Help with product maps in topological spaces
$begingroup$
Let $A, B, X, Y$ be topological spaces.
Given two functions $f : A to B$ and $g : X to Y $, let
$f times g : A × X to B times Y$, $ (f times g)(a, x) = (f(a), g(x))$.
can you help me to show that if f and g is cont. then fxg is too
and if the sets A,B is non-empty, then it holds the other way around too.
I'm pretty stuck.. can you please help me?
My solution
To show the statement, I will use Theorem 4.1. If I can show that $ftimes g(overline{Stimes T})subseteq overline{ftimes g(Stimes T)}$ for a subset $Stimes Tsubseteq Atimes X$.
Take a $(b,y)in ftimes g(overline{Stimes T})$. Then there exists a $sinoverline S$ and $tinoverline T$ such that $f(s)=b$ and $g(t)=y$.
We know that both $f$ and $g$ is continuous, so from 4.1 we get $f(s)inoverline B$ and $g(t)inoverline Y$, and then $(b,y)in overline{ftimes g (Stimes T)}$. And from 4.1 we get that $ftimes g$ is continuous.
general-topology continuity product-space
asked Dec 3 '18 at 12:02
Esteban CambiassoEsteban Cambiasso
103
$endgroup$
|
show 7 more comments
$begingroup$
Let $A, B, X, Y$ be topological spaces.
Given two functions $f : A to B$ and $g : X to Y $, let
$f times g : A × X to B times Y$, $ (f times g)(a, x) = (f(a), g(x))$.
can you help me to show that if f and g is cont. then fxg is too
and if the sets A,B is non-empty, then it holds the other way around too.
I'm pretty stuck.. can you please help me?
My solution
To show the statement, I will use Theorem 4.1. If I can show that $ftimes g(overline{Stimes T})subseteq overline{ftimes g(Stimes T)}$ for a subset $Stimes Tsubseteq Atimes X$.
Take a $(b,y)in ftimes g(overline{Stimes T})$. Then there exists a $sinoverline S$ and $tinoverline T$ such that $f(s)=b$ and $g(t)=y$.
We know that both $f$ and $g$ is continuous, so from 4.1 we get $f(s)inoverline B$ and $g(t)inoverline Y$, and then $(b,y)in overline{ftimes g (Stimes T)}$. And from 4.1 we get that $ftimes g$ is continuous.
general-topology continuity product-space
asked Dec 3 '18 at 12:02
Esteban CambiassoEsteban Cambiasso
103
$endgroup$
$begingroup$
What have you tried?
$endgroup$
– Mees de Vries
Dec 3 '18 at 12:05
$begingroup$
hi Mees. I have just added my solution to (a). Could u maybe give your opinion on that? My theoreme 4.1 says that f is continuous iff. f(closure(A)) is in Closure(f(A))
$endgroup$
– Esteban Cambiasso
Dec 3 '18 at 12:49
$begingroup$
I just saw that i miss a bar over f×g(SxT) in the last part
$endgroup$
– Esteban Cambiasso
Dec 3 '18 at 13:39
$begingroup$
For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. To help you get started, I have retyped the text from your picture.
$endgroup$
– Martin Sleziak
Dec 3 '18 at 16:21
$begingroup$
thank u. Can anybody verify my answer?
$endgroup$
– Esteban Cambiasso
Dec 3 '18 at 16:27
|
show 7 more comments
$begingroup$
Let $A, B, X, Y$ be topological spaces.
Given two functions $f : A to B$ and $g : X to Y $, let
$f times g : A × X to B times Y$, $ (f times g)(a, x) = (f(a), g(x))$.
can you help me to show that if f and g is cont. then fxg is too
and if the sets A,B is non-empty, then it holds the other way around too.
I'm pretty stuck.. can you please help me?
My solution
To show the statement, I will use Theorem 4.1. If I can show that $ftimes g(overline{Stimes T})subseteq overline{ftimes g(Stimes T)}$ for a subset $Stimes Tsubseteq Atimes X$.
Take a $(b,y)in ftimes g(overline{Stimes T})$. Then there exists a $sinoverline S$ and $tinoverline T$ such that $f(s)=b$ and $g(t)=y$.
We know that both $f$ and $g$ is continuous, so from 4.1 we get $f(s)inoverline B$ and $g(t)inoverline Y$, and then $(b,y)in overline{ftimes g (Stimes T)}$. And from 4.1 we get that $ftimes g$ is continuous.
general-topology continuity product-space
asked Dec 3 '18 at 12:02
Esteban CambiassoEsteban Cambiasso
103
$endgroup$
Let $A, B, X, Y$ be topological spaces.
Given two functions $f : A to B$ and $g : X to Y $, let
$f times g : A × X to B times Y$, $ (f times g)(a, x) = (f(a), g(x))$.
can you help me to show that if f and g is cont. then fxg is too
and if the sets A,B is non-empty, then it holds the other way around too.
I'm pretty stuck.. can you please help me?
My solution
To show the statement, I will use Theorem 4.1. If I can show that $ftimes g(overline{Stimes T})subseteq overline{ftimes g(Stimes T)}$ for a subset $Stimes Tsubseteq Atimes X$.
Take a $(b,y)in ftimes g(overline{Stimes T})$. Then there exists a $sinoverline S$ and $tinoverline T$ such that $f(s)=b$ and $g(t)=y$.
We know that both $f$ and $g$ is continuous, so from 4.1 we get $f(s)inoverline B$ and $g(t)inoverline Y$, and then $(b,y)in overline{ftimes g (Stimes T)}$. And from 4.1 we get that $ftimes g$ is continuous.
general-topology continuity product-space
general-topology continuity product-space
asked Dec 3 '18 at 12:02
Esteban CambiassoEsteban Cambiasso
103
asked Dec 3 '18 at 12:02
Esteban CambiassoEsteban Cambiasso
103
edited Dec 6 '18 at 8:56
Esteban Cambiasso
asked Dec 3 '18 at 12:02
Esteban CambiassoEsteban Cambiasso
103
asked Dec 3 '18 at 12:02
Esteban CambiassoEsteban Cambiasso
103
asked Dec 3 '18 at 12:02
Esteban CambiassoEsteban Cambiasso
103
103
$begingroup$
What have you tried?
$endgroup$
– Mees de Vries
Dec 3 '18 at 12:05
$begingroup$
hi Mees. I have just added my solution to (a). Could u maybe give your opinion on that? My theoreme 4.1 says that f is continuous iff. f(closure(A)) is in Closure(f(A))
$endgroup$
– Esteban Cambiasso
Dec 3 '18 at 12:49
$begingroup$
I just saw that i miss a bar over f×g(SxT) in the last part
$endgroup$
– Esteban Cambiasso
Dec 3 '18 at 13:39
$begingroup$
For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. To help you get started, I have retyped the text from your picture.
$endgroup$
– Martin Sleziak
Dec 3 '18 at 16:21
$begingroup$
thank u. Can anybody verify my answer?
$endgroup$
– Esteban Cambiasso
Dec 3 '18 at 16:27
|
show 7 more comments
$begingroup$
What have you tried?
$endgroup$
– Mees de Vries
Dec 3 '18 at 12:05
$begingroup$
hi Mees. I have just added my solution to (a). Could u maybe give your opinion on that? My theoreme 4.1 says that f is continuous iff. f(closure(A)) is in Closure(f(A))
$endgroup$
– Esteban Cambiasso
Dec 3 '18 at 12:49
$begingroup$
I just saw that i miss a bar over f×g(SxT) in the last part
$endgroup$
– Esteban Cambiasso
Dec 3 '18 at 13:39
$begingroup$
For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. To help you get started, I have retyped the text from your picture.
$endgroup$
– Martin Sleziak
Dec 3 '18 at 16:21
$begingroup$
thank u. Can anybody verify my answer?
$endgroup$
– Esteban Cambiasso
Dec 3 '18 at 16:27
$begingroup$
What have you tried?
$endgroup$
– Mees de Vries
Dec 3 '18 at 12:05
$begingroup$
What have you tried?
$endgroup$
– Mees de Vries
Dec 3 '18 at 12:05
$begingroup$
hi Mees. I have just added my solution to (a). Could u maybe give your opinion on that? My theoreme 4.1 says that f is continuous iff. f(closure(A)) is in Closure(f(A))
$endgroup$
– Esteban Cambiasso
Dec 3 '18 at 12:49
$begingroup$
hi Mees. I have just added my solution to (a). Could u maybe give your opinion on that? My theoreme 4.1 says that f is continuous iff. f(closure(A)) is in Closure(f(A))
$endgroup$
– Esteban Cambiasso
Dec 3 '18 at 12:49
$begingroup$
I just saw that i miss a bar over f×g(SxT) in the last part
$endgroup$
– Esteban Cambiasso
Dec 3 '18 at 13:39
$begingroup$
I just saw that i miss a bar over f×g(SxT) in the last part
$endgroup$
– Esteban Cambiasso
Dec 3 '18 at 13:39
$begingroup$
For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. To help you get started, I have retyped the text from your picture.
$endgroup$
– Martin Sleziak
Dec 3 '18 at 16:21
$begingroup$
For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. To help you get started, I have retyped the text from your picture.
$endgroup$
– Martin Sleziak
Dec 3 '18 at 16:21
$begingroup$
thank u. Can anybody verify my answer?
$endgroup$
– Esteban Cambiasso
Dec 3 '18 at 16:27
$begingroup$
thank u. Can anybody verify my answer?
$endgroup$
– Esteban Cambiasso
Dec 3 '18 at 16:27
|
show 7 more comments
1 Answer
1
active
oldest
votes
$begingroup$
By the universal property of products, a function : $F: Z to X times Y$ is continuous iff $pi_X circ F$ and $pi_Y circ F$ (the compositions with the continuous projections) are all (both) continuous.
This allows us to quickly solve the problem, applied to the product $B times Y$ and $Z = A times X$: $$pi_B circ (f times g) = fcirc pi_A$$ where the right hand side is continuous as a composition of continuous maps. The identity can be seen by evaluating both sides on an arbitary point $(a,x)$ and both expressions become $f(a)$.
Also $$pi_Y circ (f times g) = gcirc pi_X$$ and so both compositions with projections of $f times g$ are continuous and so $f times g$ is continuous.
On the other hand suppose that $f times g$ is continuous, and fix $x_0 in X$ and
define $i: A to A times X$ by $i(a) = (a,x_0)$. As $pi_A circ i$ is the identity on $A$ (hence continuous) and $pi_X circ i$ is a constant map with the value $x_0$ (also continuous) the trusted universal property tells us that $i$ is continuous, and then note that $f = pi_B circ (f times g) circ i$ and then $f$ can be written as the continuous composition of continuous maps.
That $g$ is also continuous can be seen with the same idea and another embedding (now using a fixed $a_0 in A$) from $X$ into $A times X$.
answered Dec 3 '18 at 17:51
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
$begingroup$
okay thanks for your time. Have you looked at (b)? For me, that was the hardest one
$endgroup$
– Esteban Cambiasso
Dec 4 '18 at 8:59
$begingroup$
@EstebanCambiasso it’s already part of the solution after “ on the other hand”.
$endgroup$
– Henno Brandsma
Dec 4 '18 at 9:13
add a comment |
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$begingroup$
By the universal property of products, a function : $F: Z to X times Y$ is continuous iff $pi_X circ F$ and $pi_Y circ F$ (the compositions with the continuous projections) are all (both) continuous.
This allows us to quickly solve the problem, applied to the product $B times Y$ and $Z = A times X$: $$pi_B circ (f times g) = fcirc pi_A$$ where the right hand side is continuous as a composition of continuous maps. The identity can be seen by evaluating both sides on an arbitary point $(a,x)$ and both expressions become $f(a)$.
Also $$pi_Y circ (f times g) = gcirc pi_X$$ and so both compositions with projections of $f times g$ are continuous and so $f times g$ is continuous.
On the other hand suppose that $f times g$ is continuous, and fix $x_0 in X$ and
define $i: A to A times X$ by $i(a) = (a,x_0)$. As $pi_A circ i$ is the identity on $A$ (hence continuous) and $pi_X circ i$ is a constant map with the value $x_0$ (also continuous) the trusted universal property tells us that $i$ is continuous, and then note that $f = pi_B circ (f times g) circ i$ and then $f$ can be written as the continuous composition of continuous maps.
That $g$ is also continuous can be seen with the same idea and another embedding (now using a fixed $a_0 in A$) from $X$ into $A times X$.
answered Dec 3 '18 at 17:51
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
$begingroup$
okay thanks for your time. Have you looked at (b)? For me, that was the hardest one
$endgroup$
– Esteban Cambiasso
Dec 4 '18 at 8:59
$begingroup$
@EstebanCambiasso it’s already part of the solution after “ on the other hand”.
$endgroup$
– Henno Brandsma
Dec 4 '18 at 9:13
add a comment |
$begingroup$
By the universal property of products, a function : $F: Z to X times Y$ is continuous iff $pi_X circ F$ and $pi_Y circ F$ (the compositions with the continuous projections) are all (both) continuous.
This allows us to quickly solve the problem, applied to the product $B times Y$ and $Z = A times X$: $$pi_B circ (f times g) = fcirc pi_A$$ where the right hand side is continuous as a composition of continuous maps. The identity can be seen by evaluating both sides on an arbitary point $(a,x)$ and both expressions become $f(a)$.
Also $$pi_Y circ (f times g) = gcirc pi_X$$ and so both compositions with projections of $f times g$ are continuous and so $f times g$ is continuous.
On the other hand suppose that $f times g$ is continuous, and fix $x_0 in X$ and
define $i: A to A times X$ by $i(a) = (a,x_0)$. As $pi_A circ i$ is the identity on $A$ (hence continuous) and $pi_X circ i$ is a constant map with the value $x_0$ (also continuous) the trusted universal property tells us that $i$ is continuous, and then note that $f = pi_B circ (f times g) circ i$ and then $f$ can be written as the continuous composition of continuous maps.
That $g$ is also continuous can be seen with the same idea and another embedding (now using a fixed $a_0 in A$) from $X$ into $A times X$.
answered Dec 3 '18 at 17:51
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
$begingroup$
okay thanks for your time. Have you looked at (b)? For me, that was the hardest one
$endgroup$
– Esteban Cambiasso
Dec 4 '18 at 8:59
$begingroup$
@EstebanCambiasso it’s already part of the solution after “ on the other hand”.
$endgroup$
– Henno Brandsma
Dec 4 '18 at 9:13
add a comment |
$begingroup$
By the universal property of products, a function : $F: Z to X times Y$ is continuous iff $pi_X circ F$ and $pi_Y circ F$ (the compositions with the continuous projections) are all (both) continuous.
This allows us to quickly solve the problem, applied to the product $B times Y$ and $Z = A times X$: $$pi_B circ (f times g) = fcirc pi_A$$ where the right hand side is continuous as a composition of continuous maps. The identity can be seen by evaluating both sides on an arbitary point $(a,x)$ and both expressions become $f(a)$.
Also $$pi_Y circ (f times g) = gcirc pi_X$$ and so both compositions with projections of $f times g$ are continuous and so $f times g$ is continuous.
On the other hand suppose that $f times g$ is continuous, and fix $x_0 in X$ and
define $i: A to A times X$ by $i(a) = (a,x_0)$. As $pi_A circ i$ is the identity on $A$ (hence continuous) and $pi_X circ i$ is a constant map with the value $x_0$ (also continuous) the trusted universal property tells us that $i$ is continuous, and then note that $f = pi_B circ (f times g) circ i$ and then $f$ can be written as the continuous composition of continuous maps.
That $g$ is also continuous can be seen with the same idea and another embedding (now using a fixed $a_0 in A$) from $X$ into $A times X$.
answered Dec 3 '18 at 17:51
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
By the universal property of products, a function : $F: Z to X times Y$ is continuous iff $pi_X circ F$ and $pi_Y circ F$ (the compositions with the continuous projections) are all (both) continuous.
This allows us to quickly solve the problem, applied to the product $B times Y$ and $Z = A times X$: $$pi_B circ (f times g) = fcirc pi_A$$ where the right hand side is continuous as a composition of continuous maps. The identity can be seen by evaluating both sides on an arbitary point $(a,x)$ and both expressions become $f(a)$.
Also $$pi_Y circ (f times g) = gcirc pi_X$$ and so both compositions with projections of $f times g$ are continuous and so $f times g$ is continuous.
On the other hand suppose that $f times g$ is continuous, and fix $x_0 in X$ and
define $i: A to A times X$ by $i(a) = (a,x_0)$. As $pi_A circ i$ is the identity on $A$ (hence continuous) and $pi_X circ i$ is a constant map with the value $x_0$ (also continuous) the trusted universal property tells us that $i$ is continuous, and then note that $f = pi_B circ (f times g) circ i$ and then $f$ can be written as the continuous composition of continuous maps.
That $g$ is also continuous can be seen with the same idea and another embedding (now using a fixed $a_0 in A$) from $X$ into $A times X$.
answered Dec 3 '18 at 17:51
Henno BrandsmaHenno Brandsma
106k347114
edited Dec 3 '18 at 22:04
answered Dec 3 '18 at 17:51
Henno BrandsmaHenno Brandsma
106k347114
answered Dec 3 '18 at 17:51
Henno BrandsmaHenno Brandsma
106k347114
answered Dec 3 '18 at 17:51
Henno BrandsmaHenno Brandsma
106k347114
106k347114
$begingroup$
okay thanks for your time. Have you looked at (b)? For me, that was the hardest one
$endgroup$
– Esteban Cambiasso
Dec 4 '18 at 8:59
$begingroup$
@EstebanCambiasso it’s already part of the solution after “ on the other hand”.
$endgroup$
– Henno Brandsma
Dec 4 '18 at 9:13
add a comment |
$begingroup$
okay thanks for your time. Have you looked at (b)? For me, that was the hardest one
$endgroup$
– Esteban Cambiasso
Dec 4 '18 at 8:59
$begingroup$
@EstebanCambiasso it’s already part of the solution after “ on the other hand”.
$endgroup$
– Henno Brandsma
Dec 4 '18 at 9:13
$begingroup$
okay thanks for your time. Have you looked at (b)? For me, that was the hardest one
$endgroup$
– Esteban Cambiasso
Dec 4 '18 at 8:59
$begingroup$
okay thanks for your time. Have you looked at (b)? For me, that was the hardest one
$endgroup$
– Esteban Cambiasso
Dec 4 '18 at 8:59
$begingroup$
@EstebanCambiasso it’s already part of the solution after “ on the other hand”.
$endgroup$
– Henno Brandsma
Dec 4 '18 at 9:13
$begingroup$
@EstebanCambiasso it’s already part of the solution after “ on the other hand”.
$endgroup$
– Henno Brandsma
Dec 4 '18 at 9:13
add a comment |
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$begingroup$
What have you tried?
$endgroup$
– Mees de Vries
Dec 3 '18 at 12:05
$begingroup$
hi Mees. I have just added my solution to (a). Could u maybe give your opinion on that? My theoreme 4.1 says that f is continuous iff. f(closure(A)) is in Closure(f(A))
$endgroup$
– Esteban Cambiasso
Dec 3 '18 at 12:49
$begingroup$
I just saw that i miss a bar over f×g(SxT) in the last part
$endgroup$
– Esteban Cambiasso
Dec 3 '18 at 13:39
$begingroup$
For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. To help you get started, I have retyped the text from your picture.
$endgroup$
– Martin Sleziak
Dec 3 '18 at 16:21
$begingroup$
thank u. Can anybody verify my answer?
$endgroup$
– Esteban Cambiasso
Dec 3 '18 at 16:27