Prove if $A$ is homeomorphic to $B$ then $A-\{a\}$ is homeomorphic to $B-\{b\}$ When dealing with some problems, I frequently got this statement. For example:
If $D^1\cong D^2$, then $D^1-\{0\}\cong D^2 -\{0\}$, thus $S^1\simeq S^2$...
However, why is this always true? The 1-1 correspondence is easy to construct, but how can this be the homeomorphism?
 A: If $A$ is homeomorphic to $B$, then there exists a homeomorphism $f:A\rightarrow B$.  If $f(a)=b$, then $f$ restricts to a homeomorphism from $A-\{a\}$ to $B-\{b\}$.  The important point is that $f(a)=b$.  If you can't find a homeomorphism that does this, then $A-\{a\}$ is not homeomorphic to $B-\{b\}$, for instance if $b$ is isolated, but $a$ is not.
A: It is not always true, actually. For example, let $A=[a,c)$ and $B=[x,y)$ for some $a,b,x,y\in\Bbb R$ with $a<c,$ $x<b<y.$ Then $A$ and $B$ are readily homeomorphic, but $A\setminus\{a\}=(a,c)$ and $B\setminus\{b\}=[x,b)\cup(b,y)$ are not, since one is connected and the other isn't.
If we happen to know that we have a homeomorphism $f:A\to B$ such that $f(a)=b,$ on the other hand, we can say that $A\setminus\{a\}$ and $B\setminus\{b\}$ are homeomorphic, with the restriction of $f$ to $A\setminus\{a\}$ giving a ready homeomorphism.
A: This isn't true in general. For example, let $X := S^1\vee S^1$; then certainly $X\cong X$. But if $a$ is the wedge point while $b$ is any other point, then $X\setminus\{a\} \not\cong X\setminus\{b\}$.
A true general statement would be that if $f:A\to B$ is a homeomorphism, then $A\setminus\{a\} \cong B\setminus\{f(a)\}$. The homeomorphism is just the restriction of $f$; continuity of $f$ and $f^{-1}$ follows immediately from the fact that the restriction of a continuous map to a subspace is continuous.
