[proofplan] We prove both implications directly from the definition of continuity and the defining property of a basis. If $f$ is continuous, every open subset of $Y$ has open preimage, so this applies immediately to each basis element. Conversely, if preimages of basis elements are open, then every [open set](/page/Open%20Set) in $Y$ is a union of basis elements, and its preimage is the corresponding union of open subsets of $X$. [/proofplan] [step:Use continuity to prove openness of preimages of basis elements] Assume that $f: X \to Y$ is continuous. By definition of continuity, for every set $V \in \tau_Y$, the preimage $f^{-1}(V)$ belongs to $\tau_X$. Since $\mathcal{B}_Y \subset \tau_Y$, every $B \in \mathcal{B}_Y$ is an open subset of $Y$. Therefore, for every $B \in \mathcal{B}_Y$, we have $f^{-1}(B) \in \tau_X$. [/step] [step:Express an arbitrary open set as a union of basis elements] Assume that $f^{-1}(B) \in \tau_X$ for every $B \in \mathcal{B}_Y$. Let $V \in \tau_Y$ be arbitrary. Since $\mathcal{B}_Y$ is a basis for $\tau_Y$, define the indexing set \begin{align*} \mathcal{I}_V := \{B \in \mathcal{B}_Y : B \subset V\}. \end{align*} The basis property gives \begin{align*} V = \bigcup_{B \in \mathcal{I}_V} B. \end{align*} [guided] Assume that $f^{-1}(B) \in \tau_X$ for every basis element $B \in \mathcal{B}_Y$. To prove that $f$ is continuous, we must start with an arbitrary open set in the codomain and show that its preimage is open in $X$. Let $V \in \tau_Y$ be arbitrary. Because $\mathcal{B}_Y$ is a basis for $\tau_Y$, every point of the open set $V$ lies in some basis element contained in $V$. We collect exactly those basis elements by defining \begin{align*} \mathcal{I}_V := \{B \in \mathcal{B}_Y : B \subset V\}. \end{align*} Then the defining property of a basis implies \begin{align*} V = \bigcup_{B \in \mathcal{I}_V} B. \end{align*} This representation is the central reduction: instead of checking the preimage of an arbitrary open set directly, we rewrite that open set using the basis elements whose preimages are open by hypothesis. [/guided] [/step] [step:Take preimages and use closure of a topology under unions] Using the set identity from the previous step and the fact that preimages commute with arbitrary unions, we obtain \begin{align*} f^{-1}(V) &= f^{-1}\left(\bigcup_{B \in \mathcal{I}_V} B\right) \\ &= \bigcup_{B \in \mathcal{I}_V} f^{-1}(B). \end{align*} For each $B \in \mathcal{I}_V$, we have $B \in \mathcal{B}_Y$, so $f^{-1}(B) \in \tau_X$ by hypothesis. Since $\tau_X$ is closed under arbitrary unions, it follows that \begin{align*} f^{-1}(V) \in \tau_X. \end{align*} Because $V \in \tau_Y$ was arbitrary, $f$ is continuous. [guided] We now take preimages of the basis decomposition of $V$. The preimage operation commutes with arbitrary unions: for every $x \in X$, \begin{align*} x \in f^{-1}\left(\bigcup_{B \in \mathcal{I}_V} B\right) \iff f(x) \in \bigcup_{B \in \mathcal{I}_V} B \iff \exists B \in \mathcal{I}_V \text{ such that } f(x) \in B \iff x \in \bigcup_{B \in \mathcal{I}_V} f^{-1}(B). \end{align*} Therefore \begin{align*} f^{-1}(V) &= f^{-1}\left(\bigcup_{B \in \mathcal{I}_V} B\right) \\ &= \bigcup_{B \in \mathcal{I}_V} f^{-1}(B). \end{align*} For each $B \in \mathcal{I}_V$, the definition of $\mathcal{I}_V$ gives $B \in \mathcal{B}_Y$. Hence $f^{-1}(B) \in \tau_X$ by the hypothesis of this implication. Since $\tau_X$ is a topology on $X$, it is closed under arbitrary unions, so \begin{align*} \bigcup_{B \in \mathcal{I}_V} f^{-1}(B) \in \tau_X. \end{align*} Thus $f^{-1}(V) \in \tau_X$. Since $V \in \tau_Y$ was arbitrary, the preimage of every open subset of $Y$ is open in $X$, which is exactly the statement that $f$ is continuous. [/guided] [/step]