Homeomorphism on subspace topology I have a problem with the construction of a homeomorphism. This is the task:

Suppose $f : X \to Y$ is a homeomorphism and $A$ is a subset of $X$. Prove that $X\setminus A$ is homeomorphic to $Y \setminus f(A)$ (each being given the subspace topology). 

I know that I need to construct a bijective continuous map from $X \setminus A$ to $Y\setminus f(A)$ whose inverse is also continuous but I don't know in which manner. I would be grateful for your help.
 A: Indeed. Show that $g: X\setminus A \rightarrow Y \setminus f[A]$ defined by $g(x) = f(x)$ (so just $f$ restricted) works. 
Main fact: the restriction of a continuous function is continuous. 
A: First, show that $f^*: X \setminus A \to Y \setminus f(A)$, $f^*(x) = f(x)$, is still a bijection.
Then, show that $f|_{A^c}: X \setminus A \to Y$, $f|_{A^c}(x) = f(x)$ is continuous. This might already be a theorem in your book/material; this is just the restriction of $f$ to $A^c$ (the complement of $A$).
Similarly, show that $f^{-1}|_{f(A)^c}$ is continuous.
Combine these to show that $f^*: X \setminus A \to Y \setminus f(A)$ is a homeomorphism.
A: For any sets $B, C$ with topologies  $T_B ,T_C,$ a function $f:B\to C$ is a homeomorphism iff (i) $f:B\to C$ is a bijection, and (ii) $\{f(U):U\in T_B\}=\{V:V\in T_C\}=T_C.$
For your Q: Let $B= X$ \ $A$ and $C=Y$ \ $f(A).$ 
Let $f_B$ be the restriction of $f$ to the domain $B.$ So $f_B(B)=f(B).$ 
Now $f_B:B\to f_B(B)$ is a bijection because $f$ is a bijection.
And since $f$ is a bijection and $A\subset X,$ we have $$f_B(B)=f(B)=f(X \setminus A)=f(X) \setminus f(A)=Y \setminus f(A)=C.$$
$\bullet$ . Therefore $f_B:B\to C$ is a bijection.
Let $T_X$ be the topology on $X$ and let $T_Y$ be the topology on $Y.$ The topology on $B$ as a sub-space of $X$ is $T_B=\{s\cap B: s\in T_X\}.$ The topology on $C$ as a sub-space of $Y$ is $T_C=\{t\cap C: t\in T_Y\}.$ 
Since every $b\in  T_B$ is a subset of the domain of $f_B,$ and since $f$ is a homeomorphism,  therefore $$\bullet  \quad \{f_B(U):U\in T_B\}=$$ $$=\{f_B(s\cap B):s\in T_X\}=$$ $$=\{f(s\cap B):s\in T_X\}=$$ $$=\{f(s)\cap f(B):s\in T_X\}=$$ $$=\{t\cap f(B):t\in T_Y\}=$$ $$=\{t\cap C:t\in T_Y\}=T_C.$$
So $f_B:B\to C$ is a homeomorphism. 
Remark: The short version is that $f$ maps the open sets of $X$ to the open sets of $Y$ so $f$ maps the intersections of $B$ with the open sets of $X$ to the intersections of $f(B)$ with the open sets of $Y.$
