An simple intermediate step of the proof that $partial^2 = 0$ in the case of singular homology












0












$begingroup$


Let $Delta_q$ be the standard $q$-simplex. The function $F_q^s : Delta_{q-1} to Delta_q$ is the $s$-th face of $Delta_q$, defined as the correstriction to $Delta_q$ of $(e_0,dots,e_{s-1},hat{e_s},e_{s+1},dots,e_q)$, the map $Delta_{q-1} to mathbb{R}^q$ that sends $e_i$ to itself for $i < s$ and $e_i$ to $e_{i+1}$ if $i geq s$.



I'm stuck on trying to prove the following elementary result,




Lemma. Let $0 leq j < i leq q$. Then $F_q^{i}F_{q-1}^{j} = F_q^{j}F_{q-1}^{i-1}$.




I apologize for the simplicity of the question, but I cannot seem to pinpoint where am I making the mistake, so any help is greatly appreciated.



By a direct calculation,



$$
begin{align}
F_q^{i}F_{q-1}^{j}(e_l) &= cases{F_q^{i}(e_l) text{ if $l < j$}\ F_q^{i}(e_{l+1}) text{ if $l geq j$}} \&=
cases{
e_l quad text{ if $l < j, l< i$}\
e_{l+1} text{ if $l < j, l geq i$}\
e_{l+1} text{ if $l geq j, l < i$}\
e_{l+2} text{ if $l geq j, l geq i$}} \&=
cases{
e_l quad text{ if $l < j$}\
e_{l+1} text{ if $l geq j, l < i$}\
e_{l+2} text{ if $l geq i$}}
end{align}
$$



and



$$
begin{align}
F_q^{j}F_{q-1}^{i-1}(e_l) &= cases{F_q^{j}(e_l) text{ if $l < i-1$}\ F_q^{j}(e_{l+1}) text{ if $l geq i-1$}} \ &=
cases{
e_l quad text{ if $l < i-1, l< j$}\
e_{l+1} text{ if $l < i-1, l geq j$}\
e_{l+1} text{ if $l geq i-1, l < j$}\
e_{l+2} text{ if $l geq i-1, l geq j$}} \ &=
cases{
e_l quad text{ if $l< j$}\
e_{l+1} text{ if $l < i-1, l geq j$}\
e_{l+2} text{ if $l geq i-1$}}
end{align}
$$



These don't seem to coincide. Where have I gone wrong?










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Many homology theories are described by a collection of maps called face and degeneracy maps, satisfying a set of axioms called simplicial identities (you may read more about them at the nLab). The ones you posted are precisely the ones which guarantee that their alternating sum is a differential.
    $endgroup$
    – F M
    Dec 17 '18 at 15:58
















0












$begingroup$


Let $Delta_q$ be the standard $q$-simplex. The function $F_q^s : Delta_{q-1} to Delta_q$ is the $s$-th face of $Delta_q$, defined as the correstriction to $Delta_q$ of $(e_0,dots,e_{s-1},hat{e_s},e_{s+1},dots,e_q)$, the map $Delta_{q-1} to mathbb{R}^q$ that sends $e_i$ to itself for $i < s$ and $e_i$ to $e_{i+1}$ if $i geq s$.



I'm stuck on trying to prove the following elementary result,




Lemma. Let $0 leq j < i leq q$. Then $F_q^{i}F_{q-1}^{j} = F_q^{j}F_{q-1}^{i-1}$.




I apologize for the simplicity of the question, but I cannot seem to pinpoint where am I making the mistake, so any help is greatly appreciated.



By a direct calculation,



$$
begin{align}
F_q^{i}F_{q-1}^{j}(e_l) &= cases{F_q^{i}(e_l) text{ if $l < j$}\ F_q^{i}(e_{l+1}) text{ if $l geq j$}} \&=
cases{
e_l quad text{ if $l < j, l< i$}\
e_{l+1} text{ if $l < j, l geq i$}\
e_{l+1} text{ if $l geq j, l < i$}\
e_{l+2} text{ if $l geq j, l geq i$}} \&=
cases{
e_l quad text{ if $l < j$}\
e_{l+1} text{ if $l geq j, l < i$}\
e_{l+2} text{ if $l geq i$}}
end{align}
$$



and



$$
begin{align}
F_q^{j}F_{q-1}^{i-1}(e_l) &= cases{F_q^{j}(e_l) text{ if $l < i-1$}\ F_q^{j}(e_{l+1}) text{ if $l geq i-1$}} \ &=
cases{
e_l quad text{ if $l < i-1, l< j$}\
e_{l+1} text{ if $l < i-1, l geq j$}\
e_{l+1} text{ if $l geq i-1, l < j$}\
e_{l+2} text{ if $l geq i-1, l geq j$}} \ &=
cases{
e_l quad text{ if $l< j$}\
e_{l+1} text{ if $l < i-1, l geq j$}\
e_{l+2} text{ if $l geq i-1$}}
end{align}
$$



These don't seem to coincide. Where have I gone wrong?










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Many homology theories are described by a collection of maps called face and degeneracy maps, satisfying a set of axioms called simplicial identities (you may read more about them at the nLab). The ones you posted are precisely the ones which guarantee that their alternating sum is a differential.
    $endgroup$
    – F M
    Dec 17 '18 at 15:58














0












0








0





$begingroup$


Let $Delta_q$ be the standard $q$-simplex. The function $F_q^s : Delta_{q-1} to Delta_q$ is the $s$-th face of $Delta_q$, defined as the correstriction to $Delta_q$ of $(e_0,dots,e_{s-1},hat{e_s},e_{s+1},dots,e_q)$, the map $Delta_{q-1} to mathbb{R}^q$ that sends $e_i$ to itself for $i < s$ and $e_i$ to $e_{i+1}$ if $i geq s$.



I'm stuck on trying to prove the following elementary result,




Lemma. Let $0 leq j < i leq q$. Then $F_q^{i}F_{q-1}^{j} = F_q^{j}F_{q-1}^{i-1}$.




I apologize for the simplicity of the question, but I cannot seem to pinpoint where am I making the mistake, so any help is greatly appreciated.



By a direct calculation,



$$
begin{align}
F_q^{i}F_{q-1}^{j}(e_l) &= cases{F_q^{i}(e_l) text{ if $l < j$}\ F_q^{i}(e_{l+1}) text{ if $l geq j$}} \&=
cases{
e_l quad text{ if $l < j, l< i$}\
e_{l+1} text{ if $l < j, l geq i$}\
e_{l+1} text{ if $l geq j, l < i$}\
e_{l+2} text{ if $l geq j, l geq i$}} \&=
cases{
e_l quad text{ if $l < j$}\
e_{l+1} text{ if $l geq j, l < i$}\
e_{l+2} text{ if $l geq i$}}
end{align}
$$



and



$$
begin{align}
F_q^{j}F_{q-1}^{i-1}(e_l) &= cases{F_q^{j}(e_l) text{ if $l < i-1$}\ F_q^{j}(e_{l+1}) text{ if $l geq i-1$}} \ &=
cases{
e_l quad text{ if $l < i-1, l< j$}\
e_{l+1} text{ if $l < i-1, l geq j$}\
e_{l+1} text{ if $l geq i-1, l < j$}\
e_{l+2} text{ if $l geq i-1, l geq j$}} \ &=
cases{
e_l quad text{ if $l< j$}\
e_{l+1} text{ if $l < i-1, l geq j$}\
e_{l+2} text{ if $l geq i-1$}}
end{align}
$$



These don't seem to coincide. Where have I gone wrong?










share|cite|improve this question











$endgroup$




Let $Delta_q$ be the standard $q$-simplex. The function $F_q^s : Delta_{q-1} to Delta_q$ is the $s$-th face of $Delta_q$, defined as the correstriction to $Delta_q$ of $(e_0,dots,e_{s-1},hat{e_s},e_{s+1},dots,e_q)$, the map $Delta_{q-1} to mathbb{R}^q$ that sends $e_i$ to itself for $i < s$ and $e_i$ to $e_{i+1}$ if $i geq s$.



I'm stuck on trying to prove the following elementary result,




Lemma. Let $0 leq j < i leq q$. Then $F_q^{i}F_{q-1}^{j} = F_q^{j}F_{q-1}^{i-1}$.




I apologize for the simplicity of the question, but I cannot seem to pinpoint where am I making the mistake, so any help is greatly appreciated.



By a direct calculation,



$$
begin{align}
F_q^{i}F_{q-1}^{j}(e_l) &= cases{F_q^{i}(e_l) text{ if $l < j$}\ F_q^{i}(e_{l+1}) text{ if $l geq j$}} \&=
cases{
e_l quad text{ if $l < j, l< i$}\
e_{l+1} text{ if $l < j, l geq i$}\
e_{l+1} text{ if $l geq j, l < i$}\
e_{l+2} text{ if $l geq j, l geq i$}} \&=
cases{
e_l quad text{ if $l < j$}\
e_{l+1} text{ if $l geq j, l < i$}\
e_{l+2} text{ if $l geq i$}}
end{align}
$$



and



$$
begin{align}
F_q^{j}F_{q-1}^{i-1}(e_l) &= cases{F_q^{j}(e_l) text{ if $l < i-1$}\ F_q^{j}(e_{l+1}) text{ if $l geq i-1$}} \ &=
cases{
e_l quad text{ if $l < i-1, l< j$}\
e_{l+1} text{ if $l < i-1, l geq j$}\
e_{l+1} text{ if $l geq i-1, l < j$}\
e_{l+2} text{ if $l geq i-1, l geq j$}} \ &=
cases{
e_l quad text{ if $l< j$}\
e_{l+1} text{ if $l < i-1, l geq j$}\
e_{l+2} text{ if $l geq i-1$}}
end{align}
$$



These don't seem to coincide. Where have I gone wrong?







algebraic-topology proof-explanation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 16 '18 at 6:39







Guido A.

















asked Dec 16 '18 at 6:33









Guido A.Guido A.

7,8701730




7,8701730








  • 1




    $begingroup$
    Many homology theories are described by a collection of maps called face and degeneracy maps, satisfying a set of axioms called simplicial identities (you may read more about them at the nLab). The ones you posted are precisely the ones which guarantee that their alternating sum is a differential.
    $endgroup$
    – F M
    Dec 17 '18 at 15:58














  • 1




    $begingroup$
    Many homology theories are described by a collection of maps called face and degeneracy maps, satisfying a set of axioms called simplicial identities (you may read more about them at the nLab). The ones you posted are precisely the ones which guarantee that their alternating sum is a differential.
    $endgroup$
    – F M
    Dec 17 '18 at 15:58








1




1




$begingroup$
Many homology theories are described by a collection of maps called face and degeneracy maps, satisfying a set of axioms called simplicial identities (you may read more about them at the nLab). The ones you posted are precisely the ones which guarantee that their alternating sum is a differential.
$endgroup$
– F M
Dec 17 '18 at 15:58




$begingroup$
Many homology theories are described by a collection of maps called face and degeneracy maps, satisfying a set of axioms called simplicial identities (you may read more about them at the nLab). The ones you posted are precisely the ones which guarantee that their alternating sum is a differential.
$endgroup$
– F M
Dec 17 '18 at 15:58










1 Answer
1






active

oldest

votes


















1












$begingroup$

You did not correctly calculate $F^s_q(e_{l+1})$. You have
$$
begin{align}
F_q^{i}F_{q-1}^{j}(e_l) &= cases{F_q^{i}(e_l) text{ if $l < j$}\ F_q^{i}(e_{l+1}) text{ if $l geq j$}} \&=
cases{
e_l quad text{ if $l < j, l< i$}\
e_{l+1} text{ if $l < j, l geq i$}\
e_{l+1} text{ if $l geq j, l+1 < i$}\
e_{l+2} text{ if $l geq j, l+1 geq i$}} \&=
cases{
e_l quad text{ if $l < j$}\
e_{l+1} text{ if $l geq j, l < i-1$}\
e_{l+2} text{ if $l geq i-1$}}
end{align}
$$



and



$$
begin{align}
F_q^{j}F_{q-1}^{i-1}(e_l) &= cases{F_q^{j}(e_l) text{ if $l < i-1$}\ F_q^{j}(e_{l+1}) text{ if $l geq i-1$}} \ &=
cases{
e_l quad text{ if $l < i-1, l< j$}\
e_{l+1} text{ if $l < i-1, l geq j$}\
e_{l+1} text{ if $l geq i-1, l+1< j$}\
e_{l+2} text{ if $l geq i-1, l+1 geq j$}} \ &=
cases{
e_l quad text{ if $l< j$}\
e_{l+1} text{ if $l < i-1, l geq j$}\
e_{l+2} text{ if $l geq i-1$}}
end{align}
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I see it now! I had rewritten this many times, but it seems that I had fixed my mistake already. Thanks a lot!
    $endgroup$
    – Guido A.
    Dec 17 '18 at 1:55











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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3042306%2fan-simple-intermediate-step-of-the-proof-that-partial2-0-in-the-case-of-si%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









1












$begingroup$

You did not correctly calculate $F^s_q(e_{l+1})$. You have
$$
begin{align}
F_q^{i}F_{q-1}^{j}(e_l) &= cases{F_q^{i}(e_l) text{ if $l < j$}\ F_q^{i}(e_{l+1}) text{ if $l geq j$}} \&=
cases{
e_l quad text{ if $l < j, l< i$}\
e_{l+1} text{ if $l < j, l geq i$}\
e_{l+1} text{ if $l geq j, l+1 < i$}\
e_{l+2} text{ if $l geq j, l+1 geq i$}} \&=
cases{
e_l quad text{ if $l < j$}\
e_{l+1} text{ if $l geq j, l < i-1$}\
e_{l+2} text{ if $l geq i-1$}}
end{align}
$$



and



$$
begin{align}
F_q^{j}F_{q-1}^{i-1}(e_l) &= cases{F_q^{j}(e_l) text{ if $l < i-1$}\ F_q^{j}(e_{l+1}) text{ if $l geq i-1$}} \ &=
cases{
e_l quad text{ if $l < i-1, l< j$}\
e_{l+1} text{ if $l < i-1, l geq j$}\
e_{l+1} text{ if $l geq i-1, l+1< j$}\
e_{l+2} text{ if $l geq i-1, l+1 geq j$}} \ &=
cases{
e_l quad text{ if $l< j$}\
e_{l+1} text{ if $l < i-1, l geq j$}\
e_{l+2} text{ if $l geq i-1$}}
end{align}
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I see it now! I had rewritten this many times, but it seems that I had fixed my mistake already. Thanks a lot!
    $endgroup$
    – Guido A.
    Dec 17 '18 at 1:55
















1












$begingroup$

You did not correctly calculate $F^s_q(e_{l+1})$. You have
$$
begin{align}
F_q^{i}F_{q-1}^{j}(e_l) &= cases{F_q^{i}(e_l) text{ if $l < j$}\ F_q^{i}(e_{l+1}) text{ if $l geq j$}} \&=
cases{
e_l quad text{ if $l < j, l< i$}\
e_{l+1} text{ if $l < j, l geq i$}\
e_{l+1} text{ if $l geq j, l+1 < i$}\
e_{l+2} text{ if $l geq j, l+1 geq i$}} \&=
cases{
e_l quad text{ if $l < j$}\
e_{l+1} text{ if $l geq j, l < i-1$}\
e_{l+2} text{ if $l geq i-1$}}
end{align}
$$



and



$$
begin{align}
F_q^{j}F_{q-1}^{i-1}(e_l) &= cases{F_q^{j}(e_l) text{ if $l < i-1$}\ F_q^{j}(e_{l+1}) text{ if $l geq i-1$}} \ &=
cases{
e_l quad text{ if $l < i-1, l< j$}\
e_{l+1} text{ if $l < i-1, l geq j$}\
e_{l+1} text{ if $l geq i-1, l+1< j$}\
e_{l+2} text{ if $l geq i-1, l+1 geq j$}} \ &=
cases{
e_l quad text{ if $l< j$}\
e_{l+1} text{ if $l < i-1, l geq j$}\
e_{l+2} text{ if $l geq i-1$}}
end{align}
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I see it now! I had rewritten this many times, but it seems that I had fixed my mistake already. Thanks a lot!
    $endgroup$
    – Guido A.
    Dec 17 '18 at 1:55














1












1








1





$begingroup$

You did not correctly calculate $F^s_q(e_{l+1})$. You have
$$
begin{align}
F_q^{i}F_{q-1}^{j}(e_l) &= cases{F_q^{i}(e_l) text{ if $l < j$}\ F_q^{i}(e_{l+1}) text{ if $l geq j$}} \&=
cases{
e_l quad text{ if $l < j, l< i$}\
e_{l+1} text{ if $l < j, l geq i$}\
e_{l+1} text{ if $l geq j, l+1 < i$}\
e_{l+2} text{ if $l geq j, l+1 geq i$}} \&=
cases{
e_l quad text{ if $l < j$}\
e_{l+1} text{ if $l geq j, l < i-1$}\
e_{l+2} text{ if $l geq i-1$}}
end{align}
$$



and



$$
begin{align}
F_q^{j}F_{q-1}^{i-1}(e_l) &= cases{F_q^{j}(e_l) text{ if $l < i-1$}\ F_q^{j}(e_{l+1}) text{ if $l geq i-1$}} \ &=
cases{
e_l quad text{ if $l < i-1, l< j$}\
e_{l+1} text{ if $l < i-1, l geq j$}\
e_{l+1} text{ if $l geq i-1, l+1< j$}\
e_{l+2} text{ if $l geq i-1, l+1 geq j$}} \ &=
cases{
e_l quad text{ if $l< j$}\
e_{l+1} text{ if $l < i-1, l geq j$}\
e_{l+2} text{ if $l geq i-1$}}
end{align}
$$






share|cite|improve this answer











$endgroup$



You did not correctly calculate $F^s_q(e_{l+1})$. You have
$$
begin{align}
F_q^{i}F_{q-1}^{j}(e_l) &= cases{F_q^{i}(e_l) text{ if $l < j$}\ F_q^{i}(e_{l+1}) text{ if $l geq j$}} \&=
cases{
e_l quad text{ if $l < j, l< i$}\
e_{l+1} text{ if $l < j, l geq i$}\
e_{l+1} text{ if $l geq j, l+1 < i$}\
e_{l+2} text{ if $l geq j, l+1 geq i$}} \&=
cases{
e_l quad text{ if $l < j$}\
e_{l+1} text{ if $l geq j, l < i-1$}\
e_{l+2} text{ if $l geq i-1$}}
end{align}
$$



and



$$
begin{align}
F_q^{j}F_{q-1}^{i-1}(e_l) &= cases{F_q^{j}(e_l) text{ if $l < i-1$}\ F_q^{j}(e_{l+1}) text{ if $l geq i-1$}} \ &=
cases{
e_l quad text{ if $l < i-1, l< j$}\
e_{l+1} text{ if $l < i-1, l geq j$}\
e_{l+1} text{ if $l geq i-1, l+1< j$}\
e_{l+2} text{ if $l geq i-1, l+1 geq j$}} \ &=
cases{
e_l quad text{ if $l< j$}\
e_{l+1} text{ if $l < i-1, l geq j$}\
e_{l+2} text{ if $l geq i-1$}}
end{align}
$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 16 '18 at 9:45

























answered Dec 16 '18 at 9:38









Paul FrostPaul Frost

11.5k3934




11.5k3934












  • $begingroup$
    I see it now! I had rewritten this many times, but it seems that I had fixed my mistake already. Thanks a lot!
    $endgroup$
    – Guido A.
    Dec 17 '18 at 1:55


















  • $begingroup$
    I see it now! I had rewritten this many times, but it seems that I had fixed my mistake already. Thanks a lot!
    $endgroup$
    – Guido A.
    Dec 17 '18 at 1:55
















$begingroup$
I see it now! I had rewritten this many times, but it seems that I had fixed my mistake already. Thanks a lot!
$endgroup$
– Guido A.
Dec 17 '18 at 1:55




$begingroup$
I see it now! I had rewritten this many times, but it seems that I had fixed my mistake already. Thanks a lot!
$endgroup$
– Guido A.
Dec 17 '18 at 1:55


















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%2f3042306%2fan-simple-intermediate-step-of-the-proof-that-partial2-0-in-the-case-of-si%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