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It is known that any compact metric space $Y$ is a continuous image under some surjective map $$f: C \to Y,$$

where $C$ is the Cantor set. I want to prove that if $f$ maps $C$ to $([0,1],|\cdot|)$, then $f$ cannot be bijective. So I think the idea is to show that there is some $y\in [0,1]$ such that the pullback of $y$ under $f$, i.e. $f^{-1}(y)$ contains more than one element.

If we look at a proof of the fact that any compact metric space $Y$ is a continuous image under some surjective map $f: C \to Y$, then one approach is to subdivide $Y$ into infinitely many compact sets of diameter zero and then use a ternary sequence mapped by $f$ into $Y$. But I don't yet see what to extract from this.

Is it best to prove this by contradiction? Would appreciate some hints.

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    $\begingroup$ You seem to be leaving out the word "continuous" when discussing these maps. $\endgroup$ – zhw. Nov 17 '17 at 4:04
  • $\begingroup$ $C$ is defined iteratively (its elements are leafs at the end of an infinite binary tree), thus you need to define $f(C)$ iteratively. Keeping track of $\inf, \sup f(C_i)$ could help. $\endgroup$ – reuns Nov 17 '17 at 4:06
  • $\begingroup$ @zhw. Thanks, updated. $\endgroup$ – sequence Nov 17 '17 at 4:12
  • $\begingroup$ @zhw.: you changed the title to a phrase that makes no sense. The image of something can't be bijective. The surjection that realizes that image can. $\endgroup$ – symplectomorphic Nov 17 '17 at 5:07
  • $\begingroup$ @symplectomorphic Yes you're right. Thanks. Is it better now? $\endgroup$ – zhw. Nov 17 '17 at 5:23
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A continuous bijection from a compact space (which the Cantor set is) to a Hausdorff space (which the standard unit interval is) is a homeomorphism.

But $C$ isn't connected and the unit interval is.

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