Checking My Topological Result: Is f^{-1}(S') \subset T?

[quasar987]

I just discovered the following. But since half the things I find in topology turn out to be wrong, I feel I better check with you guys.

What I convinced myself of this time is that if you have a function f:(X,T)-->(Y,S) btw topological spaces, and S' is a basis for S, then to show f is continuous, is suffices to show that $f^{-1}(S') \subset T$.

In words, that is because every open set of S can be written as a union of sets of S', and the operations of f^{-1} and union commute. (and that a union of open sets is open)

 

[Hurkyl]

That looks right. In fact, it's rather often used -- e.g. look at the definition of continuity you learned in calc I.

 

[SiddharthM]

it's a fact!

 

[quasar987]

Thanks.