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I'm going through a fairly involved proof in Algebraic Topology, and am stumbling at the last hurdle because my point-set topology is rusty.

Suppose I have a map $f : Z \to Y$, where $Y$ and $Z$ are topological spaces. If I've shown that $f^{-1}(A)$, where $A$ is any open set in a basis for the topology on $Y$, contains a set open in $Z$, does it follow that $f$ is continuous?

Thanks

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    This looks a bit like the neighbourhood definition of continuous: $f$ is continuous at a point $z \in Z$ iff for any neighbourhood $V$ of $f(z)$, there is a neighbourhood $U$ of $z$ such that $f(U) \subseteq V$, but not quite. Have you written the details correctly?2012-01-17
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    Note: the empty set is open, so your condition is always trivially satisfied...2012-01-17

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