Suppose $f:X\to Y$ is a continuous map between two metric spaces. Can we extend $f$ to a function $f':X'\to Y'$ in such a way that $f'$ is also continuous ($X'$ and $Y'$ are also metric spaces), where $X$ and $Y$ are open subsets of $X'$ and $Y'$ respectively and the restriction of $f'$ in $X$ is $f$ and $(f')^{-1}(Y)=X$?
Asked
Active
Viewed 77 times
1
-
Sure just embed $X$ into $X'$ as a connected component and do the same for $Y$. – Gregory Grant Mar 31 '15 at 16:25
2 Answers
1
Not in general.
Embed the real line $\mathbb{R}$ in the extended real line $\bar{\mathbb{R}}=\mathbb{R}\cup\{-\infty,\infty\}$ in the usual way and also as $\mathbb{R}'=\mathbb{R}\cup\{\omega\}$ (the one point compactification).
Then the map $f\colon \mathbb{R}\to\mathbb{R}$ defined by $f(x)=x$ cannot be extended to a continuous map $f'\colon \mathbb{R}'\to\bar{\mathbb{R}}$.
egreg
- 238,574
-
Suppose there is no such f initially between X and Y but can we construct any f' between X' and Y' with the above condition? – SAUVIK Mar 31 '15 at 16:47
-
-
Suppose X and Y are two metric spaces.Can you find X' and Y' such that X and Y are open in X' and Y' respectively and there exist some f' such that f' inverse Y is X? – SAUVIK Mar 31 '15 at 16:53
-
-
@user155928 Not in general. Take $Y$ a disconnected open subset of $Y'$ and $X$ a connected open subset of $X'$. – egreg Mar 31 '15 at 16:57
1
As I currently understand your question, the answer is no. For example, let $X' = \mathbb{R}, X = (0,\infty)$, and $Y = Y' = \mathbb{R}$. The natural log is a continuous map from $X$ to $Y$ that has no extension to $X'$.
Maybe you can impose further restrictions on your question?
PeterJL
- 1,160