Definition of a homeomorphism between topological spaces

RedX

The definition of a homeomorphism between topological spaces X, Y, is that there exists a function Y=f(X) that is continuous and whose inverse X=f-1(Y) is also continuous.

Can I assume that the function f is a bijection, since inverses only exist for bijections?

Also, I thought that if a function f is continuous, then its inverse is automatically continuous? So why is there a need to mention that the inverse is also continuous in the definition I have above?

Also, a manifold is defined to be a topological space that is locally homeomorphic to R^n. So take the interval [a,b] on R. Would this be a manifold? [a,b] is compact so is not homeomorphic to R, but locally it can be done with an open covering of [a,b]. If there exists just one open covering where each of the open sets is homeomorphic to R^n, then is that sufficient to say the space is a manifold?

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Hurkyl

Staff Emeritus
Gold Member
Re: homeomorphism

The definition of a homeomorphism between topological spaces X, Y, is that there exists a function Y=f(X) that is continuous and whose inverse X=f-1(Y) is also continuous.

Can I assume that the function f is a bijection, since inverses only exist for bijections?
If there exists a function whose inverse is continuous, then the function must be invertible. Also, I thought that if a function f is continuous, then its inverse is automatically continuous?
Nope. If the inverse was continuous, then f would map open sets to open sets. Can you think of functions without that property?

Also, both f and f-1 would have to map connected sets to connected sets.

Also, a manifold is defined to be a topological space that is locally homeomorphic to R^n. So take the interval [a,b] on R. Would this be a manifold? [a,b] is compact so is not homeomorphic to R, but locally it can be done with an open covering of [a,b].
Try it. Obviously, the probems would be at a and b -- what open subset of [a,b] containing a is homeomorphic to R?

RedX

Re: homeomorphism

Nope. If the inverse was continuous, then f would map open sets to open sets. Can you think of functions without that property?
I can't think of one. The book mentions f(x)=x^2 on (-a,a) maps to [0,a^2) which is not open. However, the problem with this is that f(x)=x^2 is not a bijection, so it doesn't make sense to even speak about an inverse.

Can you have a continuous bijection whose inverse is not continuous?

dx

Homework Helper
Gold Member
Re: homeomorphism

Can you have a continuous bijection whose inverse is not continuous?
Sure. The function

$$\phi : [0,1) \rightarrow S^{1}$$

defined by

$$\phi(x) = e^{2\pi i x}$$

is a continuous bijection, but its inverse is not continuous. (S1 is the unit circle in the complex plane)

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Bacle

Re: homeomorphism

"Also, a manifold is defined to be a topological space that is locally homeomorphic to R^n. So take the interval [a,b] on R. Would this be a manifold? [a,b] is compact so is not homeomorphic to R, but locally it can be done with an open covering of [a,b]. "

Try finding neighborhoods ( I assume you are using the subspace topology)
of either a or b that are homeomorphic to R.

lavinia

Gold Member
Re: homeomorphism

[a,b] is not homeomorphic to R.

qspeechc

Re: homeomorphism

A homeomorphism is a continuous bijection which has a continuous inverse. If a function has an inverse then it is a bijection and vice versa. Recall the definitions of the two terms. A bijection is a map that is both surjective and injective. An inverse $f^{-1}$ of a function $f:X\rightarrow Y$ is another function $f^{-1}:Y\rightarrow X$ such that $f\circ f^{-1}$ is the identity map on Y and $f^{-1}\circ f$ is the identity on X. You then have to prove that a function is a biection if and only if it has an inverse.

Just because a bijection is continuous it does not mean its inverse is continuous. Consider the function $f:[0,1]\cup (2,3]\rightarrow [0,2]$ defined by

f(x) = x for x in [0,1], and
f(x) = x-1 for x in (2,3]

Then f is a continuous bijection, but its inverse is not continuous (check these statements).

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RedX

Re: homeomorphism

qspeechc said:
f(x) = x for x in [0,1], and
f(x) = x-1 for x in (2,3]

That's a good one. So the point {1} is actually an open set on the domain: [0,1]$\cup (2,3]$. So if you take an open set on this domain to be say {1}$$\cup(2,2.5)$$ then it maps to [1,1.5) which is not open on the codomain [0,2]. So I guess this is an example of where Hurkyl said that f has to map connected sets to connected sets.

dx said:
$$\phi : [0,1) \rightarrow S^{1}$$
defined by
$$\phi(x) = e^{2\pi i x}$$
This is an example where the mapping is from a connected set to connected. You have a definite boundary point x=0 in the domain that will correspond to a definite point on S^1, so f([0,a))=not open, unless a=1.

Bacle

Re: homeomorphism

Lavinia wrote:
"[a,b] is not homeomorphic to R. "

True, but I think the OP was talking about manifolds, and [a,b] is an example
of a non-manifold, since neither a nor b have neighborhoods homeomorphic to R.

Jamma

Re: homeomorphism

I can't think of one. The book mentions f(x)=x^2 on (-a,a) maps to [0,a^2) which is not open. However, the problem with this is that f(x)=x^2 is not a bijection, so it doesn't make sense to even speak about an inverse.

Can you have a continuous bijection whose inverse is not continuous?
Surjectivity is never a problem; we could have just assumed that the map was only to the image of the map i.e. [0,a^2) with the subspace topology.

For me, I always think of homeomorphisms as maps between things which have the same shape. All a bijection is is a map between sets which associates each point to a new distinict point. So a bijective continuous map which is not a homeomorphism maps every point of some object into another one, but alters its shape somehow.

So these will usually be mappings which take the boundary points of the old object to the interior of the new one, such as wrapping something up, or twisting a line into a circle for example.

TMM

Re: homeomorphism

The trivial function (R->{0}) maps (non-empty) open sets to closed sets, but the preimage of every open set is open, namely the empty set.

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