How do we formally prove that the fundamental group of any Lie group is always commutative?



Commutativity of the fundamental group of any Lie Group 
As Vahid says, it is true for any topological group. Here is a proof. I'm sure there are nicer, more conceptual ones out there, but here goes.


Let $G$ be your topological group. Take two loops $\sigma$ and $\gamma$ in $G$, based at the identity of $G$, which we will denote by $e$. Let $\sigma \cdot \gamma$ be the concatenation of the two loops. This is given by
$$ (\sigma \cdot \gamma) (t) = 
\begin{cases} \sigma(2t) & \quad \text{ if } 0 \le t \le 1/2 \\\
 \gamma(2t-1) &\quad \text{ if } 1/2 \le t \le 1 \end{cases} $$
(Sorry, couldn't manage to format that any better. Feel free to edit if you know how to put a nice brace bracket to the left of that definition.)


The idea is this. We will show that $\sigma \cdot \gamma$ is homotopic to to the loop given by the pointwise product of $\sigma$ and $\gamma$. Let's call that loop $\rho$, so
$$ \rho(t) = \sigma(t)\gamma(t).$$


Now define an auxiliary function $P : [0,1] \times [0,1] \to G$ by
$$ P(s,t) =
\begin{cases} \sigma\left( \frac{2t}{1+s} \right) & \quad \text{ if } 0 \le t \le \frac{1+s}{2} \\\
 e &\quad \text{ if } \frac{1+s}{2} \le t \le 1 \end{cases}$$


At $s=0$, this function does the whole loop $\sigma$ as $t$ goes from $0$ to $1/2$, then sits at $e$. In other words, at $s=0$ this is the first half of the loop $\sigma \cdot \gamma$. As $s$ gets larger, $P$ does the whole loop $\sigma$ as $t$ goes from $0$ to $\frac{1+s}{2}$. At $s=1$, $P$ does the loop $\sigma$ at normal speed.


Then similarly define a function $Q : [0,1] \times [0,1] \to G$ by
$$ Q(s,t) =
\begin{cases} e & \quad \text{ if } 0 \le t \le \frac{1-s}{2} \\\
 \gamma \left( \frac{2t-1+s}{1+s} \right) &\quad \text{ if } \frac{1-s}{2} \le t \le 1 \end{cases}$$


At $s=0$ this is just the second half of the loop $\sigma\cdot\gamma$, while at $s=1$ it is exactly the loop $\gamma$.


So finally, define
$$ H(s,t) = P(s,t) \cdot Q(s,t). $$
At $s=0$ this is $\sigma \cdot \gamma$, while at $s=1$ it is the pointwise product loop $\rho$. $H$ is clearly continuous, and $H(s,0) = e = H(s,1)$ for all $s$, so this is a homotopy of loops between $\sigma \cdot \gamma$ and $\rho$.


Now we can redo that process and show that $\rho$ is homotopic to the other concatenation $\gamma \cdot \sigma$. So this shows that $\pi\_1(G)$ is abelian.


