If $X$ and $Y$ are homeomorphic then $X-{x}$ and $Y-{y}$ are homeomorphic? I wanna know,  If $f:X \to Y$ is a homeomorphism where $X,Y \subset \mathbb{R}^n$, then $f:X-\{x\} \to Y-\{y\}$ is a homeomorphism too??
I was thinking about it and if $y=f(x)$ then the statement is true, but if not? there is a conter example?
 A: Note That if $y \neq f(x)$ then $f: X-\{x\} \to Y-\{y\}$ is not well defined. 
If you are asking if there could be a different homeomorphism, a counterexample is 
$$Id :[0,1] \to [0,1]$$
$x=0, y=\frac{1}{2}$.
A: If $y=f(x)$, we can just take the restriction of $f$ to domain $X\setminus \{x\}$ and codomain $Y\setminus \{y\}$, and note that this is still a bijection (we removed one point and its image, so the map is still surjective, and injectiveness does not disappear by taking a smaller domain) and as $f$ amd $f^{-1}$ are continuous between $X$ and $Y$, they're still continuous when restricted (this is a standard fact about continuous maps in general). So we also have a homeomorphism between the spaces minus those points.
This idea is often used to show spaces to be non-homeomorphic by contradiction:
Suppose that $f:[0,+\infty) \to (0,1)$ were a homeomorphism. For $x =0$ we have that $X\setminus \{0\}=(0,+\infty)$ is connected and whatever $y \in (0,1)$ we take, $(0,1)\setminus\{y\}$ is not connected. This shows that whatever $f(0)$ is we can apply the previous argument that the punctured spaces should be homomorphic but they cannot be (as they differ in the topological property of connectedness). So $[0,+\infty)$ and $(0,1)$ are not homeomorphic.
