1
$\begingroup$

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

  • 3
    Note: the empty set is open, so your condition is always trivially satisfied...2012-01-17

2 Answers 2

1

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.

0

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.

  • 0
    @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.2012-01-18