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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

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Um: what do you mean by "homeomorphic"? The word "homeomorphic" is an adjective describing a relationship between two objects. So it doesn't make sense to say that "[A] function ... is homeomorphic." A function between two objects may be a "homeomorphism", if through it we can see that the two objects are homeomorphic. But given the third paragraph of your question, perhaps the adjective you are looking for is "continuous"? –  Willie Wong Sep 12 '11 at 13:09
    
It is not true that every map $f:X\rightarrow T $ with $X$ discrete is a homeomorphism; for one thing, the discrete topology is metrizable, so Hausdorff, and Hausdorff is a topological property, so if $T$ has any non-Hausdorff topology, then $f:X\rightarrow T$ cannot be a homeomorphism. –  gary Sep 12 '11 at 16:52
    
People seem to be going to an awful lot of trouble to describe situations where $f:X\rightarrow T$ cannot be a homeomorphism when $X$ is discrete... isn't "T does not have the discrete topology" sufficient? –  MartianInvader Sep 12 '11 at 18:15
    
Well, I gave specific examples because I like to see specific examples in areas I'm not familiar with, and I imagined others in similar conditions would too. –  gary Sep 12 '11 at 18:22
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I am speculating here, but based on the OP's last paragraph I think he is confusing the terms "continuous function" and "homeomorphism". –  Miha Habič Sep 12 '11 at 20:03

2 Answers 2

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.

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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.

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