Show composition mapping is continuous with compact-open topology
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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 at 6:03
Henno Brandsma
103k345112
asked Nov 20 at 0:34
The math god
1897
add a comment |
up vote
1
down vote
favorite
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 at 6:03
Henno Brandsma
103k345112
asked Nov 20 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 at 0:39
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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 at 6:03
Henno Brandsma
103k345112
asked Nov 20 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 at 6:03
Henno Brandsma
103k345112
asked Nov 20 at 0:34
The math god
1897
edited Nov 20 at 6:03
Henno Brandsma
103k345112
asked Nov 20 at 0:34
The math god
1897
edited Nov 20 at 6:03
Henno Brandsma
103k345112
edited Nov 20 at 6:03
Henno Brandsma
103k345112
edited Nov 20 at 6:03
Henno Brandsma
103k345112
103k345112
asked Nov 20 at 0:34
The math god
1897
asked Nov 20 at 0:34
The math god
1897
asked Nov 20 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 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 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 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 at 0:39
add a comment |
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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 at 0:49
Robert Thingum
7141316
add a comment |
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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 at 0:49
Robert Thingum
7141316
add a comment |
up vote
0
down vote
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 at 0:49
Robert Thingum
7141316
add a comment |
up vote
0
down vote
up vote
0
down vote
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 at 0:49
Robert Thingum
7141316
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 at 0:49
Robert Thingum
7141316
answered Nov 20 at 0:49
Robert Thingum
7141316
answered Nov 20 at 0:49
Robert Thingum
7141316
answered Nov 20 at 0:49
Robert Thingum
7141316
7141316
add a comment |
add a comment |
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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 at 0:39