Topological continuity and open sets I have a question about the answer in this thread below:
Proving restriction of function is continuous
The author of the answer says "$f^{-1}(U) \cap A$ is open in $A$ ".
We have
$f^{-1}_A(U) = \{a \in A: f_A(a) \in U\} \underbrace{=}_{f = f_A \text{ on }A} \{a \in A: f(a) \in U\} = f^{-1}(U)$
By definition of $f_A$, we have $f^{-1}(U) \subseteq A.$ So, what's the point of saying (over-specifying?) $f^{-1}(U) \cap A$ is open instead of just saying $f^{-1}(U)$ is open?

The difference is$$f^{-1}(U)=\{\color{blue}{x\in X}:f(x)\in U\}\\f^{-1}_A(U)=\{x\in A:f(x)\in U\}=f^{-1}(U)\cap A$$So $f^{-1}(U)$ is not necessarily a subset of $A$. It is actually $f^{-1}_A(U)$ that is a subset of $A$.
