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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?

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@Did: Not quite, strictly speaking. – 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 $\bigcup_{i\in I}V_i\in\tau$.

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