Continuous Bijection f:X->X not a Homeo.

Continuous Bijection f:X-->X not a Homeo.

Hi, All:

A standard example of a continuous bijection that is not a homeomorphism is the

map f:[0,1)-->S^1 : x-->(cosx,sinx) ; for one, S^1 is compact, but [0,1) is not,so

they cannot be homeomorphic to each other.

Now, I wonder if it is possible to do this for a continuous bijection of a space to itself,

(with different topologies if necessary) and, if it is possible from a space with itself ,

but a map g: (X,T)-->(X,T) , i.e., with the same topology for domain and codomain.

Yes. Let X be the integers with the topology created by declaring a set to be open if either it is a subset of $\mathbb{N}$ or it is the entire set $\mathbb{Z}$. Then the map $f:X \rightarrow X : x \mapsto x-1$ is a continuous bijection from X to itself, but it is not a homeomorphism, since the image of $\mathbb{N}$ is not open.

There are several other examples given on this stack exchange topic: http://math.stackexchange.com/quest...bijections-of-connected-spaces-homeomorphisms

mathwonk
Homework Helper

changing the topology means it is no longer the same space. Take any set at all with more than one point and give it the topology with only two open sets, (the empty set and the whole space), and then give it the topology where every set is open. You can probably figure out which way the map is continuous.

It does seem however that every continuous bijection of a finite space with itself isa homeomorphism.

I am a bit confused; {1} is an open set , as a subset of N, but its preimage

is {0}, which is not, so I don't see how f is continuous.

Mathwonk: I guess the fewer the open sets, the smaller the chances

that some property will fail.

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The preimage of {1} is {2}, not {0}.

It is quite easy to see that we can find such a map when we change the topology, but then the space doesn't really have anything to do with the other one, except having the same cardinality of its set of points. The question when the topology is the same though is interesting.

"It does seem however that every continuous bijection of a finite space with itself isa homeomorphism."

Every bijective map from a compact space to a Hausdorff space is a homeomorphism. So we need one that is either non-compact or non-Hausdorff.

Usual examples of bijective maps that are not homeomorphisms involve "curling an infinity around" to join onto itself and form a loop, or some distinctive topological feature; maybe there is an example of a space where we can do this onto itself?

We have an infinite wedge of circles and real lines (all wedged at the same point). Let's say countably infinite.

Our map maps the first circle to the second, the second to the third..... our 2nd line to the first, our 3rd to the 2nd.... and out 1st line wraps around the 1st circle.

I reckon this will do the trick!

Sorry, I meant to say "half real lines" i.e. [0,\infty). So we have an infinite wedge of circles with an infinite number of infinitely long "sticks" coming out (haha!). We can curl one of the lines around one of the circles, like in your original example of a bijective continuous map which isn't a homeomorphism, and shift all the rest so that the map of the whole space is bijective and continuous onto itself, but the inverse map won't be continuous, since we are "splitting apart" close points on one of the circles.

it's interesting to find a pair X,Y of homeomorphic topological spaces and a continous biyection f:X -> Y that is not homeomorphism. An answer to this question is the following:

Consider $X=Y= \mathbb{R} - \cup_{\text{k odd positive}} \left[k,k+1 \right)$ and $f:X\rightarrow X$ diven by

$$f(x)=\begin{cases} x/2 & \text{if } x \in \left[ 0,1 \right) \\ (x-1)/2 & \text{if } x\in \left[ 2,3 \right) \\ x-2 & \text{otherwise} \end{cases}$$

You can see that f is continous in X, and f^-1 is not in x=1/2.

That's a really nice example! (and probably a bit easier to visualise than mine).