My approach :

Suppose there exists such a continuous surjective map.

Then since $[0,1]$ is compact, $SL_2(\Bbb R)$ is also compact.

So essentially what I have to prove is that $SL_n(\Bbb R)$ is compact i.e. closed and bounded.

I know that determinant map is continuous. As $SL_2(\Bbb R)=\det^{-1}\{1\}$ and $\{1\}$ is a closed set, hence $SL_2(\Bbb R)$ is a closed set.

I don't know how to go about boundedness of $SL_2(\Bbb R)$. Nevertheless, if it turns out to be bounded set, then it would mean that such $f$ exists(correct me if I'm wrong here). and if it doesn't, then such a $f$ doesn't exist.

Also how to construct such an $f$ if it does exist?

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    $\begingroup$ Note that even if $SL_2(\Bbb R)$ was compact, it would not prove that there exists a continuous surjective map $[0,1]\to SL_2(\Bbb R)$. For example $O_2(\mathbb{R})$ is compact, but there is no surjection $[0,1]\to O_2(\Bbb R)$ because it is not connected. $\endgroup$ – Arnaud D. Jan 22 '17 at 11:01
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    $\begingroup$ Why did you put group theory tag ?? So I suppose the answer of that particular question should be none of this .. I really got confused yesterday. . Finally figured that out .. :) $\endgroup$ – Illuminata Jan 22 '17 at 11:03
  • $\begingroup$ @ArnaudD. This was an eye-opener! I was indeed thinking in a much narrow way. Thank you so much. $\endgroup$ – Error 404 Jan 22 '17 at 17:25
  • $\begingroup$ @Illuminata I put group theory tag as $SL_2(\Bbb R)$ is present in the problem. I am most familiar with it as a group. Yeah none of this. ;D $\endgroup$ – Error 404 Jan 22 '17 at 17:27

You are trying to prove a false result, since $SL_n(\mathbb R)$ (and even $SL_2(\mathbb R)$) is not a compact set because it is not bounded.

In finite dimension, all norms are equivalent, so let's consider without loss of generality a norm on $M_n(\mathbb R)$:

$$\vert \vert M \vert \vert=\sup_{i,j}\vert m_{i,j}\vert.$$

Then consider the matrix (for $k\geqslant 1$):

$$M:=\begin{pmatrix} 1 & k \\ 0 & 1\end{pmatrix}\in SL_2(\mathbb R)$$


$$\vert \vert M \vert \vert=k\xrightarrow[k\to\infty]{} +\infty$$

so $SL_2(\mathbb R)$ is not bounded.

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    $\begingroup$ Thank you so much for this clear clarification. I wasn't trying to prove the result. I was searching for a counter-example, if any. You can see I started my proof from the word 'Suppose'. I feel a bit shaky while trying to solve problems related to Matrices in topology as I can not imagine them. Your answer helps me getting familiar with the stuff. :) $\endgroup$ – Error 404 Jan 22 '17 at 17:22
  • $\begingroup$ Sorry I meant 'contradiction', not 'counter-example'. :) $\endgroup$ – Error 404 Jan 23 '17 at 4:21

You can even derive the result from qualitative properties about the space without going into detailed proof.

Recall that if you have a continuous bijection $f:X\rightarrow Y$, where $X$ is compact and $Y$ is Hausdorff, then we have a homeomorphism. We can take the contrapositive and argue that if two spaces $X,Y$ are not homeomorphic, then there is no continuous bijective map between them. Since $X=[0,1]$ is compact, and $Y=SL_{2}{(\mathbb{R})}$ is not, the two spaces are not homeomorphic, since compactness is a topological invariant (spaces that are homeomorphic are either both compact or both non-compact). Since they are not homeomorphic, there cannot be a continuous bijection between $[0,1]$ and $SL_{2}{(\mathbb{R}})$.

However, the map could still be injective, but it can't be surjective because if it were surjective, then the image would need to be bounded, and since we can find some element of $SL_{n}{(\mathbb{R})}$ which is larger than the maximum element of the image, the map can't be surjective, since not every element in codomain has a pre-image.

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    $\begingroup$ Surjections aren't a subset of bijections; the fact that there doesn't exist a bijection doesn't imply that there doesn't exist a surjection without further arguments. $\endgroup$ – Steven Stadnicki Jan 23 '17 at 22:49

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