Originally Posted by

**hatsoff** Thanks, but in your sketch you assume that $\displaystyle g(S)$ is open, which may not be the case.

I didn't mean but that's what must be deduced from my typos. I assume $\displaystyle g(S)$ that it is the union (oops! and I should have written *disjoint *union) of two open sets in $\displaystyle g(S)$ (another typo here), which means both $\displaystyle U,V$ are the intersection of $\displaystyle g(S)$ with open subsets of $\displaystyle Y$, and NOT necessarily open in $\displaystyle Y$. Sorry for that.

Tonio

If on the other hand we say simply that $\displaystyle g(S)\subseteq U\cup V$, then I won't be able to show that $\displaystyle g^{-1}(U)$ and $\displaystyle g^{-1}(V)$ are open.

Or am I misunderstanding something---always a possibility!---?