Let $X$ and $Y$ be two topological spaces, $A \subset X$, $B \subset Y$ and $f: X \longrightarrow Y$ a function, such that $f(A) \subset B$. Then $f$ determines a function $f|_{A,B} : A \longrightarrow B$. Show that:
- if $f$ is continuous, then $f|_{A,B}$ is also continuous
- $f$ is continuous $\Leftrightarrow$ $f|_{X,f(X)}$ is continuous
This is one of the problems in my topology textbook. I gave it a try but I could not succed. Can anyone help me?
Fact (2.) follows from the fact we just saw applied to $A=f[X]$.
And 1. is easy from definitions: let $O$ be open in $B$, i.e. $O = O' \cap B$ for $O' \subseteq Y$ open.
Then $$(f\restriction_{A,B})^{-1}[O] = f^{-1}[O'] \cap A$$
which is open in $A$ by continuiy of $f$ and the definition of the subspace topology on $A$.
Or a bit slicker: Note that: $$i_B \circ f\restriction_{A,B} = f \circ i_A$$ for the respective embeddings of $A$ into $X$ and $B$ into $Y$ which are continuous. Then apply the linked fact for $i_B$.
