# Definition of two homeomorphic spaces

1. Nov 26, 2013

### Streltsy

Given this definition of two homeomorphic spaces,

Deﬁnition 1.7.2. Two topological spaces X and Y are said to be homeomorphic if there are
continuous map f : X → Y and g : Y → X such that
f ° g = IY and g ° f = IX.

Suppose I know f and g are both continuous. Would it be safe to assume then, that if
f ° (g ° f) = f, X and Y are homeomorphic?

Here's my reasoning:
f ° (g ° f) = f implies g ° f = IX and due to the associativity of a composition,
f ° (g ° f) = (f ° g) ° f = f or f ° g = IY.

2. Nov 26, 2013

### jgens

Nope. Take X = [-1,1] and Y = (-1,1). These spaces are not homeomorphic since one is compact and the other is not. Now let f:X→Y be the constant function zero and let g:Y→X be any continuous function. Then f(g(f(x))) = f(x) = 0 for all x in X.

3. Nov 26, 2013

### fzero

This statement is untrue. In order for $g\circ f = I_X$, we must have $h\circ (g\circ f) = h$ for all continuous maps $h: X \rightarrow X$.

In addition to jgens example, we could consider $X=S^2$ and $Y=S^1$, with $f$ the map $(\theta,\phi)\mapsto \phi$ and $g$ the map $\phi\rightarrow (\pi/2,\phi)$. This illustrates the fact that it can be that $f\circ(g\circ f)=f$ even when the image of $g\circ f$ does not even include the whole space $X$.

4. Nov 26, 2013

### Streltsy

Sorry, I haven't gotten to the definition of compactness. Although I believe I understand why my assumption is wrong.

I have another question; though I'm not sure if it would belong in this section.

Suppose I have a function f(x,y) = 2x/(1-y), that maps the sphere (minus a pole), S1 = {(x, y) ∈ ℝ2 | x2+(y−1/2)2 = 1/4} \ {(0,1)} onto ℝ, and its inverse, f−1(x) = (2x/(x2 + 4), x2/(x2 + 4)).

How can I verify that f−1(x) ° f(x,y) is an identity map?
because, I know that f(x,y) ° f−1(x) returns x, and that is how I verified f ° f-1 was an identity map, but the other returns a coordinate which isn't just (x,y).

5. Nov 27, 2013

### fzero

This example is closely related to jgens' example and the spaces are not homeomorphic for the same reason. The circle is a compact manifold, but $\mathbb{R}$ is not. Compactness is a concept that needs some notions of topology to define, but here it boils down to the fact that we only need a finite number of open intervals to describe the circle. For $\mathbb{R}$, no finite number of open intervals can cover it, since the point at infinity is always outside of whatever finite collection of open intervals we come up with.

It is precisely the point $(0,1)$ where $f$ was not defined that is the problem (so $f$ is not continuous). Points on the circle approaching this point are mapped towards the point at infinity on $\mathbb{R}$. Since $f^{-1}\circ f$ is not defined everywhere on the circle, it cannot be the identity map.

There is a notion of "adding the point at infinity" to the real line, called the one-point compactification. The resulting space is homeomorphic to the circle.

6. Nov 27, 2013

### Streltsy

With respect to the continuity of f; doesn't the fact that f: S1 \ {(0,1)} → ℝ safeguard against discontinuity? The domain of f does not include y = 1 so, from my understanding, f-1 ° f would be defined everywhere in the domain of f.

Still, I suppose I have to arrive at the definition of compactness and attempt to understand it.

Also, the purpose of my question was rather to acquire a procedure, through which I can verify that if a function composition g ° f returns an ordered pair (x,y) for example, g ° f is an identity map.

7. Nov 27, 2013

### fzero

We can delete the point $(0,1)$, but then we are mapping an open interval to $\mathbb{R}$. Also, in either case, $f\circ f^{-1}$ is not defined at the point at infinity, since that point should be mapped to $(0,1)$.

It's very important to understand, especially for the intuition it gives in these cases. It tells us that the circle or open interval cannot be homeomorphic to the real line. Explicitly the obstruction is due to the behavior of $f$ and $f^{-1}$ at the special points and it's also very useful to see that directly.

Well this boils down to the behavior of the function and it's inverse at the special points. Also, I understand where it comes from in the construction, but it's important to remember that the ordered pair $(x,y)$ satisfies a constraint, so there is really a single parameter. In terms of the single parameter the circle is the interval $[0,1)$ where points are identified under $x \sim x +1$, while the domain of $f$ is homeomorphic to the interval $(0,1)$.

8. Nov 27, 2013

### jgens

I am confused. Both S1-* and R are homeomorphic (where * is any one-point set) but it looks like fzero is saying otherwise. Hopefully I am just misunderstanding something here.

9. Nov 27, 2013

### fzero

You're right. My logic above was faulty. The tangent function provides another homeomorphism from an open interval to the real line.

10. Nov 27, 2013

### Streltsy

Which of your claims should I still consider? (With all due respect).