All open maps homeomorphisms?

[1MileCrash]

I was told to prove that

f: X -> Y is a homeomorphism iff it is an open map

While I see that if f is a homeomorphism, it is certainly an open map, but is the implication in the other direction even true? Because I see no reason to believe it is.

 

[WannabeNewton]

That is false. I suspect they meant a bijective continuous map $f$ is a homeomorphism iff it is open iff it is closed.

 

[Mandelbroth]

I was told to prove that

f: X -> Y is a homeomorphism iff it is an open map

While I see that if f is a homeomorphism, it is certainly an open map, but is the implication in the other direction even true? Because I see no reason to believe it is.

I think it should be "A bijective, continuous map $f:X\to Y$ is a homeomorphism iff it is an open map."

For fun, prove "A bijective, continuous map $f:X\to Y$ is a homeomorphism iff it is a closed map." :tongue:

Edit: WN got there first. Why you type so fast? :rofl:

 

[mathwonk]

this false without some hypotheses. if X is an open subset of Y, the inclusion map is open but not a homeomorphism.

if f is projection of the plane X onto the y - axis Y, this map is open but not a homeomorphism.

If X and Y are the same set equipped with different topologies so that Y has more open sets than X, then the identity map is open but not a homeomorphism.

etc...

 

[lavinia]

I was told to prove that

f: X -> Y is a homeomorphism iff it is an open map

While I see that if f is a homeomorphism, it is certainly an open map, but is the implication in the other direction even true? Because I see no reason to believe it is.

take a look at the open mapping theorem for complex analysis.

Let X be any space and Y a point. Map X to Y.

 

[1MileCrash]

I think it should be "A bijective, continuous map $f:X\to Y$ is a homeomorphism iff it is an open map."

For fun, prove "A bijective, continuous map $f:X\to Y$ is a homeomorphism iff it is a closed map." :tongue:

Edit: WN got there first. Why you type so fast? :rofl:

Oh, ok. But isn't that kind of trivial? If it is an open map then that is exactly the same (from what I can see) as saying that f-inverse is continuous. So basically that means I'm asked to show that f is a homeomorphism if f is continuous, f-inverse is continuous, and f is a bijection, but that's just what a homeomorphism is.