Let $G$ be a topological group. I want to show that, up to isomorphism, $\pi_1(G, g)$ is independent of the choice of base point $g \in G$.
Here is what I have so far:
Since $ G $ is a topological group, we know that the multiplication map $ m \colon G \times G \to G $, $ (g,h) \mapsto gh $ is continuous. Thus, for any fixed $ g,h \in G $, there is an induced homomorphism $ m^* \colon \pi_1(G \times G, (g,h)) \to \pi_1(G, gh) $, $ [f] \mapsto [m \circ f] $.
Now consider $$ m^* \colon \pi_1(G\times G, (g,g^{-1})) \to \pi_1(G,
1). $$ This should be an isomorphism since $$\ker(m^*) = \{[f] \in
\pi_1(G\times G, (g,g^{-1})) : [m\circ f] = [c_1]\}, $$ where $[c_1]$
is the homotopy class of the constant loop at $1 \in G$, but $[m\circ
f] = [c_1] \iff f \sim c_{(g,g^{-1})}$.
I believe this last line is true because if we take a non-constant loop at $(g,g^{-1})$, then it may not be the case that "multiplying elements along this loop" gets us the constant map in $\pi_1(G, 1)$. That is, if $x,y \in f([0,1])$, $x \ne y^{-1}$, then $m(x,y) \ne 1$ and so $[m \circ f] \ne [c_1]$.
If this last line is correct, then I would apply the same argument to show $$\pi_1(G, 1) \cong \pi_1(G \times G, (g,g^{-1})) \cong \pi_1(G \times G, (g,g^{-1}h)) \cong \pi_1(G ,h).$$
From here, I think the result would follow since then the fundamental group at any base point is isomorphic to that at another base point via the multiplication map.
Best Answer
Because topological groups are homogeneous (for each g in G the map sending h to gh is a homeomorphism and in particular id maps to g).
For more details.
1) As a warm up, prove that, up to isomorphism, the fundamental group of a path connected space is unique. There is no `natural' isomorphism - it depends on your choice of path class connecting the distinct base points.
2) Easier, but a similar idea applies, prove that, up to isomorphism, the fundamental group of a homogeneous space is unique, even if the space is not path connected. Here, we again do not have a natural isomorphism. It depends on the choice of global homeomorphism sending one base point to the other.
3) However, if G is a topological group, we have now a natural translation sending id to g.