The Stacks project

Proof. The set $\mathcal{T} = \{ f^{-1}(U) | U \subset X \text{ open}\}$ is a topology on $Y$. Firstly, $\emptyset = f^{-1}(\emptyset )$ and $f^{-1}(X) = Y$. So $\mathcal{T}$ contains $\emptyset$ and $Y$.

Now let $\{ V_ i\} _{i \in I}$ be a collection of open subsets where $V_ i \in \mathcal{T}$ and write $V_ i = f^{-1}(U_ i)$ where $U_ i$ is an open subset of $X$, then

So $\bigcup _{i\in I} V_ i \in \mathcal{T}$ as $\bigcup _{i\in I} U_ i$ is open in $X$. Now let $V_1, V_2 \in \mathcal{T}$. We have $U_1, U_2$ open in $X$ such that $V_1 = f^{-1}(U_1)$ and $V_2 = f^{-1}(U_2)$. Then

So $V_1 \cap V_2 \in \mathcal{T}$ because $U_1 \cap U_2$ is open in $X$.

Any topology on $Y$ such that $f$ is continuous contains $\mathcal{T}$ according to the definition of a continuous map. Thus $\mathcal{T}$ is indeed the weakest topology on $Y$ such that $f$ is continuous. This proves that (1) and (2) are equivalent.

The equivalence of (2) and (3) follows from the equality $f^{-1}(X \setminus E) = Y \setminus f^{-1}(E)$ for all subsets $E \subset X$. $\square$