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Prove that every continuous one-to-one mapping of a compact space is topological.

Does this problem statement refer to a mapping of a compact space to itself?

If so, suppose the mapping is $f:A\rightarrow A$, where $A$ is compact and $f$ is continuous and 1-1. We must show that $f$ is also onto (and hence bijective), and the inverse image $f^{-1}$ is continuous.

For any compact subset $B\subseteq A$, the set $f(B)$ is compact. Since closed subsets of a compact set are compact, for any closed subset $B\subseteq A$ , the set $f(B)$ is compact (and so, closed and bounded). How might that help?

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  • $\begingroup$ What does it mean that a mapping is topological? Did you mean to say embedding? $\endgroup$ – Martin Sleziak Jul 7 '15 at 9:43
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This means that if $f:A\to B$ is continuous and one-to-one, and if $A$ is compact, then $f$ maps $A$ homeomorphically to $f[A]$. As written, though, it is incorrect. We need to know that $B$ has the property that compact subsets are closed. That's what will allow the proof to work.

It should be clear that such an $f$ maps $A$ bijectively and continuously to $f[A]$ (considered as a subspace of $B$). It remains to show, then, that $f$ maps closed subsets of $A$ to (relatively) closed subsets of $f[A],$ since this is equivalent (why?) to showing that $g:f[A]\to A$ given by $g(x)=f^{-1}(x)$ is a continuous function.

As a side note, "bounded" doesn't necessarily make sense for general spaces. It does suggest that you may be working with metric spaces, however, rather than arbitrary topological spaces. If that is the case, then there's no need to make the additional assumption that compact subsets of $B$ are closed, because this is always true in metric spaces.

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Let $f:[0,1]\to [0,1]$ be defined as $f(X) = X/2$. Then $f$ is continuous and one-one but not onto.

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This question comes from an exercise (Ex 3, pg 66) of Ahlfor "complex Analysis, 2nd edition".

In Rudin "Principle of Mathematical Analysis, 3rd ed" there is a theorem (Theorem 4.17, page 90) state that:

Suppose $f$ is a continuous $1$-$1$ mapping of a compact metric space $X$ onto a metric space $Y$. Then the inverse mapping $f$ inverse is a continuous mapping of $Y$ onto $X$.

According to this theorem we need onto in the statement of the exercise or just show for the function $f\colon A \to f[A]$.

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So there is a general result from Hall and Spencer's Elementary Topology, pg. 76, to the effect that a 1-1 continuous map from a general compact space onto a Hausdorff space is a homeomorphism. For example. a Cantor perfect set is not homeomorphic with a bounded closed interval of the real line. That's because a continuous map of a connected set is comnected. A fortiori there is no 1-1 continuous map of a Cantor set onto a closed interval.

Submitted by Edgar Cohen, Ph. D.

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