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.
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.