Mixing
Inverse function of a product space
$begingroup$
I want to prove the continuity of a function $f: (X_1,tau_1) times (X_2,tau_2) rightarrow (X'_1,tau'_1) times (X'_2,tau'_2)$ where $f(x,y) = (f_1(x),f_2(y))$ and my question is:
What is $f^{-1}(bigcuplimits_{lambda_1,lambda_2 in Lambda} U_{lambda_1} times U_{lambda_2})$? Is it $(f_1^{-1}(bigcuplimits_{lambda1 in Lambda}U_{lambda_1}) times (f_2^{-1}bigcuplimits_{lambda_2 in Lambda} U_{lambda_2}))$ ?
EDIT:
I believe that the answer to my question is $(f_1^{-1}(bigcuplimits_ {lambda_1inLambda}U_{lambda_1})times X_2) cap (X_1 times f_2^{-1}(bigcuplimits_{lambda_2inLambda}U_{lambda_2}))$
Am i correct?
Bonus question: This is the first time I'm writting something in LaTeX, and it's honestly really cool. However, how can I write something like U_lambda_1, instead of having to write U_lambda'? Also, what is the code for uppercase Tau? Tau doesn't work
general-topology functions elementary-set-theory continuity products
asked Jan 22 at 0:04
J. DionisioJ. Dionisio
11011
$endgroup$
add a comment |
$begingroup$
I want to prove the continuity of a function $f: (X_1,tau_1) times (X_2,tau_2) rightarrow (X'_1,tau'_1) times (X'_2,tau'_2)$ where $f(x,y) = (f_1(x),f_2(y))$ and my question is:
What is $f^{-1}(bigcuplimits_{lambda_1,lambda_2 in Lambda} U_{lambda_1} times U_{lambda_2})$? Is it $(f_1^{-1}(bigcuplimits_{lambda1 in Lambda}U_{lambda_1}) times (f_2^{-1}bigcuplimits_{lambda_2 in Lambda} U_{lambda_2}))$ ?
EDIT:
I believe that the answer to my question is $(f_1^{-1}(bigcuplimits_ {lambda_1inLambda}U_{lambda_1})times X_2) cap (X_1 times f_2^{-1}(bigcuplimits_{lambda_2inLambda}U_{lambda_2}))$
Am i correct?
Bonus question: This is the first time I'm writting something in LaTeX, and it's honestly really cool. However, how can I write something like U_lambda_1, instead of having to write U_lambda'? Also, what is the code for uppercase Tau? Tau doesn't work
general-topology functions elementary-set-theory continuity products
asked Jan 22 at 0:04
J. DionisioJ. Dionisio
11011
$endgroup$
$begingroup$
You can do a double-subscript (with braces appropriately) such as $U_{lambda_1}$. No your inverse is not like that. For a function $f:Arightarrow B$ the inverse $f^{-1}(C)$ (where $C subseteq B$) is the set of all $a in A$ such that $f(a) in C$. You cannot take individual inverses of each dimension. For example suppose your function $f$ maps to either $(1,1)$ or $(2,2)$ depending on the input. Then $f^{-1}({(1,2)})$ is empty but your individual-dimension inverse $f_1^{-1}({1}) times f_2^{-1}({2})$ is not empty.
$endgroup$
– Michael
Jan 22 at 0:52
$begingroup$
@Michael yeah, I figured it was wrong, thank you :) what is it, then?
$endgroup$
– J. Dionisio
Jan 22 at 9:49
add a comment |
$begingroup$
I want to prove the continuity of a function $f: (X_1,tau_1) times (X_2,tau_2) rightarrow (X'_1,tau'_1) times (X'_2,tau'_2)$ where $f(x,y) = (f_1(x),f_2(y))$ and my question is:
What is $f^{-1}(bigcuplimits_{lambda_1,lambda_2 in Lambda} U_{lambda_1} times U_{lambda_2})$? Is it $(f_1^{-1}(bigcuplimits_{lambda1 in Lambda}U_{lambda_1}) times (f_2^{-1}bigcuplimits_{lambda_2 in Lambda} U_{lambda_2}))$ ?
EDIT:
I believe that the answer to my question is $(f_1^{-1}(bigcuplimits_ {lambda_1inLambda}U_{lambda_1})times X_2) cap (X_1 times f_2^{-1}(bigcuplimits_{lambda_2inLambda}U_{lambda_2}))$
Am i correct?
Bonus question: This is the first time I'm writting something in LaTeX, and it's honestly really cool. However, how can I write something like U_lambda_1, instead of having to write U_lambda'? Also, what is the code for uppercase Tau? Tau doesn't work
general-topology functions elementary-set-theory continuity products
asked Jan 22 at 0:04
J. DionisioJ. Dionisio
11011
$endgroup$
I want to prove the continuity of a function $f: (X_1,tau_1) times (X_2,tau_2) rightarrow (X'_1,tau'_1) times (X'_2,tau'_2)$ where $f(x,y) = (f_1(x),f_2(y))$ and my question is:
What is $f^{-1}(bigcuplimits_{lambda_1,lambda_2 in Lambda} U_{lambda_1} times U_{lambda_2})$? Is it $(f_1^{-1}(bigcuplimits_{lambda1 in Lambda}U_{lambda_1}) times (f_2^{-1}bigcuplimits_{lambda_2 in Lambda} U_{lambda_2}))$ ?
EDIT:
I believe that the answer to my question is $(f_1^{-1}(bigcuplimits_ {lambda_1inLambda}U_{lambda_1})times X_2) cap (X_1 times f_2^{-1}(bigcuplimits_{lambda_2inLambda}U_{lambda_2}))$
Am i correct?
Bonus question: This is the first time I'm writting something in LaTeX, and it's honestly really cool. However, how can I write something like U_lambda_1, instead of having to write U_lambda'? Also, what is the code for uppercase Tau? Tau doesn't work
general-topology functions elementary-set-theory continuity products
general-topology functions elementary-set-theory continuity products
asked Jan 22 at 0:04
J. DionisioJ. Dionisio
11011
asked Jan 22 at 0:04
J. DionisioJ. Dionisio
11011
edited Jan 23 at 17:49
J. Dionisio
asked Jan 22 at 0:04
J. DionisioJ. Dionisio
11011
asked Jan 22 at 0:04
J. DionisioJ. Dionisio
11011
asked Jan 22 at 0:04
J. DionisioJ. Dionisio
11011
11011
$begingroup$
You can do a double-subscript (with braces appropriately) such as $U_{lambda_1}$. No your inverse is not like that. For a function $f:Arightarrow B$ the inverse $f^{-1}(C)$ (where $C subseteq B$) is the set of all $a in A$ such that $f(a) in C$. You cannot take individual inverses of each dimension. For example suppose your function $f$ maps to either $(1,1)$ or $(2,2)$ depending on the input. Then $f^{-1}({(1,2)})$ is empty but your individual-dimension inverse $f_1^{-1}({1}) times f_2^{-1}({2})$ is not empty.
$endgroup$
– Michael
Jan 22 at 0:52
$begingroup$
@Michael yeah, I figured it was wrong, thank you :) what is it, then?
$endgroup$
– J. Dionisio
Jan 22 at 9:49
add a comment |
$begingroup$
You can do a double-subscript (with braces appropriately) such as $U_{lambda_1}$. No your inverse is not like that. For a function $f:Arightarrow B$ the inverse $f^{-1}(C)$ (where $C subseteq B$) is the set of all $a in A$ such that $f(a) in C$. You cannot take individual inverses of each dimension. For example suppose your function $f$ maps to either $(1,1)$ or $(2,2)$ depending on the input. Then $f^{-1}({(1,2)})$ is empty but your individual-dimension inverse $f_1^{-1}({1}) times f_2^{-1}({2})$ is not empty.
$endgroup$
– Michael
Jan 22 at 0:52
$begingroup$
@Michael yeah, I figured it was wrong, thank you :) what is it, then?
$endgroup$
– J. Dionisio
Jan 22 at 9:49
$begingroup$
You can do a double-subscript (with braces appropriately) such as $U_{lambda_1}$. No your inverse is not like that. For a function $f:Arightarrow B$ the inverse $f^{-1}(C)$ (where $C subseteq B$) is the set of all $a in A$ such that $f(a) in C$. You cannot take individual inverses of each dimension. For example suppose your function $f$ maps to either $(1,1)$ or $(2,2)$ depending on the input. Then $f^{-1}({(1,2)})$ is empty but your individual-dimension inverse $f_1^{-1}({1}) times f_2^{-1}({2})$ is not empty.
$endgroup$
– Michael
Jan 22 at 0:52
$begingroup$
You can do a double-subscript (with braces appropriately) such as $U_{lambda_1}$. No your inverse is not like that. For a function $f:Arightarrow B$ the inverse $f^{-1}(C)$ (where $C subseteq B$) is the set of all $a in A$ such that $f(a) in C$. You cannot take individual inverses of each dimension. For example suppose your function $f$ maps to either $(1,1)$ or $(2,2)$ depending on the input. Then $f^{-1}({(1,2)})$ is empty but your individual-dimension inverse $f_1^{-1}({1}) times f_2^{-1}({2})$ is not empty.
$endgroup$
– Michael
Jan 22 at 0:52
$begingroup$
@Michael yeah, I figured it was wrong, thank you :) what is it, then?
$endgroup$
– J. Dionisio
Jan 22 at 9:49
$begingroup$
@Michael yeah, I figured it was wrong, thank you :) what is it, then?
$endgroup$
– J. Dionisio
Jan 22 at 9:49
add a comment |
1 Answer
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$begingroup$
$$f^{-1}left(bigcuplimits_{lambda_1,lambda_2 in Lambda} U_{lambda_1} times U_{lambda_2}right)= bigcuplimits_{lambda_1,lambda_2 in Lambda} f^{-1}left(U_{lambda_1} times U_{lambda_2}right)=
bigcuplimits_{lambda_1,lambda_2 in Lambda} f_1^{-1} left(U_{lambda_1}right) times f_2^{-1}left(U_{lambda_2}right).$$
answered Mar 2 at 2:53
Alex RavskyAlex Ravsky
42.5k32383
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add a comment |
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$begingroup$
$$f^{-1}left(bigcuplimits_{lambda_1,lambda_2 in Lambda} U_{lambda_1} times U_{lambda_2}right)= bigcuplimits_{lambda_1,lambda_2 in Lambda} f^{-1}left(U_{lambda_1} times U_{lambda_2}right)=
bigcuplimits_{lambda_1,lambda_2 in Lambda} f_1^{-1} left(U_{lambda_1}right) times f_2^{-1}left(U_{lambda_2}right).$$
answered Mar 2 at 2:53
Alex RavskyAlex Ravsky
42.5k32383
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add a comment |
$begingroup$
$$f^{-1}left(bigcuplimits_{lambda_1,lambda_2 in Lambda} U_{lambda_1} times U_{lambda_2}right)= bigcuplimits_{lambda_1,lambda_2 in Lambda} f^{-1}left(U_{lambda_1} times U_{lambda_2}right)=
bigcuplimits_{lambda_1,lambda_2 in Lambda} f_1^{-1} left(U_{lambda_1}right) times f_2^{-1}left(U_{lambda_2}right).$$
answered Mar 2 at 2:53
Alex RavskyAlex Ravsky
42.5k32383
$endgroup$
add a comment |
$begingroup$
$$f^{-1}left(bigcuplimits_{lambda_1,lambda_2 in Lambda} U_{lambda_1} times U_{lambda_2}right)= bigcuplimits_{lambda_1,lambda_2 in Lambda} f^{-1}left(U_{lambda_1} times U_{lambda_2}right)=
bigcuplimits_{lambda_1,lambda_2 in Lambda} f_1^{-1} left(U_{lambda_1}right) times f_2^{-1}left(U_{lambda_2}right).$$
answered Mar 2 at 2:53
Alex RavskyAlex Ravsky
42.5k32383
$endgroup$
$$f^{-1}left(bigcuplimits_{lambda_1,lambda_2 in Lambda} U_{lambda_1} times U_{lambda_2}right)= bigcuplimits_{lambda_1,lambda_2 in Lambda} f^{-1}left(U_{lambda_1} times U_{lambda_2}right)=
bigcuplimits_{lambda_1,lambda_2 in Lambda} f_1^{-1} left(U_{lambda_1}right) times f_2^{-1}left(U_{lambda_2}right).$$
answered Mar 2 at 2:53
Alex RavskyAlex Ravsky
42.5k32383
answered Mar 2 at 2:53
Alex RavskyAlex Ravsky
42.5k32383
answered Mar 2 at 2:53
Alex RavskyAlex Ravsky
42.5k32383
answered Mar 2 at 2:53
Alex RavskyAlex Ravsky
42.5k32383
42.5k32383
add a comment |
add a comment |
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$begingroup$
You can do a double-subscript (with braces appropriately) such as $U_{lambda_1}$. No your inverse is not like that. For a function $f:Arightarrow B$ the inverse $f^{-1}(C)$ (where $C subseteq B$) is the set of all $a in A$ such that $f(a) in C$. You cannot take individual inverses of each dimension. For example suppose your function $f$ maps to either $(1,1)$ or $(2,2)$ depending on the input. Then $f^{-1}({(1,2)})$ is empty but your individual-dimension inverse $f_1^{-1}({1}) times f_2^{-1}({2})$ is not empty.
$endgroup$
– Michael
Jan 22 at 0:52
$begingroup$
@Michael yeah, I figured it was wrong, thank you :) what is it, then?
$endgroup$
– J. Dionisio
Jan 22 at 9:49