Extending homeomorphism between two subsets Let $X$ be a topological space (compact, Hausdorff).
Let $U, V$ be two open subsets in $X$, and $f:U\to V$ be a homeomorphism. It is known that in general one can not extend $f$ to a homeomorphism on $X$.
However, I don't know wether there exists a topological space $Y$, containing $U,V$ as open subsets, such that $f$ extends to a homeomorphism on $Y$.
Example: Take $X=\mathbb{N}$. Let $U=\mathbb{N}\setminus \{0\}$, let $V=\mathbb{N}$ and $f$ to be left translation. Then $f$ can not be extended to a homeomorphism on $\mathbb{N}$, but we can extend $f$ to a left translation defined on $\mathbb{Z}$.
Thanks in advance.
I would be also happy to get related comments.

You can do it this way:
Let $Y:=U\bigcup V$
where the union is disjoint. Then you can fix the following Topology $\sigma$ on $Y$:
$\sigma:=\{Y,\emptyset\}\cup \tau_U\cup \tau_V$
This way you can define the map
$F: Y\to Y$ such that for each $x\in Y$
if $x\in U$ then $F(x):=f(x)$ while if $x\in V$ then $F(x):=f^{-1}(x)$.
This map is an extension of $f$ and it is an homeomorphism.
In your example you do the same thing because $\mathbb{N}\bigcup \mathbb{N}/\{0\}$ is homeomorphic to $\mathbb{Z}$.
