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Is there a continuous onto function $f:\Bbb{D} \rightarrow [-1,1]$ ,

where $\Bbb{D}$ is a closed unit disk in $\Bbb{R}^2$ ?

I think the answer is no, otherwise $f(\Bbb{D})=[-1,1]$, so removing one point in the domain is still connected but the image is not. Am I right? Any hint?

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    $\begingroup$ Try $f(x,y) = x.$ $\endgroup$ – zhw. Jul 27 '18 at 5:40
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    $\begingroup$ Because you don't assume $f$ to be injective, "removing one point" in $[-1,1]$ may make you remove way more than just one point in $\mathbb{D}$. It could for instance make you remove a diameter of the disk, which would then not be connected anymore, and your argument would fail. $\endgroup$ – Suzet Jul 27 '18 at 5:42
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Your argument fails.

The problem is, that because $f$ is not injective, we cannot assert that since $f(\mathbb D) = [-1,1]$, we have for all $x \in [-1,1]$ that $f(\mathbb D \backslash \{y\}) = [-1,1] \backslash \{x\}$ for some $y$. Indeed, what may happen is that many points of $\mathbb D$ might map to $x$, and when we remove those points, we may get exactly two connected components in whatever remains, thus preserving the number of connected components.

As you have seen, the first projection serves as an onto map.


If we assert that $f$ is injective, then your argument does work, because then we can say this : there is unique $x$ such that $f(x) = 0$, so we must have $f(D \setminus \{x\}) = [-1,1] \setminus \{0\}$, and this is contradiction because $D \setminus \{x\}$ is connected regardless of what $x$ is , but the image is disconnected.


In fact, if $f$ is injective, then since $\mathbb D$ is compact and $[-1,1]$ is Hausdorff, then $f$ is actually a homeomorphism, from a common result.

Note that $\mathbb D$ is a subset of $\mathbb R^2$ and $[-1,1]$ is a subset of $\mathbb R$. Because $2 \neq 1$, we are inclined to believe that a homeomorphism is not possible.

This result, called the invariance of domain, holds in more generality : if a set in $\mathbb R^m$ is homeomorphic to some other set in $\mathbb R^n$, then $m=n$ must hold. We use this with $m=2,n=1$ to see your result, but the general result is more difficult to prove and requires better tools than connectedness.

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Sure.

If you map each point to its radius $(x,y)\mapsto\sqrt{x^2+y^2}$, you have a continuous function $\mathbb{D}\to[0,1]$. A slight modification to $(x,y)\mapsto2\sqrt{x^2+y^2}-1$ should be what you want.

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  • $\begingroup$ Thank you sir! this one helps too! $\endgroup$ – user444830 Jul 27 '18 at 5:56

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