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Which of the followings are TRUE ?

(a) Let , $f:\mathbb Z \to \mathbb Z^2$ be a bijection. Then there exists a continuous function from $\mathbb R$ to $\mathbb R^2$ which extends $f$.

(b) Let, $D$ be a closed unit disc in $\mathbb R^2$ . Then $\exists$ a continuous map $$f:D\setminus \{(0,0)\}\to \{x\in \mathbb R:|x|\le 1\}$$which is onto.

(c) Let, $D$ be a closed unit disc in $\mathbb R^2$. Then $\exists$ a continuous map $$f:D\setminus\{(0,0)\}\to \{x\in \mathbb R:|x|>1\}$$ which is onto.

(c) FALSE. As $D\setminus \{(0,0)\}$ is path-connected & hence connected , but $\{x\in \mathbb R:|x|>1\}$ is disconnected.

I think (b) is TRUE , but I can not find an example. I want an example or a proper justification from which we can say that such a function exists.

I have no idea about option (a).

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    $\begingroup$ For (a): Do you mean $f:\mathbb Z\to\mathbb Z$ or $f:\mathbb R\to\mathbb R^2$? $\endgroup$ – sranthrop Feb 27 '15 at 20:30
  • $\begingroup$ Sorry..!! It was a typing mistake. I edited it..See now... $\endgroup$ – Empty Feb 28 '15 at 2:26
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For (b) Consider the map $f:D\setminus\{(0,0)\}\to [-1,1]$ where $f(x,y)=x$.

For (a) let $f(n)=a_n$. For each $n\in\Bbb Z$, let $f_n:\Bbb R\to\Bbb R^2$ be given by $$f_n(t)= \begin{cases} (1-t)a_n+ta_{n+1}, & 0\le t<1\\ 0, & \text{otherwise}. \end{cases}$$
This $f_n(t)$ traces the line segment from $a_n$ to $a_{n+1}$ for $0\le t<1$ and is $0$ every where else.

Now, let $$F(t)=\sum_{n\in\Bbb Z}f_n(t-n)$$

We see that $F(n)=f_n(0)=a_n$. Also, $F(t)$ is continuous on $\Bbb R$.

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  • $\begingroup$ Why does this not answer part b)? $\endgroup$ – Tim Raczkowski Feb 27 '15 at 20:57
  • $\begingroup$ My apologies, I see that the OP asked for an example. Let me delete the comment. $\endgroup$ – daOnlyBG Feb 27 '15 at 21:01
  • $\begingroup$ OK..What about option (a)? $\endgroup$ – Empty Feb 28 '15 at 2:24
  • $\begingroup$ That is true also. I'll edit my answer. $\endgroup$ – Tim Raczkowski Feb 28 '15 at 2:27
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For (a). More generally, any map $f \colon \mathbb Z \to T$, where $T$ is arcwise connected, extends to a continuous map $f \colon \mathbb R \to T$.

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  • $\begingroup$ How we can show $\mathbb Z\times\mathbb Z$ is path connected? $\endgroup$ – Empty Feb 28 '15 at 3:10
  • $\begingroup$ No, use $T = \mathbb R^2$, which is path connected. $\endgroup$ – GEdgar Feb 28 '15 at 3:21
  • $\begingroup$ In My question $f:\mathbb Z\to \mathbb Z^2$ $\endgroup$ – Empty Feb 28 '15 at 3:28
  • $\begingroup$ So first compose with inclusion $\mathbb Z^2 \to \mathbb R^2$. $\endgroup$ – GEdgar Feb 28 '15 at 14:18
  • $\begingroup$ @GEdgar do you have a proof that if T is path-connected then we can continuously extends our map to reals????if you have then please provide it I am unable to prove it! $\endgroup$ – Ibs Mar 30 at 14:56

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