ineq k$.
Then as $p$ is surjective, find $tilde{x} in X(n+1)_m$ such that for all $ineq k$, $d_i(tilde{x}) sim_n x_i$.
If $m-1 leq n$, then this implies $d_i(tilde{x}) = x_i$ for $ineq k$, so $tilde{x}$ is an appropriate filling. But we must now deal with the case $m-1> n$, i.e. $m-1geq n+1$.
Now this $n+1$ sort of makes me happy because in $X(n+1)$ we're only dealing with things up to their $(n+1)$-skeleton; but I don't really know what to do with it...
If we try to take $Delta^{n+1}overset{alpha}{to} Delta^{m-1} overset{x_i}{underset{d_i(tilde{x})}{rightrightarrows}} X(n+1)$ and show that they are equal, then if $alpha$ in its normal form has some codegeneracies, then we are done because then we pass to the $n$-skeleton, but if $alpha$ is only made up of cofaces, I don't know what to do.. I don't even know whether the same $tilde{x}$ will work, or if I somehow have to cook up another $z$ (and if I have to, this makes me think that perhaps I actually need $X$ to be a Kan complex, because otherwise I don't know how to "make simplices appear"); so that's why I need help.
EDIT : I noticed that I didn't mention that I knew how to prove that $Xto X(n)$ was a Kan fibration if $X$ is a Kan complex. Maybe that can help ?
EDIT 2 : Update, thanks to the first edit, I know it suffices to prove that if $qcirc p$ and $p$ are surjective fibrations, then $q$ is a fibration. For that I reduced it to the following statement (which I saw online and so think is true) : if $p: Xto Y$ is a surjective fibration, then any map $Lambda^n_k to Y$ lifts to $Lambda^n_k to X$; I'll try to prove this; but unless you have a better proof for the old question, this should be the new question : Why is this lifting property true ?
algebraic-topology homotopy-theory simplicial-stuff fibration
asked Dec 27 '18 at 12:29
Max
12.9k11040
If you already have the model structure, your fact in edit 2 just follows from the fact that $Delta^0 to Lambda^n_k$ is an acyclic cofibration.
– Justin Young
Dec 30 '18 at 17:21
@JustinYoung : thanks for your answer, but I would like to prove it from some more basic facts if possible (though I understand your argument )
– Max
Dec 30 '18 at 18:53
@JustinYoung : but if you know how to prove that it has the lifting property against fibrations, say by exhibiting a sequence of pushouts/transfinite compositions of $Lambda^n_k to Delta^n$, then that would be fine by me
– Max
Dec 30 '18 at 18:55
add a comment |
I am trying to understand a specific construction of Postnikov towers for simplicial sets, as explained for instance here (under "absolute Postnikov tower")
So you start with a simplicial set $X$ (I don't know if it works for all simplicial sets, I don't mind assuming it's a Kan complex if it's necessary; or even if it makes the argument substantially easier) and define, for all $ngeq 0$ the following relation on simplices of $X$: if $alpha,beta in X_q$, seen as $alpha,beta :Delta^q to X$ , then $alphasim_n beta$ if and only if $sk_n(alpha) = sk_n(beta) : sk_nDelta^q to X$.
It is quite clear that this relation is compatible with boundaries and degeneracies, because if you look at $alpha in X_q$ as $alpha: Delta^qto X$, then $d_i(alpha) = alphacirc delta_i$, and $sk_n$ is a functor so everything works well here; and we may define $X(n) = X/sim_n$ as a simplicial set; and we have canonical projection maps $Xto X(n), X(n+1)to X(n)$ for all $ngeq 0$.
It's easy to see that the $X(n)$ satisfy the homotopy groups part of the requirements for a Postnikov tower; but I'm struggling to see why $X(n+1)to X(n)$ should be a Kan fibration. My question is :
Why is it a Kan fibration ?
Here's what I worked out : Let $p:X(n+1)to X(n)$ the wannabe-fibration; and start with $(m-1)$-simplices $x_0,...,widehat{x_k},...,x_min X(n+1)_{m-1}$ with compatible boundaries, and $yin X(n)_m$ such that $d_i(y) = p(x_i)$ for $ineq k$.
Then as $p$ is surjective, find $tilde{x} in X(n+1)_m$ such that for all $ineq k$, $d_i(tilde{x}) sim_n x_i$.
If $m-1 leq n$, then this implies $d_i(tilde{x}) = x_i$ for $ineq k$, so $tilde{x}$ is an appropriate filling. But we must now deal with the case $m-1> n$, i.e. $m-1geq n+1$.
Now this $n+1$ sort of makes me happy because in $X(n+1)$ we're only dealing with things up to their $(n+1)$-skeleton; but I don't really know what to do with it...
If we try to take $Delta^{n+1}overset{alpha}{to} Delta^{m-1} overset{x_i}{underset{d_i(tilde{x})}{rightrightarrows}} X(n+1)$ and show that they are equal, then if $alpha$ in its normal form has some codegeneracies, then we are done because then we pass to the $n$-skeleton, but if $alpha$ is only made up of cofaces, I don't know what to do.. I don't even know whether the same $tilde{x}$ will work, or if I somehow have to cook up another $z$ (and if I have to, this makes me think that perhaps I actually need $X$ to be a Kan complex, because otherwise I don't know how to "make simplices appear"); so that's why I need help.
EDIT : I noticed that I didn't mention that I knew how to prove that $Xto X(n)$ was a Kan fibration if $X$ is a Kan complex. Maybe that can help ?
EDIT 2 : Update, thanks to the first edit, I know it suffices to prove that if $qcirc p$ and $p$ are surjective fibrations, then $q$ is a fibration. For that I reduced it to the following statement (which I saw online and so think is true) : if $p: Xto Y$ is a surjective fibration, then any map $Lambda^n_k to Y$ lifts to $Lambda^n_k to X$; I'll try to prove this; but unless you have a better proof for the old question, this should be the new question : Why is this lifting property true ?
algebraic-topology homotopy-theory simplicial-stuff fibration
asked Dec 27 '18 at 12:29
Max
12.9k11040
I am trying to understand a specific construction of Postnikov towers for simplicial sets, as explained for instance here (under "absolute Postnikov tower")
So you start with a simplicial set $X$ (I don't know if it works for all simplicial sets, I don't mind assuming it's a Kan complex if it's necessary; or even if it makes the argument substantially easier) and define, for all $ngeq 0$ the following relation on simplices of $X$: if $alpha,beta in X_q$, seen as $alpha,beta :Delta^q to X$ , then $alphasim_n beta$ if and only if $sk_n(alpha) = sk_n(beta) : sk_nDelta^q to X$.
It is quite clear that this relation is compatible with boundaries and degeneracies, because if you look at $alpha in X_q$ as $alpha: Delta^qto X$, then $d_i(alpha) = alphacirc delta_i$, and $sk_n$ is a functor so everything works well here; and we may define $X(n) = X/sim_n$ as a simplicial set; and we have canonical projection maps $Xto X(n), X(n+1)to X(n)$ for all $ngeq 0$.
It's easy to see that the $X(n)$ satisfy the homotopy groups part of the requirements for a Postnikov tower; but I'm struggling to see why $X(n+1)to X(n)$ should be a Kan fibration. My question is :
Why is it a Kan fibration ?
Here's what I worked out : Let $p:X(n+1)to X(n)$ the wannabe-fibration; and start with $(m-1)$-simplices $x_0,...,widehat{x_k},...,x_min X(n+1)_{m-1}$ with compatible boundaries, and $yin X(n)_m$ such that $d_i(y) = p(x_i)$ for $ineq k$.
Then as $p$ is surjective, find $tilde{x} in X(n+1)_m$ such that for all $ineq k$, $d_i(tilde{x}) sim_n x_i$.
If $m-1 leq n$, then this implies $d_i(tilde{x}) = x_i$ for $ineq k$, so $tilde{x}$ is an appropriate filling. But we must now deal with the case $m-1> n$, i.e. $m-1geq n+1$.
Now this $n+1$ sort of makes me happy because in $X(n+1)$ we're only dealing with things up to their $(n+1)$-skeleton; but I don't really know what to do with it...
If we try to take $Delta^{n+1}overset{alpha}{to} Delta^{m-1} overset{x_i}{underset{d_i(tilde{x})}{rightrightarrows}} X(n+1)$ and show that they are equal, then if $alpha$ in its normal form has some codegeneracies, then we are done because then we pass to the $n$-skeleton, but if $alpha$ is only made up of cofaces, I don't know what to do.. I don't even know whether the same $tilde{x}$ will work, or if I somehow have to cook up another $z$ (and if I have to, this makes me think that perhaps I actually need $X$ to be a Kan complex, because otherwise I don't know how to "make simplices appear"); so that's why I need help.
EDIT : I noticed that I didn't mention that I knew how to prove that $Xto X(n)$ was a Kan fibration if $X$ is a Kan complex. Maybe that can help ?
EDIT 2 : Update, thanks to the first edit, I know it suffices to prove that if $qcirc p$ and $p$ are surjective fibrations, then $q$ is a fibration. For that I reduced it to the following statement (which I saw online and so think is true) : if $p: Xto Y$ is a surjective fibration, then any map $Lambda^n_k to Y$ lifts to $Lambda^n_k to X$; I'll try to prove this; but unless you have a better proof for the old question, this should be the new question : Why is this lifting property true ?
algebraic-topology homotopy-theory simplicial-stuff fibration
algebraic-topology homotopy-theory simplicial-stuff fibration
asked Dec 27 '18 at 12:29
Max
12.9k11040
asked Dec 27 '18 at 12:29
Max
12.9k11040
edited Dec 27 '18 at 17:15
asked Dec 27 '18 at 12:29
Max
12.9k11040
asked Dec 27 '18 at 12:29
Max
12.9k11040
asked Dec 27 '18 at 12:29
Max
12.9k11040
12.9k11040
If you already have the model structure, your fact in edit 2 just follows from the fact that $Delta^0 to Lambda^n_k$ is an acyclic cofibration.
– Justin Young
Dec 30 '18 at 17:21
@JustinYoung : thanks for your answer, but I would like to prove it from some more basic facts if possible (though I understand your argument )
– Max
Dec 30 '18 at 18:53
@JustinYoung : but if you know how to prove that it has the lifting property against fibrations, say by exhibiting a sequence of pushouts/transfinite compositions of $Lambda^n_k to Delta^n$, then that would be fine by me
– Max
Dec 30 '18 at 18:55
add a comment |
If you already have the model structure, your fact in edit 2 just follows from the fact that $Delta^0 to Lambda^n_k$ is an acyclic cofibration.
– Justin Young
Dec 30 '18 at 17:21
@JustinYoung : thanks for your answer, but I would like to prove it from some more basic facts if possible (though I understand your argument )
– Max
Dec 30 '18 at 18:53
@JustinYoung : but if you know how to prove that it has the lifting property against fibrations, say by exhibiting a sequence of pushouts/transfinite compositions of $Lambda^n_k to Delta^n$, then that would be fine by me
– Max
Dec 30 '18 at 18:55
If you already have the model structure, your fact in edit 2 just follows from the fact that $Delta^0 to Lambda^n_k$ is an acyclic cofibration.
– Justin Young
Dec 30 '18 at 17:21
If you already have the model structure, your fact in edit 2 just follows from the fact that $Delta^0 to Lambda^n_k$ is an acyclic cofibration.
– Justin Young
Dec 30 '18 at 17:21
@JustinYoung : thanks for your answer, but I would like to prove it from some more basic facts if possible (though I understand your argument )
– Max
Dec 30 '18 at 18:53
@JustinYoung : thanks for your answer, but I would like to prove it from some more basic facts if possible (though I understand your argument )
– Max
Dec 30 '18 at 18:53
@JustinYoung : but if you know how to prove that it has the lifting property against fibrations, say by exhibiting a sequence of pushouts/transfinite compositions of $Lambda^n_k to Delta^n$, then that would be fine by me
– Max
Dec 30 '18 at 18:55
@JustinYoung : but if you know how to prove that it has the lifting property against fibrations, say by exhibiting a sequence of pushouts/transfinite compositions of $Lambda^n_k to Delta^n$, then that would be fine by me
– Max
Dec 30 '18 at 18:55
add a comment |
0
active
oldest
votes
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%2f3053874%2fabout-a-kan-fibration-postnikov-towers-for-simplicial-sets%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
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%2f3053874%2fabout-a-kan-fibration-postnikov-towers-for-simplicial-sets%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
If you already have the model structure, your fact in edit 2 just follows from the fact that $Delta^0 to Lambda^n_k$ is an acyclic cofibration.
– Justin Young
Dec 30 '18 at 17:21
@JustinYoung : thanks for your answer, but I would like to prove it from some more basic facts if possible (though I understand your argument )
– Max
Dec 30 '18 at 18:53
@JustinYoung : but if you know how to prove that it has the lifting property against fibrations, say by exhibiting a sequence of pushouts/transfinite compositions of $Lambda^n_k to Delta^n$, then that would be fine by me
– Max
Dec 30 '18 at 18:55