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Let $X$ be a compact metric space and let $f:X\to X$ be a continuous injective map. Which of the following is true?

a) $f(X)$ is dense in $X$

b) $X$ and $f(X)$ are homeomorphic.

c) There exist $x\in X$ such that $f(x)=x$

For a) $X=[0,1]$ defined by $f(x)=\frac{x}{2}$

c) Is Continuous function on a compact set to compact set have a fixed point? b) Continuous function on a compact set is compact. Is the image homemomorphic to $X$?

THIRUMAL 5688
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    $(c)$ is false, consider $f:\Bbb S^1\to \Bbb S^1$ given by $f(z)=-z$ for all $|z|=1$. $(b)$ is true, see here why $X\simeq f(X)$ https://math.stackexchange.com/questions/2578558/why-isnt-a-continuous-bijection-from-a-locally-compact-space-to-a-hausdorff-spa – Sumanta Sep 30 '20 at 03:55
  • What are you saying for c)? For b) you are not asking the right question yet. What should that question be? – Ted Shifrin Sep 30 '20 at 03:55
  • @Thiru Thirumal $(a)$ is certainly not true, see your own example. – Sumanta Sep 30 '20 at 03:59

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Answer for b): YES. $X$ and $f(X)$ are compact metric spaces. The inverse is automatically continuous: If $U$ is open in $X$ then $X\setminus U$ is closed, hence compact. So the image $f(X\setminus U)$ is compact (by continuity) hence closed. Its complement is $f(U)$ so $f(U)$ is open. This proves that $f^{-1}$ is continuous.

Kavi Rama Murthy
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