TITLE: Prove that $\operatorname{SL}(n,\Bbb R)$ is connected.
QUESTION [20 upvotes]: Prove that $\operatorname{SL}(n, \Bbb R)$ is connected.
The problem is I know only topological groups from Munkres only. Again Just started fundamental groups. So if anyone can explain to me how it is true in a lucid language and in an easy way such that it remains in my boundary of knowledge then it would be a great help. I have mentioned what I know.
Again if tag this in the wrong field. Please forgive me.

REPLY [5 votes]: $\textbf{Lemma: }$Let $Y$ be connected and $f\colon X\to Y$ be surjective continuous map having connected fibers. If $f$ is open then $X$ also connected.
Proof. We can prove this by contradiction. If possible, write $X=U\bigsqcup V$ where $U,V$ are non-empty open subsets of $X$. Then, $f(U),f(V)$ are open subsets of $Y$ such that $f(U)\cup f(V)=Y$. Now, $Y$ is connected implies $f(U)\cap f(V)\not=\emptyset.$ Take, $y\in f(U)\cap f(V)$, then $f^{-1}(y)\cap U\not =\emptyset$ and $f^{-1}(y)\cap V\not=\emptyset$, contradicts to the fact that fibers are connected sets. $\square$
$\textbf{Theorem.}$ $\text{GL}^+(n,\Bbb R):=\big\{A\in \text{M}(n,\Bbb R): \det(A)>0\big\}$ is connected.
Proof. We will prove $\text{GL}^+(n,\Bbb R)$ is connected by induction on $n$.
Consider the map $p\colon \text{M}(n,\Bbb R)=\Bbb R^n\times \text{M}\big(n\times (n-1),\Bbb R\big)\longrightarrow \Bbb R^n$ given by $p(A)=Ae_1$. That is $p$ sends $A\in \text{M}(n,\Bbb R)$ to its first column. Note that $p$ is a projection map, hence open and continuous.
Note that, $\text{GL}^+(1,\Bbb R)=(0,\infty)$. Now, let $f\colon \text{GL}^+(n,\Bbb R)\longrightarrow \Bbb R^n-\{0\}$ be the restriction of $p$ to the $\text{GL}^+(n,\Bbb R)\subseteq_{\text{open}}\text{M}(n,\Bbb R)$. So, $f$ is also open continuous.
Next, $f^{-1}(e_1)=\Bbb R^{n-1}\times \text{GL}^+(n-1,\Bbb R)$, which is a connected set by induction. For $y\in \Bbb R^n-\{0\}$ choose $B\in \text{GL}^+(n,\Bbb R)$ with $f(B)=y$. Then, $f^{-1}(y)=\big\{B\cdot C:C\in f^{-1}(e_1)\big\}$. Hence each fibre of $f$ is connected. So, $\text{GL}^+(n,\Bbb R)$ is connected by the above Lemma. $\square$
$\textbf{Theorem.}$ $\text{SL}(n,\Bbb R)$ is connected.
Proof. To prove $\text{SL}(n,\Bbb R)$ is connected, consider the continuous surjective map $$\Psi\colon \text{GL}^+(n,\Bbb R)\ni A\longmapsto \frac{A}{\det A}\in \text{SL}(n,\Bbb R).$$ Recall that continuous image of a connected set is connected, so we are done by the previous theorem. $\square$