Proposition

Prove that up to isomorphism, $\pi_1(G, g)$ is independent of the choice of the base point $g \in G$.
Prove that $\pi_1(G, g)$ is abelian.

Solution

1

Let $g_1, g_2 \in G$. Let $\phi: G \rightarrow G$ be defined such that $\phi(g) = g_2g_1^{-1}g$. We claim that $\phi$ is a homeomorphism.

Is $\phi$ continuous?
- Let $f_1:G \rightarrow G \times G$ be defined such that $f_1(g) = (g, g)$. Then $f_1$ is continuous since each coordinate function is the identity map. Let $f_2: G \times G \rightarrow G \times G$ be defined such that $f_2(g, h) = (g_2g_1^{-1}, h)$. Then $f_1$ is continuous since the first coordinate function is a constant function, and the second coordinate function is the identity. The multiplication operation is continuous because $G$ is a topological group. Then a composition of those functions must be continuous, and $\phi$ is precisely the composition of multiplication, $f_2$ and $f_1$.
Is $\phi$ injective?
- Let $g, h \in G$. Suppose $\phi(g) = \phi(h)$. Then $g_2g_1^{-1}g = g_2g_1^{-1}h$, so $g = h$.
Is $\phi$ surjective?
- Let $g \in G$. Then $g_1g_2^{-1}g \in G$, and $\phi(g_1g_2^{-1}g) = (g_2g_1^{-1})(g_1g_2^{-1})g = g$.
Is $\phi^{-1}$ continuous?
- Since $\phi$ is bijective, $\phi$ indeed has an inverse. The inverse is defined by $\psi(g) = g_1g_2^{-1}g$. (This can be verified by confirming that $\phi \circ \psi$ and $\psi \circ \phi$ are both the identity map.) Using the same logic as above, we can confirm that $\psi$ is continuous.

Therefore, $\phi$ is indeed a homeomorphism. It is known that a homeomorphism induces an isomorphism between the fundamental groups. For instance, see Proposition 7.26 of Introduction to Topological Manifolds.

2

By 1, it suffices to show that $\pi_1(G, e)$ is abelian. Let $[f], [g] \in \pi_1(G, e)$.

Then both $f$ and $g$ map $I$ into $G$ continuously.

$(s, t) \mapsto (f(s), g(t))$ is continuous since each component function is continuous.
$(f(s), g(t)) \mapsto f(s)g(t)$ is continuous since multiplication is continuous.

Therefore, $F(s, t) = f(s)g(t)$ is continuous.

$F(s, 0) = f(s)e = f(s)$.
$F(1, s) = f(1)g(s) = g(s)$.
$F(0, s) = f(0)g(s) = g(s)$.
$F(s, 1) = f(s)g(1) = f(s)$.

Therefore, by the square lemma, $[f \cdot g] = [g \cdot f]$. Hence, $[f][g] = [g][f]$, so $\pi_1(G, e)$ is abelian.