$ \tau _{f} = \lbrace A \subset Y: f^{-1}(A) \in \tau \rbrace $. Show that $ A $ is open in $ Y $ if and only if $ f^{-1}(A) $ is open in $ X $ .

Question

Let $ (X,\tau) $ be a topology space, $ f : X \rightarrow Y $ a surjective function, $ A \subset Y $, and consider the topology in $ Y $: $\,$ $ \tau _{f} = \lbrace A \subset Y: f^{-1}(A) \in \tau \rbrace$. Show that: $ A $ is closed in $ Y $ if and only if $ f^{-1}(A) $ is closed in $ X.$

I got stuck with the converse: suppose that $ f^{-1}(A) $ is closed in $ X $. Then $ X-f^{-1} (A)$ is open in $ X $. That is, $ X-f^{-1} (A) \in \tau $. This implies that

$ X-f^{-1}(A) \stackrel{?}{=} f^{-1} (f(X)) - f^{-1}(A) = f^{-1}(f(X)-A) = f^{-1} (Y-A) \in \tau$.

Since $ Y-A \subset Y$ and $ f^{-1} (Y-A) \in \tau $, we conclude that $ Y-A \in \tau_{f} $ and therefore $ A $ is closed in $ Y $.

I know that $ \stackrel{?}{=} $ is not true, unless $f$ is injective. So I got stuck in that part.

Answer 1

The surjectivity of $f$ is not needed at all.

If $B$ is closed in $Y$ then $Y - B$ is open in $Y$. So by the definition of $\tau_Y$, we have that $f^{-1}[Y-B] = X-f^{-1}[B]$ is open in $X$. Hence $f^{-1}[B]$ is closed in $X$. All implications are reversible, so the reverse also holds.

The identity $f^{-1}[Y-B] = X-f^{-1}[B]$ holds for all $B$ by virtue of the definition of $f^{-1}[\cdot]$: $x \in f^{-1}[Y-B]$ iff $f(x) \in Y-B$ iff $f(x) \notin B$ iff $x \notin f^{-1}[B]$ iff $x \in X-f^{-1}[B]$.

Answer 2

$X=f^{-1}f(X)$ is always true: First of all, we must have $f^{-1}f(X)\subseteq X$. Now given $x\in X$, then $f(x)\in f(X)$, so $x\in f^{-1}f(X)$, so $X\subseteq f^{-1}f(X)$.

Note that if we have $f(x)\in S$, then $x\in f^{-1}(S)$, now put $S=f(X)$.

Answer 3

On base of:$$x\in f^{-1}(B)\iff f(x)\in B$$ it can be deduced that for every function we have $f^{-1}(A^{\complement})=f^{-1}(A)^{\complement}$.

Just note that the following statements are equivalent:

• $x\in f^{-1}(A^{\complement})$
• $f(x)\in A^{\complement}$
• $f(x)\notin A$
• $x\notin f^{-1}(A)$
• $x\in f^{-1}(A)^{\complement}$

Then in the context of your question the following statements are equivalent:

• $f^{-1}(A)$ is closed in $X$
• $f^{-1}(A)^{\complement}\in\tau_X$
• $f^{-1}(A^{\complement})\in\tau_X$
• $A^{\complement}\in\tau_f$
• $A$ is closed in $Y$