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I am currently reading an introduction to topological and metric spaces and want to know whether the following statement is true:
Consider the Euclidean space $\mathbb{R}^n$ endowed with the Euclidean metric. Any function that maps an open ball in $\mathbb{R}^n$ to another open ball is homeomorphic.
It is clear to me that any function $f:X\rightarrow T$, with $X$ the discrete topology and T an arbitrary topology, is homeomorphic. Is the beforementioned statement somehow linked to this?
Thx
Consider the map $z \mapsto z^2$ in $\mathbb C = \mathbb R^2$. This maps the open unit ball continuously onto itself but is not a homeomorphism because it is not injective.
I think you may be referring to invariance of domain http://en.wikipedia.org/wiki/Invariance_of_domain , which says that if $U$ is ( U are?) an injective and continuous, then $U$ and $f(U)$ are homeomorphic.
And your second statement that any map $f: X\rightarrow T$ where $X$ is discrete, and $T$ has any topology is a homeomorphism , can fail in many ways: i)The discrete topology is totally-disconnected: take any p,q in your space X: then {p} and {X-p} is a disconnection. So if your $T$ is a connected topology, then f cannot be a homeomorphism. ii)Hausdorff: The discrete metric is metrizable, using the discrete metric $d(x,y)=1$ if $x \neq y$, and $0$ otherwise. iii)Metrizability: by metrizability of the discrete metric, if $T$ is not metrizable, $f$ cannot be a homeomorphism.
And then there is the fact that the underlying spaces need to have the same cardinality for the map to be a homeomorphism.
