# F is an open mapping implies f inverse cont.

1. Nov 21, 2008

### Unassuming

We have a continuous bijection f:X-->Y.

Prove that if f is open, then f inverse is continuous.

I can't figure it out.
"Proof". For V open in Y, there exists W open in X such that $$f[W] \subseteq V$$. Where does the f is open definition apply?

2. Nov 21, 2008

### Unassuming

Ah! I got it. Posting always helps the blood get flowing. Trying to explain what I know to others.