Preimage of a set in the product topology
$begingroup$
Claim: When $X ×Y$ is endowed with the product topology $T_{X×Y}$ , the projection
maps $p_X : X × Y → X , p_X(x, y) = x$, and $p_Y : X × Y → Y , p_Y (x, y) = y$ , are continuous.
Proof: Indeed for any open set $U$ in $X$ , $p^{−1}_X (U) = U × Y$
This is a claim from my lecture notes. However, the projection map is clearly not injective. How can we define the preimage of a non-injective map? Clearly, $p_X(x,y_1)=x=p_X(x, y_2)$ for any $y_1neq y_2$
general-topology projection
asked Dec 5 '18 at 9:37
asdfasdf
3,691519
$endgroup$
add a comment |
$begingroup$
Claim: When $X ×Y$ is endowed with the product topology $T_{X×Y}$ , the projection
maps $p_X : X × Y → X , p_X(x, y) = x$, and $p_Y : X × Y → Y , p_Y (x, y) = y$ , are continuous.
Proof: Indeed for any open set $U$ in $X$ , $p^{−1}_X (U) = U × Y$
This is a claim from my lecture notes. However, the projection map is clearly not injective. How can we define the preimage of a non-injective map? Clearly, $p_X(x,y_1)=x=p_X(x, y_2)$ for any $y_1neq y_2$
general-topology projection
asked Dec 5 '18 at 9:37
asdfasdf
3,691519
$endgroup$
add a comment |
$begingroup$
Claim: When $X ×Y$ is endowed with the product topology $T_{X×Y}$ , the projection
maps $p_X : X × Y → X , p_X(x, y) = x$, and $p_Y : X × Y → Y , p_Y (x, y) = y$ , are continuous.
Proof: Indeed for any open set $U$ in $X$ , $p^{−1}_X (U) = U × Y$
This is a claim from my lecture notes. However, the projection map is clearly not injective. How can we define the preimage of a non-injective map? Clearly, $p_X(x,y_1)=x=p_X(x, y_2)$ for any $y_1neq y_2$
general-topology projection
asked Dec 5 '18 at 9:37
asdfasdf
3,691519
$endgroup$
Claim: When $X ×Y$ is endowed with the product topology $T_{X×Y}$ , the projection
maps $p_X : X × Y → X , p_X(x, y) = x$, and $p_Y : X × Y → Y , p_Y (x, y) = y$ , are continuous.
Proof: Indeed for any open set $U$ in $X$ , $p^{−1}_X (U) = U × Y$
This is a claim from my lecture notes. However, the projection map is clearly not injective. How can we define the preimage of a non-injective map? Clearly, $p_X(x,y_1)=x=p_X(x, y_2)$ for any $y_1neq y_2$
general-topology projection
general-topology projection
asked Dec 5 '18 at 9:37
asdfasdf
3,691519
asked Dec 5 '18 at 9:37
asdfasdf
3,691519
asked Dec 5 '18 at 9:37
asdfasdf
3,691519
asked Dec 5 '18 at 9:37
asdfasdf
3,691519
asked Dec 5 '18 at 9:37
asdfasdf
3,691519
3,691519
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For any function $f$ an any set $A$ in the range the notation $f^{-1}(A)$ stands for all points $x$ in the domain of $f$ such that $f(x)in A$. If $f$ happens to have an inverse then $f^{-1}(A)$ becomes the image of $A$ under $f^{-1}$.
answered Dec 5 '18 at 9:39
Kavi Rama MurthyKavi Rama Murthy
57.7k42160
$endgroup$
$begingroup$
I see. So we are defining a map $f^{-1}:X rightarrow X$ x $Y$ by $f^{-1}(t)=t$ x $Y$ rather than taking the inverse of $f$, is that correct? Will mark the answer as correct when the website allows me to!
$endgroup$
– asdf
Dec 5 '18 at 9:42
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%2f3026873%2fpreimage-of-a-set-in-the-product-topology%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
$begingroup$
For any function $f$ an any set $A$ in the range the notation $f^{-1}(A)$ stands for all points $x$ in the domain of $f$ such that $f(x)in A$. If $f$ happens to have an inverse then $f^{-1}(A)$ becomes the image of $A$ under $f^{-1}$.
answered Dec 5 '18 at 9:39
Kavi Rama MurthyKavi Rama Murthy
57.7k42160
$endgroup$
$begingroup$
I see. So we are defining a map $f^{-1}:X rightarrow X$ x $Y$ by $f^{-1}(t)=t$ x $Y$ rather than taking the inverse of $f$, is that correct? Will mark the answer as correct when the website allows me to!
$endgroup$
– asdf
Dec 5 '18 at 9:42
add a comment |
$begingroup$
For any function $f$ an any set $A$ in the range the notation $f^{-1}(A)$ stands for all points $x$ in the domain of $f$ such that $f(x)in A$. If $f$ happens to have an inverse then $f^{-1}(A)$ becomes the image of $A$ under $f^{-1}$.
answered Dec 5 '18 at 9:39
Kavi Rama MurthyKavi Rama Murthy
57.7k42160
$endgroup$
$begingroup$
I see. So we are defining a map $f^{-1}:X rightarrow X$ x $Y$ by $f^{-1}(t)=t$ x $Y$ rather than taking the inverse of $f$, is that correct? Will mark the answer as correct when the website allows me to!
$endgroup$
– asdf
Dec 5 '18 at 9:42
add a comment |
$begingroup$
For any function $f$ an any set $A$ in the range the notation $f^{-1}(A)$ stands for all points $x$ in the domain of $f$ such that $f(x)in A$. If $f$ happens to have an inverse then $f^{-1}(A)$ becomes the image of $A$ under $f^{-1}$.
answered Dec 5 '18 at 9:39
Kavi Rama MurthyKavi Rama Murthy
57.7k42160
$endgroup$
For any function $f$ an any set $A$ in the range the notation $f^{-1}(A)$ stands for all points $x$ in the domain of $f$ such that $f(x)in A$. If $f$ happens to have an inverse then $f^{-1}(A)$ becomes the image of $A$ under $f^{-1}$.
answered Dec 5 '18 at 9:39
Kavi Rama MurthyKavi Rama Murthy
57.7k42160
answered Dec 5 '18 at 9:39
Kavi Rama MurthyKavi Rama Murthy
57.7k42160
answered Dec 5 '18 at 9:39
Kavi Rama MurthyKavi Rama Murthy
57.7k42160
answered Dec 5 '18 at 9:39
Kavi Rama MurthyKavi Rama Murthy
57.7k42160
57.7k42160
$begingroup$
I see. So we are defining a map $f^{-1}:X rightarrow X$ x $Y$ by $f^{-1}(t)=t$ x $Y$ rather than taking the inverse of $f$, is that correct? Will mark the answer as correct when the website allows me to!
$endgroup$
– asdf
Dec 5 '18 at 9:42
add a comment |
$begingroup$
I see. So we are defining a map $f^{-1}:X rightarrow X$ x $Y$ by $f^{-1}(t)=t$ x $Y$ rather than taking the inverse of $f$, is that correct? Will mark the answer as correct when the website allows me to!
$endgroup$
– asdf
Dec 5 '18 at 9:42
$begingroup$
I see. So we are defining a map $f^{-1}:X rightarrow X$ x $Y$ by $f^{-1}(t)=t$ x $Y$ rather than taking the inverse of $f$, is that correct? Will mark the answer as correct when the website allows me to!
$endgroup$
– asdf
Dec 5 '18 at 9:42
$begingroup$
I see. So we are defining a map $f^{-1}:X rightarrow X$ x $Y$ by $f^{-1}(t)=t$ x $Y$ rather than taking the inverse of $f$, is that correct? Will mark the answer as correct when the website allows me to!
$endgroup$
– asdf
Dec 5 '18 at 9:42
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%2f3026873%2fpreimage-of-a-set-in-the-product-topology%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
