I want to prove the following:

Let $(Y,\tau)$ be a topological space and $X\neq \emptyset$ a set. Let $f$ be a function from $X$ to $Y$. Then, $\tau_1=\{f^{-1}(S): S\in \tau\}$ is a topology on $X$

My attempt:

  1. $\emptyset \in \tau$ and $Y\in \tau$. $\emptyset=f^{-1}(\emptyset)\in \tau_1$ and $X=f^{-1}(Y)\in \tau_1$

  2. Let $\{U_i: i\in I\}$ be an indexed family of elements of $\tau$. Then $$\bigcup_{i\in I}{f^{-1}(U_i)}=f^{-1}(\bigcup_{i\in I}{U_i})\in \tau_1$$

  3. Let $U_1,U_2\in \tau$. Then $f^{-1}(U_1) \cap f^{-1}(U_2)=f^{-1}(U_1\cap U_2)\in \tau_1$

Is it correct?

  • $\begingroup$ @Did: Not quite, strictly speaking. $\endgroup$ – Brian M. Scott Mar 1 '13 at 13:44

The essentials are all there, but there’s one thing that you ought to do differently in (2) and (3). I’ll use (2) as my example. You want to show that $\tau_1$ is closed under arbitrary unions, so you should start with an arbitrary family $\{U_i:i\in I\}\subseteq\tau_1$, not with a family in $\tau$. Now by the definition of $\tau_1$ you know that for each $i\in I$ there is a $V_i\in\tau$ such that $U_i=f^{-1}[V_i]$, so

$$\bigcup_{i\in I}U_i=\bigcup_{i\in I}f^{-1}[V_i]=f^{-1}\left[\bigcup_{i\in I}V_i\right]\in\tau_1\;,$$

since $\bigcup_{i\in I}V_i\in\tau$.


I am facing the same proof but I have a little concern about the statement that $f^{-1}(Y)\in \tau_1$ . Well, nothing is said about $f$ being surjective, so we should have to prove, indeed, that $f^{-1}(f(X))\in \tau _{1}$. Am I correct?

So, wouldn`t it be a necessary hypothesis that $f(X)\in \tau$ for us to prove that?

  • $\begingroup$ Since $f$ is a function from $X$, $f^{-1}(Y) = X$. Surjectivity is unnecessary. $\endgroup$ – KCd May 1 at 10:31

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.