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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? –  Daniel Freedman Jan 17 '12 at 21:46
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Note: the empty set is open, so your condition is always trivially satisfied... –  Bruno Joyal Jan 17 '12 at 21:47

2 Answers 2

Even if you "don't count the empty set as open" this is wrong. For example the identity map on the set $M:=\{1,2,3,4\}$ is not continuous but satisfies your condition if we use the following topologies on $M$:

$M_1$: $\{1\},\{2,3\},\{4\}$ are a basis

$M_2$: $\{1,2\}$ and $\{3,4\}$ are a basis.

Now the identity map is a discontinuous map $M_1\rightarrow M_2$ satisfying your criterion.

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No. You need for $f^{-1}(A)$ to be open for any (necessarily open) set in the basis for the topology of $Y$. If you have this, then for any open set $U$ we have $U=\bigcup\limits_{i\in I} A_i$ for some collection of $A_i$ in the basis for $Y$, and so $$f^{-1}(U)=f^{-1}\left(\bigcup\limits_{i\in I} A_i\right)=\bigcup\limits_{i\in I} f^{-1}(A_i)$$ is open as the union of a collection of open sets is open.

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Checking that $f^{-1}[S]$ is open for every $S$ in some subbase for the topology of $Y$ suffices. –  Brian M. Scott Jan 18 '12 at 4:40
    
@BrianM.Scott That's true, but OP is working with a basis and it is certainly necessary for the preimages of basis elements be open. –  Alex Becker Jan 18 '12 at 5:06

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