Are all continuous bijections homeomorphisms?

[lugita15]

If f is a continuous bijection from a metric space M to a metric space N, is the inverse function of f necessarily continuous?

My intuition tells me no; it seems like there could be a continuous bijection f whose inverse exists but is not continuous. Intuitively, I see two spaces being homeomorphic as a stronger condition than the existence of a continuous bijection between them.

I tried proving that the inverse of f must be continuous, but I couldn't.

Any help would be greatly appreciated.
Thank you in advance.

 

[QuanPenguin]

No, take f : R discrete -> R Euclidean which maps x -> x, then f is obviously bijective, and it's continuous because every set is open in the domain, but for the inverse function, the pre-image of a singleton set is a singleton set which is not open in the euclidean topology.

 

[Landau]

The standard counter-example is the map [0,1)->S^1 given by $t\mapsto \exp(2\pi i t)$.

 

[Anthony]

You might like to prove the following: let $$X$$ be compact and $$Y$$ Hausdorff, then if $$f:X\rightarrow Y$$ is bijective and continuous, then it is a homeomorphism.

 

[lugita15]

The standard counter-example is the map [0,1)->S^1 given by $t\mapsto \exp(2\pi i t)$.

Is that a bijection? Isn't this function periodic, and hence not one-to-one?

 

[Landau]

Instead of asking, try to prove it yourself. To prove that it would not be injective, you need to find distinct t_1 and t_2 in [0,1) such that f(t_1)=f(t_2) (if I call the function f). Can you find them?

 

[lugita15]

Instead of asking, try to prove it yourself. To prove that it would not be injective, you need to find distinct t_1 and t_2 in [0,1) such that f(t_1)=f(t_2) (if I call the function f). Can you find them?

f(0)=1, f(1)=1

 

[Landau]

With [0,1) I meant the interval EXcluding 1, i.e.

$$[0,1):=\{x\in\mathbb{R}: 0\leq x<1\}.$$

 

[lugita15]

With [0,1) I meant the interval EXcluding 1, i.e.

$$[0,1):=\{x\in\mathbb{R}: 0\leq x<1\}.$$

Oh sorry, I didn't notice you had restricted the domain.

 

[Landau]

Well, that is the point of this example: the inverse of this map sends the point 1 (of the complex unit circle S^1) to 0, and and the point just 'below' 1 to 1-e for small e. In other words, the inverse 'jumps' from values arbitrarily close to 1, to 0. Hence is not continuous.

 

[mathwonk]

there are several standard examples where bijective morphisms are isomorphisms: vector spaces, banach spaces, compact hausdorff spaces, groups, rings, normal algebraic varieties.

These give rise to some examples of bijective maps which may not be isomorphisms: singular algebraic curves, and arbitrary topological spaces.

 

[lugita15]

there are several standard examples where bijective morphisms are isomorphisms: vector spaces, banach spaces, compact hausdorff spaces, groups, rings, normal algebraic varieties.

These give rise to some examples of bijective maps which may not be isomorphisms: singular algebraic curves, and arbitrary topological spaces.

I don't really understand what you're saying, although I gather it has something to do with category theory. Anyway, does what you said lead to additional examples of continuous bijections between metric spaces whose inverses are not continuous?