For help, A simple question about topology

[Xian'an Jin]

Is a continuous 1-1 and onto mapping from the Euclidean 3-space to itself a homeomorphism? i.e. Is its inverse also continous?

 

[waht]

Is a continuous 1-1 and onto mapping from the Euclidean 3-space to itself a homeomorphism? i.e. Is its inverse also continous?

Yes, a function is bijective if it is a continuous 1-1 and onto mapping. If a function is bijective it's easy to show it has an inverse. A homeomorphism is basically a bijection between topological spaces. Since euclidean space is a topological space, it is a homeomorphism.

 

[morphism]

Yes, a function is bijective if it is a continuous 1-1 and onto mapping. If a function is bijective it's easy to show it has an inverse. A homeomorphism is basically a bijection between topological spaces. Since euclidean space is a topological space, it is a homeomorphism.

A function is bijective if it's 1-1 and onto. That's all - it has nothing to do with continuity. A homeomorphism is a continuous bijection whose inverse is continuous as well. A continuous bijection between topological spaces may as well have an inverse that isn't continuous.

In calculus it's usually proved that a continuous bijection f:R->R is a homeomorphism. I don't think this generalizes to R^3 though. IIRC, there's a theorem that says if f:R^n->R^n is a continuous bijection that has first-order partials and nonzero jacobian, then its inverse is continuous.

 

[waht]

A function is bijective if it's 1-1 and onto. That's all - it has nothing to do with continuity. A homeomorphism is a continuous bijection whose inverse is continuous as well. A continuous bijection between topological spaces may as well have an inverse that isn't continuous.

Yes, that's it. My error is due to looking at the question and retyping some of it too quickly, sorry.

 

[Crosson]

In general a continuous bijection with a compact domain will have a continuous inverse.

 

[DeadWolfe]

In calculus it's usually proved that a continuous bijection f:R->R is a homeomorphism.

What theorem is this? I don't recall it from calculus.

 

[morphism]

What theorem is this? I don't recall it from calculus.

It usually comes before the inverse function theorem. It's probably stated as: a function that is continuous and 1-1 on an interval has a continuous inverse.

 

[mathwonk]

for continuous bijective maps from R to R it is easy, since they are monotone.

for R^2 it isn't coming to me immediately yet. i think you have to prove the map is proper somehow, if it is.

 

[mathwonk]

ok. let p map to q (R^2-->R^2). want to show f^(-1) continuous at q.

take a disc centered at p, and by jordan curve theorem, the circular boundary maps to a closed loop that has q in its interior.

thus the image of the compact disc around p is a compact "disc" around q, with q in the interior.

then the continuous inverse of the restriction of the map to the compact disc, is also continuous as a map on all of R^2.

so yes.

whew! what about R^3? I guess th same is true, i.e. the image of a ball around p should map to a ball around q, and use the same argument.

so this seems to use the jordan theorem to argue that the image of a sphere separates R^3 into an inside and an outside. (provided the map was defined and bijective on all of R^3.)

i hope this is true. look up the generalized jordan curve theorem. [yep, hatcher, p.169.]

so this question is simple but difficult.

 

[haiha]

Sorry, delete it. Thanks

 

[Xian'an Jin]

In general a continuous bijection with a compact domain will have a continuous inverse.

I know it, a contiunous bijection with a compact domain and a Hausdorff range is a homeomorphism. Unfortunately, here R^3 is not compact.

 

[Xian'an Jin]

In calculus it's usually proved that a continuous bijection f:R->R is a homeomorphism. [/QUOTE]

I think it can be proved by the following steps:
step 1: show that f is monotone
setp 2: a monotone continuous function has continouse inverse.

Is the sketch of proof right? and can you provide a detailed proof of this proposition.

 

[Xian'an Jin]

What is a Jordan curve

ok.
take a disc centered at p, and by jordan curve theorem, the circular boundary maps to a closed loop that has q in its interior.

Let C be the circle x^2+y^2=1 in R^2, f is continuous bijection from R^2 to R^2, is f(C) a Jordan curve? and what is a Jordan curve? what is its definition?