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Proving $f$ is continuous if $f_1$ and $f_2$ are continuous and the reciprocal
$begingroup$
For $i=1,2$ let $(X_i,tau_i)$ and $(X'_i,tau_i)$ be topological spaces, and $f_i: X_ito X'_i$ functions. Consider the function $f:X_1times X_2to X'_1times X'_2$ defined by $f(x_1,x_2)=(f_1(x_1),f_2(x_2))$.
Show that if $f_1,f_2$ are continuous then $f$ is continuous for the product topologies. Prove the reciprocal.
I tried to prove the statement in the following way:
$Rightarrow$ Let $mathscr{U}$ be an open set such that by definition of product topology $mathscr{U}=Atimes B$. As $f_1$ and $f_2$ are continuous functions, then $f_1^{-1}(A)intau'_1$ and $f_2^{-1}(B)intau'_2$.
So $f^{-1}(mathscr{U})=f^{-1}(Atimes B)=f_1^{-1}(A)times f_2^{-1}(B)$so that by definition of open sets in the product topology are open hence $f_1^{-1}(A)times f_2^{-1}(B)intau'_i$.
$Leftarrow$ Let's define the following functions $I_1: (X'_i,tau'_i)to (X_1times a_2)$ and $I_2: (X'_i,tau'_i)to (a_2times X_2)$ such that $(a_1,a_2)in X'_i $ By known theorem $I_1$ and $I_2$ are homeomorphisms.
Using the $p_1$ and $p_2$ as the projection functions. $f_1=p_1circ fcirc I_1$
and $f_2=p_2circ fcirc I_2$. As the composition of continuous functions is continuous hence $f_1$ and $f_2$ are continuous.
Questions:
1) Is my proof right? Can I give the following step $f^{-1}(Atimes B)=f_1^{-1}(A)times f_2^{-1}(B)$?
2) I was advised to define $I_1$ and $I_2$. Why are the projections not enough? I mean $f_1=p_1circ f$ and $f_2=p_2circ f$.
Thanks in advance!
general-topology proof-verification proof-writing
asked Jan 20 at 12:39
Pedro GomesPedro Gomes
1,8982721
$endgroup$
add a comment |
$begingroup$
For $i=1,2$ let $(X_i,tau_i)$ and $(X'_i,tau_i)$ be topological spaces, and $f_i: X_ito X'_i$ functions. Consider the function $f:X_1times X_2to X'_1times X'_2$ defined by $f(x_1,x_2)=(f_1(x_1),f_2(x_2))$.
Show that if $f_1,f_2$ are continuous then $f$ is continuous for the product topologies. Prove the reciprocal.
I tried to prove the statement in the following way:
$Rightarrow$ Let $mathscr{U}$ be an open set such that by definition of product topology $mathscr{U}=Atimes B$. As $f_1$ and $f_2$ are continuous functions, then $f_1^{-1}(A)intau'_1$ and $f_2^{-1}(B)intau'_2$.
So $f^{-1}(mathscr{U})=f^{-1}(Atimes B)=f_1^{-1}(A)times f_2^{-1}(B)$so that by definition of open sets in the product topology are open hence $f_1^{-1}(A)times f_2^{-1}(B)intau'_i$.
$Leftarrow$ Let's define the following functions $I_1: (X'_i,tau'_i)to (X_1times a_2)$ and $I_2: (X'_i,tau'_i)to (a_2times X_2)$ such that $(a_1,a_2)in X'_i $ By known theorem $I_1$ and $I_2$ are homeomorphisms.
Using the $p_1$ and $p_2$ as the projection functions. $f_1=p_1circ fcirc I_1$
and $f_2=p_2circ fcirc I_2$. As the composition of continuous functions is continuous hence $f_1$ and $f_2$ are continuous.
Questions:
1) Is my proof right? Can I give the following step $f^{-1}(Atimes B)=f_1^{-1}(A)times f_2^{-1}(B)$?
2) I was advised to define $I_1$ and $I_2$. Why are the projections not enough? I mean $f_1=p_1circ f$ and $f_2=p_2circ f$.
Thanks in advance!
general-topology proof-verification proof-writing
asked Jan 20 at 12:39
Pedro GomesPedro Gomes
1,8982721
$endgroup$
$begingroup$
You need a small argument for the step in 1, plus some extra reasoning. See below.
$endgroup$
– Henno Brandsma
Jan 20 at 13:18
$begingroup$
$X_1 times a_2$ is sloppy notation.
$endgroup$
– Henno Brandsma
Jan 20 at 13:19
add a comment |
$begingroup$
For $i=1,2$ let $(X_i,tau_i)$ and $(X'_i,tau_i)$ be topological spaces, and $f_i: X_ito X'_i$ functions. Consider the function $f:X_1times X_2to X'_1times X'_2$ defined by $f(x_1,x_2)=(f_1(x_1),f_2(x_2))$.
Show that if $f_1,f_2$ are continuous then $f$ is continuous for the product topologies. Prove the reciprocal.
I tried to prove the statement in the following way:
$Rightarrow$ Let $mathscr{U}$ be an open set such that by definition of product topology $mathscr{U}=Atimes B$. As $f_1$ and $f_2$ are continuous functions, then $f_1^{-1}(A)intau'_1$ and $f_2^{-1}(B)intau'_2$.
So $f^{-1}(mathscr{U})=f^{-1}(Atimes B)=f_1^{-1}(A)times f_2^{-1}(B)$so that by definition of open sets in the product topology are open hence $f_1^{-1}(A)times f_2^{-1}(B)intau'_i$.
$Leftarrow$ Let's define the following functions $I_1: (X'_i,tau'_i)to (X_1times a_2)$ and $I_2: (X'_i,tau'_i)to (a_2times X_2)$ such that $(a_1,a_2)in X'_i $ By known theorem $I_1$ and $I_2$ are homeomorphisms.
Using the $p_1$ and $p_2$ as the projection functions. $f_1=p_1circ fcirc I_1$
and $f_2=p_2circ fcirc I_2$. As the composition of continuous functions is continuous hence $f_1$ and $f_2$ are continuous.
Questions:
1) Is my proof right? Can I give the following step $f^{-1}(Atimes B)=f_1^{-1}(A)times f_2^{-1}(B)$?
2) I was advised to define $I_1$ and $I_2$. Why are the projections not enough? I mean $f_1=p_1circ f$ and $f_2=p_2circ f$.
Thanks in advance!
general-topology proof-verification proof-writing
asked Jan 20 at 12:39
Pedro GomesPedro Gomes
1,8982721
$endgroup$
For $i=1,2$ let $(X_i,tau_i)$ and $(X'_i,tau_i)$ be topological spaces, and $f_i: X_ito X'_i$ functions. Consider the function $f:X_1times X_2to X'_1times X'_2$ defined by $f(x_1,x_2)=(f_1(x_1),f_2(x_2))$.
Show that if $f_1,f_2$ are continuous then $f$ is continuous for the product topologies. Prove the reciprocal.
I tried to prove the statement in the following way:
$Rightarrow$ Let $mathscr{U}$ be an open set such that by definition of product topology $mathscr{U}=Atimes B$. As $f_1$ and $f_2$ are continuous functions, then $f_1^{-1}(A)intau'_1$ and $f_2^{-1}(B)intau'_2$.
So $f^{-1}(mathscr{U})=f^{-1}(Atimes B)=f_1^{-1}(A)times f_2^{-1}(B)$so that by definition of open sets in the product topology are open hence $f_1^{-1}(A)times f_2^{-1}(B)intau'_i$.
$Leftarrow$ Let's define the following functions $I_1: (X'_i,tau'_i)to (X_1times a_2)$ and $I_2: (X'_i,tau'_i)to (a_2times X_2)$ such that $(a_1,a_2)in X'_i $ By known theorem $I_1$ and $I_2$ are homeomorphisms.
Using the $p_1$ and $p_2$ as the projection functions. $f_1=p_1circ fcirc I_1$
and $f_2=p_2circ fcirc I_2$. As the composition of continuous functions is continuous hence $f_1$ and $f_2$ are continuous.
Questions:
1) Is my proof right? Can I give the following step $f^{-1}(Atimes B)=f_1^{-1}(A)times f_2^{-1}(B)$?
2) I was advised to define $I_1$ and $I_2$. Why are the projections not enough? I mean $f_1=p_1circ f$ and $f_2=p_2circ f$.
Thanks in advance!
general-topology proof-verification proof-writing
general-topology proof-verification proof-writing
asked Jan 20 at 12:39
Pedro GomesPedro Gomes
1,8982721
asked Jan 20 at 12:39
Pedro GomesPedro Gomes
1,8982721
asked Jan 20 at 12:39
Pedro GomesPedro Gomes
1,8982721
asked Jan 20 at 12:39
Pedro GomesPedro Gomes
1,8982721
asked Jan 20 at 12:39
Pedro GomesPedro Gomes
1,8982721
1,8982721
$begingroup$
You need a small argument for the step in 1, plus some extra reasoning. See below.
$endgroup$
– Henno Brandsma
Jan 20 at 13:18
$begingroup$
$X_1 times a_2$ is sloppy notation.
$endgroup$
– Henno Brandsma
Jan 20 at 13:19
add a comment |
$begingroup$
You need a small argument for the step in 1, plus some extra reasoning. See below.
$endgroup$
– Henno Brandsma
Jan 20 at 13:18
$begingroup$
$X_1 times a_2$ is sloppy notation.
$endgroup$
– Henno Brandsma
Jan 20 at 13:19
$begingroup$
You need a small argument for the step in 1, plus some extra reasoning. See below.
$endgroup$
– Henno Brandsma
Jan 20 at 13:18
$begingroup$
You need a small argument for the step in 1, plus some extra reasoning. See below.
$endgroup$
– Henno Brandsma
Jan 20 at 13:18
$begingroup$
$X_1 times a_2$ is sloppy notation.
$endgroup$
– Henno Brandsma
Jan 20 at 13:19
$begingroup$
$X_1 times a_2$ is sloppy notation.
$endgroup$
– Henno Brandsma
Jan 20 at 13:19
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Firstly, an open set in the product topology is not always of the form $U times V$, it is a union of such open sets, as these form a base for the product topology. So if $O$ is product open and so $O= bigcup_{i in I} (U_i times V_i)$ we indeed have that $$f^{-1}[U_i times V_i] = {(x,x'): f(x,x') in U_i times V_i} =
{(x,x'): (f_1(x), f_2(x') in U_i times V_i} = {(x,x'): f_1(x) in U_i text{ and } f_2(x') in V_i} =
{(x,x'): xin f_1^{-1}[U_i] text{ and } x' in f_2^{-1}[V_i]} =f_1^{-1}[U_i] times f_2^{-1}[V_i]$$ which is basic product open, and so $f^{-1}[O] = bigcup_i f^{-1}[U_i times V_i]$ is a union of open sets and hence open.
The reverse needs the embeddings, as $pi_1 circ f$ is a map defined on $X_1 times X_2$ and not on $X_1$, so is not equal to $f_1$. In order to use $f$ we need two arguments, so fix $a in X_1$ and $a' in X_2$ and define $j: X_1 to X_1 times X_2 $ by $j(x)=(x,a')$. It's continuous because the composition of $j$ with the two projections are either the identity on $X_1$ or a constant map, so continuous always. And then $(pi_1 circ f circ j)(x)=pi_1(f(x,a'))=pi_1(f_1(x), f_2(a'))= f_1(x)$ for all $xin X_1$ so $f_1$ is then a composition of three continuous maps, hence continuous.
$f_2$ is similarly continuous using $i: X_1 to X_1 times X_2$ defined by $i(x)=(a,x)$.
answered Jan 20 at 13:17
Henno BrandsmaHenno Brandsma
111k348120
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Firstly, an open set in the product topology is not always of the form $U times V$, it is a union of such open sets, as these form a base for the product topology. So if $O$ is product open and so $O= bigcup_{i in I} (U_i times V_i)$ we indeed have that $$f^{-1}[U_i times V_i] = {(x,x'): f(x,x') in U_i times V_i} =
{(x,x'): (f_1(x), f_2(x') in U_i times V_i} = {(x,x'): f_1(x) in U_i text{ and } f_2(x') in V_i} =
{(x,x'): xin f_1^{-1}[U_i] text{ and } x' in f_2^{-1}[V_i]} =f_1^{-1}[U_i] times f_2^{-1}[V_i]$$ which is basic product open, and so $f^{-1}[O] = bigcup_i f^{-1}[U_i times V_i]$ is a union of open sets and hence open.
The reverse needs the embeddings, as $pi_1 circ f$ is a map defined on $X_1 times X_2$ and not on $X_1$, so is not equal to $f_1$. In order to use $f$ we need two arguments, so fix $a in X_1$ and $a' in X_2$ and define $j: X_1 to X_1 times X_2 $ by $j(x)=(x,a')$. It's continuous because the composition of $j$ with the two projections are either the identity on $X_1$ or a constant map, so continuous always. And then $(pi_1 circ f circ j)(x)=pi_1(f(x,a'))=pi_1(f_1(x), f_2(a'))= f_1(x)$ for all $xin X_1$ so $f_1$ is then a composition of three continuous maps, hence continuous.
$f_2$ is similarly continuous using $i: X_1 to X_1 times X_2$ defined by $i(x)=(a,x)$.
answered Jan 20 at 13:17
Henno BrandsmaHenno Brandsma
111k348120
$endgroup$
add a comment |
$begingroup$
Firstly, an open set in the product topology is not always of the form $U times V$, it is a union of such open sets, as these form a base for the product topology. So if $O$ is product open and so $O= bigcup_{i in I} (U_i times V_i)$ we indeed have that $$f^{-1}[U_i times V_i] = {(x,x'): f(x,x') in U_i times V_i} =
{(x,x'): (f_1(x), f_2(x') in U_i times V_i} = {(x,x'): f_1(x) in U_i text{ and } f_2(x') in V_i} =
{(x,x'): xin f_1^{-1}[U_i] text{ and } x' in f_2^{-1}[V_i]} =f_1^{-1}[U_i] times f_2^{-1}[V_i]$$ which is basic product open, and so $f^{-1}[O] = bigcup_i f^{-1}[U_i times V_i]$ is a union of open sets and hence open.
The reverse needs the embeddings, as $pi_1 circ f$ is a map defined on $X_1 times X_2$ and not on $X_1$, so is not equal to $f_1$. In order to use $f$ we need two arguments, so fix $a in X_1$ and $a' in X_2$ and define $j: X_1 to X_1 times X_2 $ by $j(x)=(x,a')$. It's continuous because the composition of $j$ with the two projections are either the identity on $X_1$ or a constant map, so continuous always. And then $(pi_1 circ f circ j)(x)=pi_1(f(x,a'))=pi_1(f_1(x), f_2(a'))= f_1(x)$ for all $xin X_1$ so $f_1$ is then a composition of three continuous maps, hence continuous.
$f_2$ is similarly continuous using $i: X_1 to X_1 times X_2$ defined by $i(x)=(a,x)$.
answered Jan 20 at 13:17
Henno BrandsmaHenno Brandsma
111k348120
$endgroup$
add a comment |
$begingroup$
Firstly, an open set in the product topology is not always of the form $U times V$, it is a union of such open sets, as these form a base for the product topology. So if $O$ is product open and so $O= bigcup_{i in I} (U_i times V_i)$ we indeed have that $$f^{-1}[U_i times V_i] = {(x,x'): f(x,x') in U_i times V_i} =
{(x,x'): (f_1(x), f_2(x') in U_i times V_i} = {(x,x'): f_1(x) in U_i text{ and } f_2(x') in V_i} =
{(x,x'): xin f_1^{-1}[U_i] text{ and } x' in f_2^{-1}[V_i]} =f_1^{-1}[U_i] times f_2^{-1}[V_i]$$ which is basic product open, and so $f^{-1}[O] = bigcup_i f^{-1}[U_i times V_i]$ is a union of open sets and hence open.
The reverse needs the embeddings, as $pi_1 circ f$ is a map defined on $X_1 times X_2$ and not on $X_1$, so is not equal to $f_1$. In order to use $f$ we need two arguments, so fix $a in X_1$ and $a' in X_2$ and define $j: X_1 to X_1 times X_2 $ by $j(x)=(x,a')$. It's continuous because the composition of $j$ with the two projections are either the identity on $X_1$ or a constant map, so continuous always. And then $(pi_1 circ f circ j)(x)=pi_1(f(x,a'))=pi_1(f_1(x), f_2(a'))= f_1(x)$ for all $xin X_1$ so $f_1$ is then a composition of three continuous maps, hence continuous.
$f_2$ is similarly continuous using $i: X_1 to X_1 times X_2$ defined by $i(x)=(a,x)$.
answered Jan 20 at 13:17
Henno BrandsmaHenno Brandsma
111k348120
$endgroup$
Firstly, an open set in the product topology is not always of the form $U times V$, it is a union of such open sets, as these form a base for the product topology. So if $O$ is product open and so $O= bigcup_{i in I} (U_i times V_i)$ we indeed have that $$f^{-1}[U_i times V_i] = {(x,x'): f(x,x') in U_i times V_i} =
{(x,x'): (f_1(x), f_2(x') in U_i times V_i} = {(x,x'): f_1(x) in U_i text{ and } f_2(x') in V_i} =
{(x,x'): xin f_1^{-1}[U_i] text{ and } x' in f_2^{-1}[V_i]} =f_1^{-1}[U_i] times f_2^{-1}[V_i]$$ which is basic product open, and so $f^{-1}[O] = bigcup_i f^{-1}[U_i times V_i]$ is a union of open sets and hence open.
The reverse needs the embeddings, as $pi_1 circ f$ is a map defined on $X_1 times X_2$ and not on $X_1$, so is not equal to $f_1$. In order to use $f$ we need two arguments, so fix $a in X_1$ and $a' in X_2$ and define $j: X_1 to X_1 times X_2 $ by $j(x)=(x,a')$. It's continuous because the composition of $j$ with the two projections are either the identity on $X_1$ or a constant map, so continuous always. And then $(pi_1 circ f circ j)(x)=pi_1(f(x,a'))=pi_1(f_1(x), f_2(a'))= f_1(x)$ for all $xin X_1$ so $f_1$ is then a composition of three continuous maps, hence continuous.
$f_2$ is similarly continuous using $i: X_1 to X_1 times X_2$ defined by $i(x)=(a,x)$.
answered Jan 20 at 13:17
Henno BrandsmaHenno Brandsma
111k348120
answered Jan 20 at 13:17
Henno BrandsmaHenno Brandsma
111k348120
answered Jan 20 at 13:17
Henno BrandsmaHenno Brandsma
111k348120
answered Jan 20 at 13:17
Henno BrandsmaHenno Brandsma
111k348120
111k348120
add a comment |
add a comment |
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$begingroup$
You need a small argument for the step in 1, plus some extra reasoning. See below.
$endgroup$
– Henno Brandsma
Jan 20 at 13:18
$begingroup$
$X_1 times a_2$ is sloppy notation.
$endgroup$
– Henno Brandsma
Jan 20 at 13:19