Topological space, manifold, open sets Consider a homeomorphism $f$ sending an open set $X$ of one manifold to an open set $Y$ of another manifold:
$$
f\,:~X \longrightarrow Y~~.
$$
Being a homeomorphism, $f$ is continuous -- and so is its inverse $g = f^{-1}~$:
$$
g\,:~   Y\longrightarrow X ~~.
$$
Then a (not necessarily open) subset
$$
A \subset X 
$$
is mapped by $f$ to a (not necessarily open) subset $f A\subset Y$.
The resulting restrictions of $f$ and $g = f^{-1}$ are:
$$
f_{~|A}\,: ~~~~ A \longrightarrow f(A)~~,
$$
$$
g_{~|f(A)}\,: ~~~ f(A) \longrightarrow A  ~~.
$$
Now, if I postulate that the image $f(A)~$ ${\underline{\mbox{is}}}$ open, will that imply that the preimage $A$ is open also?
Stated shortly: is a restriction of a continuous map continuous too?
/It certainly is in calculus -- but how to show this in topology?/

If $f:X\to Y$ is a homeomorphism and $A\subseteq X$, then $f_{|A}:A\to f(A)$ is a homeomorphism as well. Regardless of what $A$ is, open or not.
Indeed, being a homeomorphism means that there is a continuous inverse $g:Y\to X$. Now a restriction of continuous function is continuous and so $f_{|A}$ is continuous. Obviously $g_{|f(A)}:f(A)\to A$ is well defined (because $g(f(A))=A$) and continuous as well. This $g_{|f(A)}$ is the inverse of $f_{|A}$.
