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Which of the following statements are false ??

a) There exists a continuous bijection $f:[0,1]\to[0,1]\times[0,1]$

b) There exists a continuous map $f:S^1\to \mathbb{R}$ which is injective, where $S^1$ stands for the unit circle in the plane

c) There exists a continuous map $f:[0,1]\to SL_2(\mathbb{R})$ which is surjective.

I think a) is false as it is similar to construct a continuous bijective map from side to square (Space-filling curve which is continuous but not bijective) not possible, but how do I prove that?

For b) i have no any idea

For c) as we know any continuous image of a compact space is compact and here $[0,1]$ is compact where $SL_2(\mathbb{R})$ is not compact as it is not bounded

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  • $\begingroup$ In case b, $\mathbb{R}$ is not compact. $\endgroup$ – hemu Nov 20 '17 at 18:05
  • $\begingroup$ @shrinit why is that relevant? $\endgroup$ – Andres Mejia Nov 20 '17 at 18:12
  • $\begingroup$ @AndresMejia I think since $f(S_1)$ is compact and it should look like a closed and bounded in $\mathbb{R}$ and then we use inverse image of an open set will be open as f is continuous. I think we will get contradiction. $\endgroup$ – hemu Nov 20 '17 at 18:35
  • $\begingroup$ $S^1$ is open. The whole set is open in any topology. $\endgroup$ – Andres Mejia Nov 20 '17 at 19:25
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For (b), assume $f\colon S^1\to\Bbb R$ is injective. Then $g(z):=f(z)-f(-z)$ is a continuous map $S^1\to \Bbb R\setminus\{0\}$ that takes positive as well as negative value. This contradicts connectedness of $S^1$.

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    $\begingroup$ Not sure why I found this solution so appealing, but +1 :) $\endgroup$ – Andres Mejia Nov 20 '17 at 18:13
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A: if there were a continuous bijection, then since $[0,1]$ is compact, and $[[0,1]^2$ is hausdorff, it would be a homeomoprhism, which it is not since any homeomorphism $h$ would restrict to $\tilde{h}:[0,1]\setminus\{.5\} \to[0,1] \times [0,1] \setminus \{h(.5)\}$, but the image of the inverse map would be disconnected, while the domain is connected.

This is just keeping track of cut points

B: Suppose that $f:S^1 \hookrightarrow \mathbb R$. Then $f(S^1) \subset \mathbb R$ is a compact, connected component (if you know what these look like, stop here.) Otherwise, use the theorem again, so that $f(S^1)$ is homeomorphic to $S^1$, but then examine cut points, and deduce a contradiction.

C: Looks good to me.

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