**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

\[ \bigcup \nolimits _{i\in I} V_ i = \bigcup \nolimits _{i\in I} f^{-1}(U_ i) = f^{-1}\left(\bigcup \nolimits _{i\in I} U_ i\right) \]

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

\[ V_1 \cap V_2 = f^{-1}(U_1) \cap f^{-1}(U_2) = f^{-1}(U_1 \cap U_2) \]

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 $E \subset X$.
$\square$

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