Homeomorphism $f : (–1,1) \rightarrow \mathbb{R}$

Question

Working through "Topology Without Tears", and Sidney Morris defines homeomorphism on p 91:

Let $(X,T)$ and $(Y,T_1)$ be topological spaces. Then they are said to be homeomorphic if there exists a function $f: X \rightarrow Y$ which has the following properties:
(i) $f$ is one-to-one (that is $f(x_1)=f(x_2)$ implies $x_1=x_2$)
(ii) $f$ is onto (that is, for any $y\in Y$ there exists an $x\in X$ such that $f(x)=y$)
(iii) for each $U \in T_1, f^{-1} (U) \in T$, and
(iv) for each $V \in T, f(V) \in T_1$

For an example, there are functions that are homeomorphic from $(-1,1) \rightarrow \mathbb{R}$, like $f(x)=e^x – e^{–x}$ and $f(x)=ln(1+x) – ln(1–x)$. Alternatively, the map $f(x)=\frac{x^2}{1-x^2}$ does not allow for a homeomophism as it is not onto (ii from the definition).

Graphically, the map that doesn't allow for the homemorphism looks like a parabola (even) in that domain while the map that does allow for the homeomorphism looks odd.

Are my functions (and logic) respectively correct? Thanks.

Answer 1

If you're looking for homeomorphisms of $\mathbb{R}$, then certainly even maps are a no-no, because they violate injectivity.

That said, an homeomorphism of $\mathbb{R}$ need not have some special kind of symmetry (so it need not 'look odd'). Any continuous, strictly monotonic map on $(-1,1)$ which blows up near each endpoint defines an homeomorphism of $\mathbb{R}$.

Answer 2

Observe that if $(X_1,T_1)$ is homeomorphic to $(X_2,T_2)$ and $(X_2,T_2)$ is homeomorphic to $(X_3,T_3)$ then $(X_1,T_1)$ is homeomorphic to $(X_3,T_3).$

The interval $(0,1)$ is homeomorphic to $(-1,1).$ E.g. let $g(x)=2x-1.$ A geometric way to get a homeomorphism from $(-1,1)$ to $\mathbb R$ is to homeomorphically "bend" $(-1,1)$ to the semi-circle $C=\{(x,y)\in \mathbb R^2: |x|<1\land y<1\land x^2+(y-1)^2=1\}....$

... And now map $C$ to $\mathbb R$ by (i). $f((0,0))=0.$ (ii). For $(0,0)\ne p\in C,$ let the line thru $p$ and $(1,0)$ (which is the center of $C$) meet the real axis at $(f(p),0).$ Then $f:C\to \mathbb R$ is a homeomorphism.

The most common textbook example of a homeomorphism from $(-1,1) \to \mathbb R$ is probably $f(x)=\tan (\pi x/2)$.

In general $f:(-1,1)\to \mathbb R$ is a homeomorphism iff $f$ is continuous, strictly monotonic, and $\infty =\lim_{|x|\to 1}|f(x)|$ .