Mixing
Show composition mapping is continuous with compact-open topology
Let $X$ be a compact Hausdorff space, and $H(X)$ be the set of homeomorphisms from $X$ to $X$, with the compact-open topology.
Prove that the mapping $h:H(X)times H(X)rightarrow H(X)$, $h(f,g)=fcirc g$ is continuous.
Note, if $C(X,X)$ is the set of all continuous mappings from $X$ to $X$, the compact-open topology on $C(X,X)$ is generated by subsets of the form $B(K,U)={f:f(K)subset U}$ where $K$ is compact in $X$ and $U$ is open in $X$.
I honestly have no clue how to work with the compact open topology and would appreciate any hints.
Let's take $U$ open in $H(X)$. I want to show that $h^{-1}(U)$ is open in $H(X)times H(X)$.
I believe I am overthinking this, and apologize for the lack of work, I am just really confused how to show this. Any help would be much appreciated.
general-topology
edited Nov 20 '18 at 6:03
Henno Brandsma
105k346113
asked Nov 20 '18 at 0:34
The math god
1897
add a comment |
Let $X$ be a compact Hausdorff space, and $H(X)$ be the set of homeomorphisms from $X$ to $X$, with the compact-open topology.
Prove that the mapping $h:H(X)times H(X)rightarrow H(X)$, $h(f,g)=fcirc g$ is continuous.
Note, if $C(X,X)$ is the set of all continuous mappings from $X$ to $X$, the compact-open topology on $C(X,X)$ is generated by subsets of the form $B(K,U)={f:f(K)subset U}$ where $K$ is compact in $X$ and $U$ is open in $X$.
I honestly have no clue how to work with the compact open topology and would appreciate any hints.
Let's take $U$ open in $H(X)$. I want to show that $h^{-1}(U)$ is open in $H(X)times H(X)$.
I believe I am overthinking this, and apologize for the lack of work, I am just really confused how to show this. Any help would be much appreciated.
general-topology
edited Nov 20 '18 at 6:03
Henno Brandsma
105k346113
asked Nov 20 '18 at 0:34
The math god
1897
Two things. First, the function whose continuity you are proving is $h$, not $f$. Second, for proving continuity, you don't always have to prove that the inverse image of every open set is open; you can instead prove that the inverse image of every basis element is open.
– Lee Mosher
Nov 20 '18 at 0:39
add a comment |
Let $X$ be a compact Hausdorff space, and $H(X)$ be the set of homeomorphisms from $X$ to $X$, with the compact-open topology.
Prove that the mapping $h:H(X)times H(X)rightarrow H(X)$, $h(f,g)=fcirc g$ is continuous.
Note, if $C(X,X)$ is the set of all continuous mappings from $X$ to $X$, the compact-open topology on $C(X,X)$ is generated by subsets of the form $B(K,U)={f:f(K)subset U}$ where $K$ is compact in $X$ and $U$ is open in $X$.
I honestly have no clue how to work with the compact open topology and would appreciate any hints.
Let's take $U$ open in $H(X)$. I want to show that $h^{-1}(U)$ is open in $H(X)times H(X)$.
I believe I am overthinking this, and apologize for the lack of work, I am just really confused how to show this. Any help would be much appreciated.
general-topology
edited Nov 20 '18 at 6:03
Henno Brandsma
105k346113
asked Nov 20 '18 at 0:34
The math god
1897
Let $X$ be a compact Hausdorff space, and $H(X)$ be the set of homeomorphisms from $X$ to $X$, with the compact-open topology.
Prove that the mapping $h:H(X)times H(X)rightarrow H(X)$, $h(f,g)=fcirc g$ is continuous.
Note, if $C(X,X)$ is the set of all continuous mappings from $X$ to $X$, the compact-open topology on $C(X,X)$ is generated by subsets of the form $B(K,U)={f:f(K)subset U}$ where $K$ is compact in $X$ and $U$ is open in $X$.
I honestly have no clue how to work with the compact open topology and would appreciate any hints.
Let's take $U$ open in $H(X)$. I want to show that $h^{-1}(U)$ is open in $H(X)times H(X)$.
I believe I am overthinking this, and apologize for the lack of work, I am just really confused how to show this. Any help would be much appreciated.
general-topology
general-topology
edited Nov 20 '18 at 6:03
Henno Brandsma
105k346113
asked Nov 20 '18 at 0:34
The math god
1897
edited Nov 20 '18 at 6:03
Henno Brandsma
105k346113
asked Nov 20 '18 at 0:34
The math god
1897
edited Nov 20 '18 at 6:03
Henno Brandsma
105k346113
edited Nov 20 '18 at 6:03
Henno Brandsma
105k346113
edited Nov 20 '18 at 6:03
Henno Brandsma
105k346113
105k346113
asked Nov 20 '18 at 0:34
The math god
1897
asked Nov 20 '18 at 0:34
The math god
1897
asked Nov 20 '18 at 0:34
The math god
1897
1897
Two things. First, the function whose continuity you are proving is $h$, not $f$. Second, for proving continuity, you don't always have to prove that the inverse image of every open set is open; you can instead prove that the inverse image of every basis element is open.
– Lee Mosher
Nov 20 '18 at 0:39
add a comment |
Two things. First, the function whose continuity you are proving is $h$, not $f$. Second, for proving continuity, you don't always have to prove that the inverse image of every open set is open; you can instead prove that the inverse image of every basis element is open.
– Lee Mosher
Nov 20 '18 at 0:39
Two things. First, the function whose continuity you are proving is $h$, not $f$. Second, for proving continuity, you don't always have to prove that the inverse image of every open set is open; you can instead prove that the inverse image of every basis element is open.
– Lee Mosher
Nov 20 '18 at 0:39
Two things. First, the function whose continuity you are proving is $h$, not $f$. Second, for proving continuity, you don't always have to prove that the inverse image of every open set is open; you can instead prove that the inverse image of every basis element is open.
– Lee Mosher
Nov 20 '18 at 0:39
add a comment |
1 Answer
1
active
oldest
votes
We will use the characterization of continuity that says that a map $f:Xrightarrow Y$ is continuous if for all $xin X$ and open $Usubseteq Y$ such that $f(x)in U$ we have that there is an open $Vsubseteq X$ such that $xin V$ and $f(V)subseteq U$.
Let $B(K,U)subseteq H(X)$ be given where $Ksubseteq X$ is closed (compact) and $Usubseteq X$ is open. Then, if $gcirc fin B(K,U)$ we have that $gf(K)subseteq U$. Because $g$ is continuous we have that $g^{-1}(U)$ is open in $X$ can contains $f(K)$. Because $X$ is normal (recall that compact Hausdorff spaces are normal) there is an open $Vsubseteq X$ such that $f(K)subseteq Vsubseteqoverline{V}subseteq g^{-1}(U)$. Because $X$ is compact we have that $overline{V}$ is compact. Moreover it is clear that $g(overline{V})subseteq U$. We then claim the following:
$$(B(overline{V},U)circ B(K,V))subseteq B(K,U)$$
To see this we simply let $(k,l)in B(overline{V},U)times B(K,V)$. Then, by definition $l(K)subseteq V$ and $k(overline{V})subseteq U$. Then we can easily see that $(kcirc l)(K)subseteq U$. It is also clear that $gcirc fin B(overline{V},U)times B(K,V)$. Therefore composition is continuous.
Note: This result generalizes quite easily to the following situation as seen in an exercise of Munkres. Let $Y$ be locally compact Hausdorff, and $X$ and $Z$ general spaces. Also let $mathcal{C}(X,Y),,mathcal{C}(Y,Z),$ and $mathcal{C}(X,Z)$ denote the spaces of continuous functions from the respective spaces with the compact open topology. Then the composition map
$$mathcal{C}(X,Y)timesmathcal{C}(Y,Z)rightarrowmathcal{C}(X,Z)$$
is continuous.
answered Nov 20 '18 at 0:49
Robert Thingum
7641316
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%2f3005754%2fshow-composition-mapping-is-continuous-with-compact-open-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
We will use the characterization of continuity that says that a map $f:Xrightarrow Y$ is continuous if for all $xin X$ and open $Usubseteq Y$ such that $f(x)in U$ we have that there is an open $Vsubseteq X$ such that $xin V$ and $f(V)subseteq U$.
Let $B(K,U)subseteq H(X)$ be given where $Ksubseteq X$ is closed (compact) and $Usubseteq X$ is open. Then, if $gcirc fin B(K,U)$ we have that $gf(K)subseteq U$. Because $g$ is continuous we have that $g^{-1}(U)$ is open in $X$ can contains $f(K)$. Because $X$ is normal (recall that compact Hausdorff spaces are normal) there is an open $Vsubseteq X$ such that $f(K)subseteq Vsubseteqoverline{V}subseteq g^{-1}(U)$. Because $X$ is compact we have that $overline{V}$ is compact. Moreover it is clear that $g(overline{V})subseteq U$. We then claim the following:
$$(B(overline{V},U)circ B(K,V))subseteq B(K,U)$$
To see this we simply let $(k,l)in B(overline{V},U)times B(K,V)$. Then, by definition $l(K)subseteq V$ and $k(overline{V})subseteq U$. Then we can easily see that $(kcirc l)(K)subseteq U$. It is also clear that $gcirc fin B(overline{V},U)times B(K,V)$. Therefore composition is continuous.
Note: This result generalizes quite easily to the following situation as seen in an exercise of Munkres. Let $Y$ be locally compact Hausdorff, and $X$ and $Z$ general spaces. Also let $mathcal{C}(X,Y),,mathcal{C}(Y,Z),$ and $mathcal{C}(X,Z)$ denote the spaces of continuous functions from the respective spaces with the compact open topology. Then the composition map
$$mathcal{C}(X,Y)timesmathcal{C}(Y,Z)rightarrowmathcal{C}(X,Z)$$
is continuous.
answered Nov 20 '18 at 0:49
Robert Thingum
7641316
add a comment |
We will use the characterization of continuity that says that a map $f:Xrightarrow Y$ is continuous if for all $xin X$ and open $Usubseteq Y$ such that $f(x)in U$ we have that there is an open $Vsubseteq X$ such that $xin V$ and $f(V)subseteq U$.
Let $B(K,U)subseteq H(X)$ be given where $Ksubseteq X$ is closed (compact) and $Usubseteq X$ is open. Then, if $gcirc fin B(K,U)$ we have that $gf(K)subseteq U$. Because $g$ is continuous we have that $g^{-1}(U)$ is open in $X$ can contains $f(K)$. Because $X$ is normal (recall that compact Hausdorff spaces are normal) there is an open $Vsubseteq X$ such that $f(K)subseteq Vsubseteqoverline{V}subseteq g^{-1}(U)$. Because $X$ is compact we have that $overline{V}$ is compact. Moreover it is clear that $g(overline{V})subseteq U$. We then claim the following:
$$(B(overline{V},U)circ B(K,V))subseteq B(K,U)$$
To see this we simply let $(k,l)in B(overline{V},U)times B(K,V)$. Then, by definition $l(K)subseteq V$ and $k(overline{V})subseteq U$. Then we can easily see that $(kcirc l)(K)subseteq U$. It is also clear that $gcirc fin B(overline{V},U)times B(K,V)$. Therefore composition is continuous.
Note: This result generalizes quite easily to the following situation as seen in an exercise of Munkres. Let $Y$ be locally compact Hausdorff, and $X$ and $Z$ general spaces. Also let $mathcal{C}(X,Y),,mathcal{C}(Y,Z),$ and $mathcal{C}(X,Z)$ denote the spaces of continuous functions from the respective spaces with the compact open topology. Then the composition map
$$mathcal{C}(X,Y)timesmathcal{C}(Y,Z)rightarrowmathcal{C}(X,Z)$$
is continuous.
answered Nov 20 '18 at 0:49
Robert Thingum
7641316
add a comment |
We will use the characterization of continuity that says that a map $f:Xrightarrow Y$ is continuous if for all $xin X$ and open $Usubseteq Y$ such that $f(x)in U$ we have that there is an open $Vsubseteq X$ such that $xin V$ and $f(V)subseteq U$.
Let $B(K,U)subseteq H(X)$ be given where $Ksubseteq X$ is closed (compact) and $Usubseteq X$ is open. Then, if $gcirc fin B(K,U)$ we have that $gf(K)subseteq U$. Because $g$ is continuous we have that $g^{-1}(U)$ is open in $X$ can contains $f(K)$. Because $X$ is normal (recall that compact Hausdorff spaces are normal) there is an open $Vsubseteq X$ such that $f(K)subseteq Vsubseteqoverline{V}subseteq g^{-1}(U)$. Because $X$ is compact we have that $overline{V}$ is compact. Moreover it is clear that $g(overline{V})subseteq U$. We then claim the following:
$$(B(overline{V},U)circ B(K,V))subseteq B(K,U)$$
To see this we simply let $(k,l)in B(overline{V},U)times B(K,V)$. Then, by definition $l(K)subseteq V$ and $k(overline{V})subseteq U$. Then we can easily see that $(kcirc l)(K)subseteq U$. It is also clear that $gcirc fin B(overline{V},U)times B(K,V)$. Therefore composition is continuous.
Note: This result generalizes quite easily to the following situation as seen in an exercise of Munkres. Let $Y$ be locally compact Hausdorff, and $X$ and $Z$ general spaces. Also let $mathcal{C}(X,Y),,mathcal{C}(Y,Z),$ and $mathcal{C}(X,Z)$ denote the spaces of continuous functions from the respective spaces with the compact open topology. Then the composition map
$$mathcal{C}(X,Y)timesmathcal{C}(Y,Z)rightarrowmathcal{C}(X,Z)$$
is continuous.
answered Nov 20 '18 at 0:49
Robert Thingum
7641316
We will use the characterization of continuity that says that a map $f:Xrightarrow Y$ is continuous if for all $xin X$ and open $Usubseteq Y$ such that $f(x)in U$ we have that there is an open $Vsubseteq X$ such that $xin V$ and $f(V)subseteq U$.
Let $B(K,U)subseteq H(X)$ be given where $Ksubseteq X$ is closed (compact) and $Usubseteq X$ is open. Then, if $gcirc fin B(K,U)$ we have that $gf(K)subseteq U$. Because $g$ is continuous we have that $g^{-1}(U)$ is open in $X$ can contains $f(K)$. Because $X$ is normal (recall that compact Hausdorff spaces are normal) there is an open $Vsubseteq X$ such that $f(K)subseteq Vsubseteqoverline{V}subseteq g^{-1}(U)$. Because $X$ is compact we have that $overline{V}$ is compact. Moreover it is clear that $g(overline{V})subseteq U$. We then claim the following:
$$(B(overline{V},U)circ B(K,V))subseteq B(K,U)$$
To see this we simply let $(k,l)in B(overline{V},U)times B(K,V)$. Then, by definition $l(K)subseteq V$ and $k(overline{V})subseteq U$. Then we can easily see that $(kcirc l)(K)subseteq U$. It is also clear that $gcirc fin B(overline{V},U)times B(K,V)$. Therefore composition is continuous.
Note: This result generalizes quite easily to the following situation as seen in an exercise of Munkres. Let $Y$ be locally compact Hausdorff, and $X$ and $Z$ general spaces. Also let $mathcal{C}(X,Y),,mathcal{C}(Y,Z),$ and $mathcal{C}(X,Z)$ denote the spaces of continuous functions from the respective spaces with the compact open topology. Then the composition map
$$mathcal{C}(X,Y)timesmathcal{C}(Y,Z)rightarrowmathcal{C}(X,Z)$$
is continuous.
answered Nov 20 '18 at 0:49
Robert Thingum
7641316
answered Nov 20 '18 at 0:49
Robert Thingum
7641316
answered Nov 20 '18 at 0:49
Robert Thingum
7641316
answered Nov 20 '18 at 0:49
Robert Thingum
7641316
7641316
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.
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%2f3005754%2fshow-composition-mapping-is-continuous-with-compact-open-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
Two things. First, the function whose continuity you are proving is $h$, not $f$. Second, for proving continuity, you don't always have to prove that the inverse image of every open set is open; you can instead prove that the inverse image of every basis element is open.
– Lee Mosher
Nov 20 '18 at 0:39