Very basic question about pre-additive category
$begingroup$
I am trying to prove whether in a pre-additive category, $0_{Mor(y,z)}circ f=0_{Mor(x,z)}$ for objects $x,y,z$ and $fin Mor(x,y)$. Now, by bi-linearity of composition maps $$ 0_{Mor(y,z)}circ f+ 0_{Mor(y,z)}circ (-f)= 0_{Mor(y,z)} circ 0_{Mor(x,y)} $$
Putting $f=0_{Mor(x,y)}; ;$ we have $; ; 0_{Mor(y,z)} circ 0_{Mor(x,y)} =0_{Mor(x,z)}; ; $and we also get $; ; 0_{Mor(y,z)}circ f=-( 0_{Mor(y,z)}circ (-f))$. But I can't seem to progress any further.
Thanks in advance!
category-theory homological-algebra
asked Dec 31 '18 at 12:19
solgaleosolgaleo
8312
$endgroup$
add a comment |
$begingroup$
I am trying to prove whether in a pre-additive category, $0_{Mor(y,z)}circ f=0_{Mor(x,z)}$ for objects $x,y,z$ and $fin Mor(x,y)$. Now, by bi-linearity of composition maps $$ 0_{Mor(y,z)}circ f+ 0_{Mor(y,z)}circ (-f)= 0_{Mor(y,z)} circ 0_{Mor(x,y)} $$
Putting $f=0_{Mor(x,y)}; ;$ we have $; ; 0_{Mor(y,z)} circ 0_{Mor(x,y)} =0_{Mor(x,z)}; ; $and we also get $; ; 0_{Mor(y,z)}circ f=-( 0_{Mor(y,z)}circ (-f))$. But I can't seem to progress any further.
Thanks in advance!
category-theory homological-algebra
asked Dec 31 '18 at 12:19
solgaleosolgaleo
8312
$endgroup$
1
$begingroup$
Bilinearity of composition implies by definition $0circ f=0$ since the map $-circ f$ is a homomorphism of groups.
$endgroup$
– asdq
Dec 31 '18 at 13:15
add a comment |
$begingroup$
I am trying to prove whether in a pre-additive category, $0_{Mor(y,z)}circ f=0_{Mor(x,z)}$ for objects $x,y,z$ and $fin Mor(x,y)$. Now, by bi-linearity of composition maps $$ 0_{Mor(y,z)}circ f+ 0_{Mor(y,z)}circ (-f)= 0_{Mor(y,z)} circ 0_{Mor(x,y)} $$
Putting $f=0_{Mor(x,y)}; ;$ we have $; ; 0_{Mor(y,z)} circ 0_{Mor(x,y)} =0_{Mor(x,z)}; ; $and we also get $; ; 0_{Mor(y,z)}circ f=-( 0_{Mor(y,z)}circ (-f))$. But I can't seem to progress any further.
Thanks in advance!
category-theory homological-algebra
asked Dec 31 '18 at 12:19
solgaleosolgaleo
8312
$endgroup$
I am trying to prove whether in a pre-additive category, $0_{Mor(y,z)}circ f=0_{Mor(x,z)}$ for objects $x,y,z$ and $fin Mor(x,y)$. Now, by bi-linearity of composition maps $$ 0_{Mor(y,z)}circ f+ 0_{Mor(y,z)}circ (-f)= 0_{Mor(y,z)} circ 0_{Mor(x,y)} $$
Putting $f=0_{Mor(x,y)}; ;$ we have $; ; 0_{Mor(y,z)} circ 0_{Mor(x,y)} =0_{Mor(x,z)}; ; $and we also get $; ; 0_{Mor(y,z)}circ f=-( 0_{Mor(y,z)}circ (-f))$. But I can't seem to progress any further.
Thanks in advance!
category-theory homological-algebra
category-theory homological-algebra
asked Dec 31 '18 at 12:19
solgaleosolgaleo
8312
asked Dec 31 '18 at 12:19
solgaleosolgaleo
8312
asked Dec 31 '18 at 12:19
solgaleosolgaleo
8312
asked Dec 31 '18 at 12:19
solgaleosolgaleo
8312
asked Dec 31 '18 at 12:19
solgaleosolgaleo
8312
8312
1
$begingroup$
Bilinearity of composition implies by definition $0circ f=0$ since the map $-circ f$ is a homomorphism of groups.
$endgroup$
– asdq
Dec 31 '18 at 13:15
add a comment |
1
$begingroup$
Bilinearity of composition implies by definition $0circ f=0$ since the map $-circ f$ is a homomorphism of groups.
$endgroup$
– asdq
Dec 31 '18 at 13:15
1
1
$begingroup$
Bilinearity of composition implies by definition $0circ f=0$ since the map $-circ f$ is a homomorphism of groups.
$endgroup$
– asdq
Dec 31 '18 at 13:15
$begingroup$
Bilinearity of composition implies by definition $0circ f=0$ since the map $-circ f$ is a homomorphism of groups.
$endgroup$
– asdq
Dec 31 '18 at 13:15
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The answer to your question is basically in asdq's comment, allow me to expand a little.
By bilinearity you have that for $f in Mor(x,y)$ the mapping
$$
begin{align*}
-circ f colon Mor(y,z) &to Mor(x,z) \
x mapsto x circ f
end{align*}
$$
is a group homomorphism, in particular it sends $0_{Mor(y,z)}$ into $0_{Mor(x,z)}$, that is that $0_{Mor(y,z)}circ f = 0_{Mor(x,z)}$.
answered Dec 31 '18 at 22:01
Giorgio MossaGiorgio Mossa
14.2k11749
$endgroup$
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%2f3057655%2fvery-basic-question-about-pre-additive-category%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
$begingroup$
The answer to your question is basically in asdq's comment, allow me to expand a little.
By bilinearity you have that for $f in Mor(x,y)$ the mapping
$$
begin{align*}
-circ f colon Mor(y,z) &to Mor(x,z) \
x mapsto x circ f
end{align*}
$$
is a group homomorphism, in particular it sends $0_{Mor(y,z)}$ into $0_{Mor(x,z)}$, that is that $0_{Mor(y,z)}circ f = 0_{Mor(x,z)}$.
answered Dec 31 '18 at 22:01
Giorgio MossaGiorgio Mossa
14.2k11749
$endgroup$
add a comment |
$begingroup$
The answer to your question is basically in asdq's comment, allow me to expand a little.
By bilinearity you have that for $f in Mor(x,y)$ the mapping
$$
begin{align*}
-circ f colon Mor(y,z) &to Mor(x,z) \
x mapsto x circ f
end{align*}
$$
is a group homomorphism, in particular it sends $0_{Mor(y,z)}$ into $0_{Mor(x,z)}$, that is that $0_{Mor(y,z)}circ f = 0_{Mor(x,z)}$.
answered Dec 31 '18 at 22:01
Giorgio MossaGiorgio Mossa
14.2k11749
$endgroup$
add a comment |
$begingroup$
The answer to your question is basically in asdq's comment, allow me to expand a little.
By bilinearity you have that for $f in Mor(x,y)$ the mapping
$$
begin{align*}
-circ f colon Mor(y,z) &to Mor(x,z) \
x mapsto x circ f
end{align*}
$$
is a group homomorphism, in particular it sends $0_{Mor(y,z)}$ into $0_{Mor(x,z)}$, that is that $0_{Mor(y,z)}circ f = 0_{Mor(x,z)}$.
answered Dec 31 '18 at 22:01
Giorgio MossaGiorgio Mossa
14.2k11749
$endgroup$
The answer to your question is basically in asdq's comment, allow me to expand a little.
By bilinearity you have that for $f in Mor(x,y)$ the mapping
$$
begin{align*}
-circ f colon Mor(y,z) &to Mor(x,z) \
x mapsto x circ f
end{align*}
$$
is a group homomorphism, in particular it sends $0_{Mor(y,z)}$ into $0_{Mor(x,z)}$, that is that $0_{Mor(y,z)}circ f = 0_{Mor(x,z)}$.
answered Dec 31 '18 at 22:01
Giorgio MossaGiorgio Mossa
14.2k11749
answered Dec 31 '18 at 22:01
Giorgio MossaGiorgio Mossa
14.2k11749
answered Dec 31 '18 at 22:01
Giorgio MossaGiorgio Mossa
14.2k11749
answered Dec 31 '18 at 22:01
Giorgio MossaGiorgio Mossa
14.2k11749
14.2k11749
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.
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%2f3057655%2fvery-basic-question-about-pre-additive-category%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
1
$begingroup$
Bilinearity of composition implies by definition $0circ f=0$ since the map $-circ f$ is a homomorphism of groups.
$endgroup$
– asdq
Dec 31 '18 at 13:15