True or False : There exists a continuous map $f:[0,1] \to SL_2(\Bbb R)$ which is surjective. (NBHM 2017 exam India) 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?
 A: 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 $k\geqslant 1$):
$$M:=\begin{pmatrix} 1 & k \\ 0 & 1\end{pmatrix}\in SL_2(\mathbb R)$$
but 
$$\vert \vert M \vert \vert=k\xrightarrow[k\to\infty]{} +\infty$$
so $SL_2(\mathbb R)$ is not bounded.
A: 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. 
