Let $f: X \to Y$ be continuous open map and $A \subset X$. Define $f_1:A \to f[A]$. Give an example where $f_1$ is not an open map.

Let $f: X \to Y$ be continuous open map and $A \subset X$. Define $f_1:A \to f[A]$. Give an example where $f_1$ is not an open map.

To give an example where $f_1$ is not open I think that every non-injective function will satisfy this as the restriction of open map is only open when $f$ is an injection? So for example when $f: \Bbb R \to \Bbb R , x \mapsto x^2$ and $A=[-2,2]$ we get that $f_1:[-2,2] \to f[-2,2]$ is not an injection as $-2 \mapsto 4$ and $2 \mapsto 4$. How can I show that this is not open? If I pick $U=(-2,2)$, then $U$ is open, but $f_1[U]=[0,4)$ is not open?

Is the statement

for open map $f: X \to Y$ the restriction of $f$ to $A \subset X$ is open if and only if $f$ is injective?

Solution 1:

Let $f : \mathbb{R}^2 \rightarrow \mathbb{R}$ defined for every $(x,y) \in \mathbb{R}^2$ by $f(x,y)=x$. It is easy to show that $f$ is continuous and open.

Let $a = (0,1) \in \mathbb{R}^2$, and let $A = \lbrace a\rbrace \cup \left( \mathbb{R} \times \lbrace 0 \rbrace\right)$.

Then $\lbrace a \rbrace$ is an open subset of $A$, but $f(\lbrace a \rbrace) = \lbrace 0 \rbrace$ is not an open subset of $f(A)= \mathbb{R}$.

So $f_{|A} : A \rightarrow f(A)$ is not open.