= Solution
A <basis of a topology>[basis for a topology] is a family of open sets such that every <open set> is a <union> of members of the family. Equivalently, for every point in an <open set> there is a basis member containing that point and contained in the <open set>.
For a basis $\mathcal B$ on $Y$, <continuity> is equivalent to \b[$f^{-1}(B)$ being open for every $B\in\mathcal B$]. Necessity follows because basis sets are open. Conversely, taking preimages of a <union> of basis sets gives a <union> of open preimages, so every <open set> has an open <preimage>.
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