Let $X$ and $Y$ be two metric spaces and $f$ a mapping from $X$ to $Y$ .
Prove or disprove: Assume that $f$ is injective. Then $f$ is continuous iff for every subset $A$ of $X$, $f(\bar A) = \overline{f(A)}$
My attempt (Trying to prove the above):
I have already proved the fact that $f$ is continuous if and only if for every subset $A$ of X, $f(\bar A) \subset \overline{f(A)}$.
Suppose $$f(\bar A) = \overline{f(A)}$$
$$\implies \bar A = f^{-1}(\overline{f(A)})$$
As $\bar A$ and $\overline{f(A)}$ are closed sets, we can say that $f$ is continuous as the inverse image of the closed set is closed.
Now to show the converse part we assume that f is continuous and injective, we need to show $f(\bar A) = \overline{f(A)}$.
By using the above fact we can show that $f(\bar A) \subset \overline{f(A)}$, but I need help to show $\overline{f(A)} \subset f(\bar A)$?
Best Answer
No, your argument for the first part is not correct. You have to start with an arbitary closed set $C$ in $Y$ and show that $f^{-1}(C)$ is closed. For this take $A=f^{-1}(C)$ in the hypothesis. You get $\overline {f(A)} =f(\overline A)$ so $f(\overline A) \subseteq C$ since $f(A)=f(f^{-1}(C)) \subseteq C$ and $C$ is closed. Hence, $\overline A \subseteq f^{-1}(C)$ which means $\overline A \subseteq A$, so $A$ is closed.
The converse is not true. Define $f: [0,1) \to S^{1}$ by $f(x)=e^{2\pi ix}$. Let $A=[\frac 1 2 ,1)$. Then $A$ is closed but $f(A)$ is not closed, so we do not have $f(\overline A)=\overline {f(A)}$ even though $f$ is continuous and injective.