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?/

OK, I think I got it now, thanks to @NDewolf and @freakish.
In my question, one detail was missing: I forgot to say that $A$ is given the subspace topology. With this detail added, the answer should look like this:
If $f$ is continuous, this means that for any open $f(A)$ its preimage $A$ must be open.
Specifically, the restriction $f_{|A}$ is continuous given that $A$ is given the subspace topology.
