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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.

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- Thread starter 1MileCrash
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In summary, the statement "f: X -> Y is a homeomorphism iff it is an open map" is true, as long as f is a bijective, continuous map. This can be proven by using the open mapping theorem for complex analysis and noting that if f is continuous and f-inverse is continuous, then f is a homeomorphism.

- #1

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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.

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1MileCrash said:

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

For fun, prove "A bijective, continuous map ##f:X\to Y## is a homeomorphism

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

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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...

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1MileCrash said:

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.

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Mandelbroth said:I think it should be "A bijective, continuous map ##f:X\to Y## is a homeomorphismiffit is an open map."

For fun, prove "A bijective, continuous map ##f:X\to Y## is a homeomorphismiffit 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.

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