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This kind of questions is perhaps very familier here but still I'm in trouble to solve it. The problem that I have asked is "Consider closed unit disc in $\mathbb{R}^2$ i.e. the set $D=\{z\in \mathbb{C}:|z|\leq 1$$\}$ and the closed interval $I=[-1,1]$ .Prove that there does not exist any continuous on to map from $D$ to $I$?"

MY TRY: If possible let $f$ be such on to continuous map from $D$ to $I$. let $f(x)\in (-1,1)$ ,For some $x\in D$.By continuty ,for some $s>0$ at $f(x)$ if we consider an open ball $B_s(f(x))$ we can find an open set at $x$ in $R^2$ say, $U_x$ such that the second ball is the inverse image of the first ball. This tells us that those $x$ cann't lie at the boundary of the disc as every open sets at the boundaries are not contained in $D$ but $f$ is defined only in $D$. So the image of the boundary is need to be either $\{$$-1,1$$\}$ or singletons$\{$$-1$$\}$ or $\{$$ 1$$\}$. The image cann't be $\{$$ -1,1 $$\}$ as the boundary of $D$ is connected. If the image is $\{$ 1$\}$ then the inverse image of $\{$$-1$$\}$ is the set set obtained by removing two sets ,the boundary of $D$ and an open set (inverse image of $(-1,1)$) which is not closed. This also can't be happen as inverse image of closed sets is closed under continuous map!By similar argument we can disprove $\{$$-1$$\}$ as image of boundary points.

Considering these arguments I'm saying such $f$ does not exist. Have I done correct? If there are any other way to do it, please give me a hint! I have always find hard to solve such problems like "existense of on to or one one continuous map between interval to interval,disc etc", can you suggest me any book or notes regerding those problems? Thanks for reading.

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  • $\begingroup$ What about the projection onto the $x$-axis? Isn't it a continous and onto map from $D$ to $I$? $\endgroup$
    – Javi
    Dec 25, 2018 at 12:45
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    $\begingroup$ and also on to indeed! oh! I have given wrong statement to prove $\endgroup$ Dec 25, 2018 at 12:49
  • $\begingroup$ So, definitely my argument is wrong, I said inverse of (-1,1) is open! Here if you consider projection map it is D excluding 1 and -1, open in D,under subspace topology! But I thaught open in $R^2$ , I guess here I made misstake. Isn't it @Javi? $\endgroup$ Dec 25, 2018 at 12:56
  • $\begingroup$ The inverse image of an open set under a continuous map is always an open subset. Indeed, if you replace $I=[-1,1]$ by its interior $(-1,1)$ the statement does hold, i.e., there is no onto continous map from $D$ to $(-1,1)$ since the continuous image of a compact set is compact ($D$ is compact and $(-1,1)$ is not). $\endgroup$
    – Javi
    Dec 25, 2018 at 13:07
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    $\begingroup$ There is a small chance that where you wrote the two words "on to" you should have written "one to one". Then your question would make sense. $\endgroup$ Dec 25, 2018 at 13:33

2 Answers 2

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The original statement is wrong. Just consider the projection onto de $x$-axis $p:D\to I$ defined as $p(x,y)=x$, which is clearly a continous onto map.

If you replace $I$ by its interior $(-1,1)$ the statement does hold since the continuous image of a compact set must be compact, but $D$ is compact and $(-1,1)$ is not.

In your argument you assumed that an open subset of $D$ cannot touch de boundary, but it can. Open subsets of $D$ with the subspace topology are intersections of open sets of $\mathbb{R}^2$ with $D$. Consider the right half-plane in $\mathbb{R}^2$. Its intersection with $D$ is the right half-disc, which obviously includes the right half-boundary. This is indeed $p^{-1}((0,1])$.

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    $\begingroup$ yes!i apologiies. thanks $\endgroup$ Dec 25, 2018 at 14:47
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Assuming you want to prove there is no continuous $one-to-one$ $f:D\to I.$ There are many ways to do this.

Assume by contradiction there exists such an $f.$ Take some (any) $p\in D$ with $-1\ne f(p)\ne 1.$ Let $f(q^-)=-1$ and $f(q^+)=1.$

There exists a continuous $g$ from $[0,1]$ into $D$ with $g(0)=q^-$ and $g(1)=q^+,$ such that $p\not \in \{g(x):x \in [0,1]\}.$

Let $\bar f$ be $f$ restricted to the domain $\{g(x):x\in [0,1]\}.$

Then $\bar f g:[0,1]\to [-1,1]$ is continuous with $\bar f g(0)=-1$ and $\bar f g(1)=1 .$

But $-1<f(p)<1$ and no $x\in [0,1]$ satisfies $\bar fg (x)=f(p).$ This is impossible by the Intermediate Value Property of continuous real functions.

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