2
$\begingroup$

I'm trying to prove that, for topological spaces $X,Y$ and $f:X\to Y$, the statement

$$"f(\text{cl}\, A)\subseteq \text{cl}\, f(A),\;\forall A\subseteq X"$$ Implies $"B$ closed in $Y\Rightarrow f^{-1}(B)$ closed in $X"$ ($\text{cl}=$ closure). This is my attempt:

Let $U\subseteq Y$ be closed. Then: $$f\big(\color{blue}{\text{cl}\,f^{-1}(U)}\big)\subseteq \text{cl}\,f\big(f^{-1}(U)\big)=\text{cl}\,U=U=f\big(\color{blue}{f^{-1}(U)}\big)$$

So $\text{cl}\, f^{-1}(U)\subseteq f^{-1}(U)\Rightarrow f^{-1}(U)$ is closed in $X$.

Is my proof correct? I am a little weary of using $f$ and $f^{-1}$ as I have.

$\endgroup$
2
$\begingroup$

The equality $\text{cl} f(f^{-1}(U)) = \text{cl}(U)$ and the subsequent one are not correct because $f(f^{-1}(U)) \neq U$ in general.

However, it holds that $f(f^{-1}(U) \subset U$, so you can argue as follows:

$$\color{blue}{f(\text{cl}(f^{-1}(U)))} \subset \text{cl}f(f^{-1}(U)) \subset \text{cl} (U )= \color{blue}{U}$$

So,

$$\text{cl}(f^{-1}(U)) \subset f^{-1}(U)$$

(Using $f(A) \subset B \implies A \subset f^{-1}(B)$)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.