# Is This Another Definition of A Homeomorphism?

Suppose we have a map $$f: X \rightarrow Y$$ where $$X$$ and $$Y$$ are topological spaces. Also suppose that $$U$$ and $$V$$ are open subsets of $$X.$$

Suppose that $$V \supset U$$ if and only if $$f(V) \supset f(U).$$ Is this another way of saying that $$f$$ is a homeomorphism? If so, why?

• It’s not. Counterexamples aren’t hard to come by. – Randall Feb 24 at 16:29
• Does this property imply anything about the map? – Math N00b Feb 24 at 16:30
• I don't think this implies anything particularly nice without further assumptions on your spaces (i.e. separation axioms) – Aweygan Feb 24 at 16:36
• It may imply that the function is an open map. – Randall Feb 24 at 16:55
• This might be it. I now see that the map in question is bijective, continuous and open. The author is attempting to show that it is also a homeomorphism, and he bases that claim on this property. – Math N00b Feb 24 at 17:08

Define $$f$$ to be the identity map from $$\mathbb{R}$$ with the standard topology to $$\mathbb{R}$$ with the discrete topology. Then for any two sets (even if they are not open) $$U,V$$ we have $$U\subset V$$ if and only if $$f(U)\subset f(V)$$. But obviously $$f$$ is not a homeomorphism. It is not even continuous.