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Let $(X,\mathcal T)$ be a compact Hausdorff topological space and $f:X\to X$ be one-to-one and continuous. Is $f$ surjective?

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Are you sure this is the question you meant to ask? Otherwise, take $X=[0,1]$ and $f(x)=x/2$. Also note that $f:X\longrightarrow X$ has automatically $f(X)\subseteq X$, whence $f^{n+1}(X)\subseteq f^n(X)$. – 1015 Apr 22 '13 at 18:04
yep, thanks. it is true in a cancellative topological semigroup when $f$ is multiplication by a constant. I just wanted to know if it has anything to do with it's algebraic structure. – user59671 Apr 22 '13 at 18:11
I am not sure what result in topological groups you mean. But for sure, any function $f:X\rightarrow X$ satisfies your containment condition for iterations. So your question amounts to asking whether any continuous injective function is surjective. That's true is $X$ is finite. – 1015 Apr 22 '13 at 18:15
when adding that condition I was remembering something in linear-algebra, when $T(A)\subseteq A$ then $T$ can be limited to $A$. yes, it is not needed in my question. – user59671 Apr 22 '13 at 18:21
I see. But now I realize the question could be: for which compact spaces does continuous+injective imply surjective? Other than the trivial finite case, I have no idea. This would be related to some indecomposability property I am not aware of. When does a compact space contain a proper copy of itself? Most likely this has been studied. Of course, this is a highly non-linear question. – 1015 Apr 22 '13 at 18:27

No. Consider $X=[0,1]$ and $f(x)=\dfrac x2$. This is clearly continuous, one-to-one and $f^{n+1}(X)\subseteq f^n(X)$, but it is certainly not surjective.

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If $X$ is metric (and compact), and if $d(f(x), f(y)) \geq d(x,y)$, then $f$ is (of course) injective, and you can show that $f$ is also surjective.

Hint :

Show that x is the limit of a subsequence of $(f^n(x))_{n \in \mathbb{N}}$. Which is, I suppose, the same idea of your proof when $X$ is a topological cancellative semigroup.

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+1) for the symbol >! – Paul May 2 '13 at 8:16

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