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

Let $(V_ i)_{i \in I}$ be a family of elements $V_ i \in \mathcal{T}$. Then

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

Thus $\bigcup _{i\in I} V_ i \in \mathcal{T}$ as $\bigcup _{i\in I}f^{-1}(V_ i)$ is open in $X$. Furthermore if $V_1, V_2 \in \mathcal{T}$ then

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

So $V_1 \cap V_2 \in \mathcal{T}$ because $f^{-1}(V_1) \cap f^{-1}(V_2)$ is open in $X$.

Finally a topology on $Y$ such that $f$ is continuous is included in $\mathcal{T}$ according to the definition of a continuous function, so $\mathcal{T}$ is the strongest topology on $Y$ such that $f$ is continuous. It proves that (1) and (2) are equivalent.

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

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