Mixing
Three questions about Properties of Ext functor.
$begingroup$
I have three questions about the Ext functor's properties.
(i) $Ext(H oplus H',G) = Ext(H,G) oplus Ext(H',G)$
(ii) $Ext(H,G) = 0$ if $H$ is free
(iii) $Ext(mathbb{Z_n}, G) = G/nG$
There is a short proof in Hatcher page 195 and here are my questions:
(i) For (i), it is say this is done ("clearly") by the free resolutions
$0 to F_2 to F_1 to F_0 to H to 0$
$oplus$
$0 to F_2' to F_1' to F_0' to H' to 0$
= $0 to F_2 oplus F_2' to F_1 oplus F_1' to F_0 oplus F_0' to Hoplus H' to 0$
Can someone write down the detail in this? I feel like we can't just write what I did, take the Ext and compare it to the Ext when we direct sum.
(ii) The proof is $0 to H to H to 0$. I am guessing this is because the Ext functor is invariant under $H$, so the $F_2 to F_1 to F_0 to dots$ doesn't matter and we can take all of that as just $H$ since $H$ is free. Unrelated, but how does the Ext functor depend/invariant on $G$?
(iii)
The proof is the map which is a dual of $0 to Z to Z to Z_n to 0$
How do we get $Hom to Ext$, the far left map? (Do not use vertical map)
algebraic-topology proof-explanation homological-algebra
asked Feb 1 at 6:50
HawkHawk
5,5751140110
$endgroup$
add a comment |
$begingroup$
I have three questions about the Ext functor's properties.
(i) $Ext(H oplus H',G) = Ext(H,G) oplus Ext(H',G)$
(ii) $Ext(H,G) = 0$ if $H$ is free
(iii) $Ext(mathbb{Z_n}, G) = G/nG$
There is a short proof in Hatcher page 195 and here are my questions:
(i) For (i), it is say this is done ("clearly") by the free resolutions
$0 to F_2 to F_1 to F_0 to H to 0$
$oplus$
$0 to F_2' to F_1' to F_0' to H' to 0$
= $0 to F_2 oplus F_2' to F_1 oplus F_1' to F_0 oplus F_0' to Hoplus H' to 0$
Can someone write down the detail in this? I feel like we can't just write what I did, take the Ext and compare it to the Ext when we direct sum.
(ii) The proof is $0 to H to H to 0$. I am guessing this is because the Ext functor is invariant under $H$, so the $F_2 to F_1 to F_0 to dots$ doesn't matter and we can take all of that as just $H$ since $H$ is free. Unrelated, but how does the Ext functor depend/invariant on $G$?
(iii)
The proof is the map which is a dual of $0 to Z to Z to Z_n to 0$
How do we get $Hom to Ext$, the far left map? (Do not use vertical map)
algebraic-topology proof-explanation homological-algebra
asked Feb 1 at 6:50
HawkHawk
5,5751140110
$endgroup$
$begingroup$
(i) you are just taking direct sums of everything. (ii) $F_i=0$ for al $i>0$. (iii) this is the connecting map in the long exact sequence, and fact that $text{Ext}(Bbb Z,G)=0$.
$endgroup$
– Lord Shark the Unknown
Feb 1 at 7:40
$begingroup$
(i) Yeah I find what I wrote is still not very rigorous. (ii) Oh right so the whole free resolution is free, so the maps $F_0 to H$ is an inclusion and so the 1st cohomology group is also $0$ as the rest of the left maps (iii) You mind referencing the commutative diagram in the book...? I can't recall which it is.
$endgroup$
– Hawk
Feb 1 at 8:40
$begingroup$
Your statements are a bit confusing. You change $H$s to $A$ and $B$, and also probably mean $mathbb Z/nmathbb Z$ in the last bullet point, and not $mathbb Z$.
$endgroup$
– Pedro Tamaroff♦
Feb 1 at 17:37
$begingroup$
@PedroTamaroff thank you. I just changed it.
$endgroup$
– Hawk
Feb 1 at 21:55
add a comment |
$begingroup$
I have three questions about the Ext functor's properties.
(i) $Ext(H oplus H',G) = Ext(H,G) oplus Ext(H',G)$
(ii) $Ext(H,G) = 0$ if $H$ is free
(iii) $Ext(mathbb{Z_n}, G) = G/nG$
There is a short proof in Hatcher page 195 and here are my questions:
(i) For (i), it is say this is done ("clearly") by the free resolutions
$0 to F_2 to F_1 to F_0 to H to 0$
$oplus$
$0 to F_2' to F_1' to F_0' to H' to 0$
= $0 to F_2 oplus F_2' to F_1 oplus F_1' to F_0 oplus F_0' to Hoplus H' to 0$
Can someone write down the detail in this? I feel like we can't just write what I did, take the Ext and compare it to the Ext when we direct sum.
(ii) The proof is $0 to H to H to 0$. I am guessing this is because the Ext functor is invariant under $H$, so the $F_2 to F_1 to F_0 to dots$ doesn't matter and we can take all of that as just $H$ since $H$ is free. Unrelated, but how does the Ext functor depend/invariant on $G$?
(iii)
The proof is the map which is a dual of $0 to Z to Z to Z_n to 0$
How do we get $Hom to Ext$, the far left map? (Do not use vertical map)
algebraic-topology proof-explanation homological-algebra
asked Feb 1 at 6:50
HawkHawk
5,5751140110
$endgroup$
I have three questions about the Ext functor's properties.
(i) $Ext(H oplus H',G) = Ext(H,G) oplus Ext(H',G)$
(ii) $Ext(H,G) = 0$ if $H$ is free
(iii) $Ext(mathbb{Z_n}, G) = G/nG$
There is a short proof in Hatcher page 195 and here are my questions:
(i) For (i), it is say this is done ("clearly") by the free resolutions
$0 to F_2 to F_1 to F_0 to H to 0$
$oplus$
$0 to F_2' to F_1' to F_0' to H' to 0$
= $0 to F_2 oplus F_2' to F_1 oplus F_1' to F_0 oplus F_0' to Hoplus H' to 0$
Can someone write down the detail in this? I feel like we can't just write what I did, take the Ext and compare it to the Ext when we direct sum.
(ii) The proof is $0 to H to H to 0$. I am guessing this is because the Ext functor is invariant under $H$, so the $F_2 to F_1 to F_0 to dots$ doesn't matter and we can take all of that as just $H$ since $H$ is free. Unrelated, but how does the Ext functor depend/invariant on $G$?
(iii)
The proof is the map which is a dual of $0 to Z to Z to Z_n to 0$
How do we get $Hom to Ext$, the far left map? (Do not use vertical map)
algebraic-topology proof-explanation homological-algebra
algebraic-topology proof-explanation homological-algebra
asked Feb 1 at 6:50
HawkHawk
5,5751140110
asked Feb 1 at 6:50
HawkHawk
5,5751140110
edited Feb 1 at 21:54
Hawk
asked Feb 1 at 6:50
HawkHawk
5,5751140110
asked Feb 1 at 6:50
HawkHawk
5,5751140110
asked Feb 1 at 6:50
HawkHawk
5,5751140110
5,5751140110
$begingroup$
(i) you are just taking direct sums of everything. (ii) $F_i=0$ for al $i>0$. (iii) this is the connecting map in the long exact sequence, and fact that $text{Ext}(Bbb Z,G)=0$.
$endgroup$
– Lord Shark the Unknown
Feb 1 at 7:40
$begingroup$
(i) Yeah I find what I wrote is still not very rigorous. (ii) Oh right so the whole free resolution is free, so the maps $F_0 to H$ is an inclusion and so the 1st cohomology group is also $0$ as the rest of the left maps (iii) You mind referencing the commutative diagram in the book...? I can't recall which it is.
$endgroup$
– Hawk
Feb 1 at 8:40
$begingroup$
Your statements are a bit confusing. You change $H$s to $A$ and $B$, and also probably mean $mathbb Z/nmathbb Z$ in the last bullet point, and not $mathbb Z$.
$endgroup$
– Pedro Tamaroff♦
Feb 1 at 17:37
$begingroup$
@PedroTamaroff thank you. I just changed it.
$endgroup$
– Hawk
Feb 1 at 21:55
add a comment |
$begingroup$
(i) you are just taking direct sums of everything. (ii) $F_i=0$ for al $i>0$. (iii) this is the connecting map in the long exact sequence, and fact that $text{Ext}(Bbb Z,G)=0$.
$endgroup$
– Lord Shark the Unknown
Feb 1 at 7:40
$begingroup$
(i) Yeah I find what I wrote is still not very rigorous. (ii) Oh right so the whole free resolution is free, so the maps $F_0 to H$ is an inclusion and so the 1st cohomology group is also $0$ as the rest of the left maps (iii) You mind referencing the commutative diagram in the book...? I can't recall which it is.
$endgroup$
– Hawk
Feb 1 at 8:40
$begingroup$
Your statements are a bit confusing. You change $H$s to $A$ and $B$, and also probably mean $mathbb Z/nmathbb Z$ in the last bullet point, and not $mathbb Z$.
$endgroup$
– Pedro Tamaroff♦
Feb 1 at 17:37
$begingroup$
@PedroTamaroff thank you. I just changed it.
$endgroup$
– Hawk
Feb 1 at 21:55
$begingroup$
(i) you are just taking direct sums of everything. (ii) $F_i=0$ for al $i>0$. (iii) this is the connecting map in the long exact sequence, and fact that $text{Ext}(Bbb Z,G)=0$.
$endgroup$
– Lord Shark the Unknown
Feb 1 at 7:40
$begingroup$
(i) you are just taking direct sums of everything. (ii) $F_i=0$ for al $i>0$. (iii) this is the connecting map in the long exact sequence, and fact that $text{Ext}(Bbb Z,G)=0$.
$endgroup$
– Lord Shark the Unknown
Feb 1 at 7:40
$begingroup$
(i) Yeah I find what I wrote is still not very rigorous. (ii) Oh right so the whole free resolution is free, so the maps $F_0 to H$ is an inclusion and so the 1st cohomology group is also $0$ as the rest of the left maps (iii) You mind referencing the commutative diagram in the book...? I can't recall which it is.
$endgroup$
– Hawk
Feb 1 at 8:40
$begingroup$
(i) Yeah I find what I wrote is still not very rigorous. (ii) Oh right so the whole free resolution is free, so the maps $F_0 to H$ is an inclusion and so the 1st cohomology group is also $0$ as the rest of the left maps (iii) You mind referencing the commutative diagram in the book...? I can't recall which it is.
$endgroup$
– Hawk
Feb 1 at 8:40
$begingroup$
Your statements are a bit confusing. You change $H$s to $A$ and $B$, and also probably mean $mathbb Z/nmathbb Z$ in the last bullet point, and not $mathbb Z$.
$endgroup$
– Pedro Tamaroff♦
Feb 1 at 17:37
$begingroup$
Your statements are a bit confusing. You change $H$s to $A$ and $B$, and also probably mean $mathbb Z/nmathbb Z$ in the last bullet point, and not $mathbb Z$.
$endgroup$
– Pedro Tamaroff♦
Feb 1 at 17:37
$begingroup$
@PedroTamaroff thank you. I just changed it.
$endgroup$
– Hawk
Feb 1 at 21:55
$begingroup$
@PedroTamaroff thank you. I just changed it.
$endgroup$
– Hawk
Feb 1 at 21:55
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%2f3095902%2fthree-questions-about-properties-of-ext-functor%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.
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%2f3095902%2fthree-questions-about-properties-of-ext-functor%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
$begingroup$
(i) you are just taking direct sums of everything. (ii) $F_i=0$ for al $i>0$. (iii) this is the connecting map in the long exact sequence, and fact that $text{Ext}(Bbb Z,G)=0$.
$endgroup$
– Lord Shark the Unknown
Feb 1 at 7:40
$begingroup$
(i) Yeah I find what I wrote is still not very rigorous. (ii) Oh right so the whole free resolution is free, so the maps $F_0 to H$ is an inclusion and so the 1st cohomology group is also $0$ as the rest of the left maps (iii) You mind referencing the commutative diagram in the book...? I can't recall which it is.
$endgroup$
– Hawk
Feb 1 at 8:40
$begingroup$
Your statements are a bit confusing. You change $H$s to $A$ and $B$, and also probably mean $mathbb Z/nmathbb Z$ in the last bullet point, and not $mathbb Z$.
$endgroup$
– Pedro Tamaroff♦
Feb 1 at 17:37
$begingroup$
@PedroTamaroff thank you. I just changed it.
$endgroup$
– Hawk
Feb 1 at 21:55