# An example: continuous map which is not homeomorphic

Reference: Topology by Hocking and Young. Dover Publications.

Let us present an example of a continuous mapping (one-to-one) which is not a homeomorphism:

Let S be the set of all non-negative real numbers with their usual metric topology, and let T be the unit circle in its metric topology. Consider $f: S \rightarrow T$. For each x in S, let $f(x) = (1, \frac{2\pi x^{2}}{(1+x^{2})})$, a point in polar coordinates on T.

It is easily shown that f is continuous and one-to-one.

But the set of all x in S such that $x<1$ is open in S while its image is not open in T. Hence, f is not interior. (meaning that: A transformation $f: S \rightarrow T$ of the space S into the space T is said to be interior if f is continuous and and if the image of every open subset of S is open in T) , and is not a homeomorphism (because of the following theorem: A necessary and sufficient condition that the one-to-one continuous map $f: S \rightarrow T$ of the space S onto the space S be a homeomorphism is that f be interior).

Regards,

Nalin Pithwa.