Finding a continuous function
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Let $X$ be a normal space and $Y$ be a closed subspace of $X$ and let $f$ be a continuous function from the subset $Y$ to $Z^2 = {(x,y) in mathbb{R^2}:0 leq x,y leq1 }$. Show that there exist a continuous function $phi:X rightarrow Z^2$ such that the restriction of $phi$ to $Y$ is $f$.
Cannot find a way to show this, any help would be appreciated.
real-analysis continuity
edited Dec 9 '18 at 13:37
GNUSupporter 8964民主女神 地下教會
12.9k72445
asked Dec 9 '18 at 13:26
jaz laykjaz layk
91
$endgroup$
add a comment |
$begingroup$
Let $X$ be a normal space and $Y$ be a closed subspace of $X$ and let $f$ be a continuous function from the subset $Y$ to $Z^2 = {(x,y) in mathbb{R^2}:0 leq x,y leq1 }$. Show that there exist a continuous function $phi:X rightarrow Z^2$ such that the restriction of $phi$ to $Y$ is $f$.
Cannot find a way to show this, any help would be appreciated.
real-analysis continuity
edited Dec 9 '18 at 13:37
GNUSupporter 8964民主女神 地下教會
12.9k72445
asked Dec 9 '18 at 13:26
jaz laykjaz layk
91
$endgroup$
$begingroup$
This is just the Tietze Extension Theorem. (Applied to $f_1$ and $f_2$, if $f=(f_1,f_2)$.)
$endgroup$
– David C. Ullrich
Dec 9 '18 at 13:54
add a comment |
$begingroup$
Let $X$ be a normal space and $Y$ be a closed subspace of $X$ and let $f$ be a continuous function from the subset $Y$ to $Z^2 = {(x,y) in mathbb{R^2}:0 leq x,y leq1 }$. Show that there exist a continuous function $phi:X rightarrow Z^2$ such that the restriction of $phi$ to $Y$ is $f$.
Cannot find a way to show this, any help would be appreciated.
real-analysis continuity
edited Dec 9 '18 at 13:37
GNUSupporter 8964民主女神 地下教會
12.9k72445
asked Dec 9 '18 at 13:26
jaz laykjaz layk
91
$endgroup$
Let $X$ be a normal space and $Y$ be a closed subspace of $X$ and let $f$ be a continuous function from the subset $Y$ to $Z^2 = {(x,y) in mathbb{R^2}:0 leq x,y leq1 }$. Show that there exist a continuous function $phi:X rightarrow Z^2$ such that the restriction of $phi$ to $Y$ is $f$.
Cannot find a way to show this, any help would be appreciated.
real-analysis continuity
real-analysis continuity
edited Dec 9 '18 at 13:37
GNUSupporter 8964民主女神 地下教會
12.9k72445
asked Dec 9 '18 at 13:26
jaz laykjaz layk
91
edited Dec 9 '18 at 13:37
GNUSupporter 8964民主女神 地下教會
12.9k72445
asked Dec 9 '18 at 13:26
jaz laykjaz layk
91
edited Dec 9 '18 at 13:37
GNUSupporter 8964民主女神 地下教會
12.9k72445
edited Dec 9 '18 at 13:37
GNUSupporter 8964民主女神 地下教會
12.9k72445
edited Dec 9 '18 at 13:37
GNUSupporter 8964民主女神 地下教會
12.9k72445
12.9k72445
asked Dec 9 '18 at 13:26
jaz laykjaz layk
91
asked Dec 9 '18 at 13:26
jaz laykjaz layk
91
asked Dec 9 '18 at 13:26
jaz laykjaz layk
91
91
$begingroup$
This is just the Tietze Extension Theorem. (Applied to $f_1$ and $f_2$, if $f=(f_1,f_2)$.)
$endgroup$
– David C. Ullrich
Dec 9 '18 at 13:54
add a comment |
$begingroup$
This is just the Tietze Extension Theorem. (Applied to $f_1$ and $f_2$, if $f=(f_1,f_2)$.)
$endgroup$
– David C. Ullrich
Dec 9 '18 at 13:54
$begingroup$
This is just the Tietze Extension Theorem. (Applied to $f_1$ and $f_2$, if $f=(f_1,f_2)$.)
$endgroup$
– David C. Ullrich
Dec 9 '18 at 13:54
$begingroup$
This is just the Tietze Extension Theorem. (Applied to $f_1$ and $f_2$, if $f=(f_1,f_2)$.)
$endgroup$
– David C. Ullrich
Dec 9 '18 at 13:54
add a comment |
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$begingroup$
This is just the Tietze Extension Theorem. (Applied to $f_1$ and $f_2$, if $f=(f_1,f_2)$.)
$endgroup$
– David C. Ullrich
Dec 9 '18 at 13:54