Topology: homeomorphism with restricted domain In topology, if I restrict the domain of a homeomorphism and find who's the image of the restricted homeomorphism, the domain and the image are still homeomorphs?
For example... Prove that if $X$ and $Y$are homeomorph topological spaces => $\forall x \in X, \exists y \in Y $such that X\{x} $\cong$ Y\{y}. 
I began saying that if $X$ and $Y$are homeomorphs, then there exists a homeomorphism between them, let's call it $h$. Let's restrict the domain of $h$ to X\{x}.
Since $h$ is a bijection, the $x$ I'm taking out has the property that if $h(x) = y$, there can be no other $x' \in X$ such that $h(x')=y$. So the image of the restricted $h$ becomes Y\{y}.
Since I took out only one point of $X$ and one point of $Y$, I didn't break the rules of homeomorphism, the result should still be a homeomorphism. But I'm not sure about that, because I haven't considered any topology over X\{x} and Y\{y}... 
 A: If $A\subset X$ and $B\subset Y$ are related by $B=f(A)$, where $f:X\to Y$ is a homeomorphism, then $A$ is homeomorphic to $B$ (where both subsets are endowed with the subspace topology). This is because of the more general fact stated below

Let $X,Y$ be spaces related by a continuous map $f:X\to Y$, and let $A\subset X$, $B\subset Y$ be subsets such that $f(A)\subset B$. Endow both $A$ and $B$ with the subspace topology. Then the induced map $\overline{f}=f\big|^B_A:A\to B$ is continuous.

To get the result you want you apply this fact to the homeomorphism $f:X\to Y$ and its inverse $g:Y\to X$. The result in the highlighted paragraph tells us that they induce continuous maps $A\to B$ and $B\to A$, and from $g\circ f=\mathrm{id}_X$ and $f\circ g=\mathrm{id}_Y$ it follows that $\overline{g}\circ\overline{f}=\mathrm{id}_A$ and $\overline{f}\circ\overline{g}=\mathrm{id}_B$ it follows that $\overline{g},\overline{f}$ are a pair of inverse homeomorphisms.
A: If you have the subspace topology on your restricted domain, the restriction doesn't change the continuity of the function: Let $f:X\rightarrow Y$ be a continuous function and $f|_{D}:D\rightarrow Y$ its restriction to a subset $D\subseteq X$ equipped with the subspace topology.
 Take an open set $O\subseteq Y$ and consider its inverse image $$f^{-1}(O)=f^{-1}\left(\left(O\cap f(D)\right)\cup \left(O\cap f\left(D\right)^C\right)\right)=f^{-1}\left( O\cap f(D) \right)=f^{-1}(O)\cap D,$$ which is open by the definition of the subspace topology and the continuity of $f$.
Therefore, your restricted homeomorphism is still continuous on $D$ and the same is true for its inverse function restricted to $f(D)$. Since these are still inverse to each other, the restrictios is a homeomorphism again.
