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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.

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    $\begingroup$ The fact that $G$ is a topological group is a red herring. Show 1/ that the fundamental group of a path-connected space doesn't depend on the base point 2/ all the path components of a topological group are homeomorphic. $\endgroup$ – Najib Idrissi Nov 19 '15 at 20:24
  • $\begingroup$ I'm having a difficult time seeing your second claim. Is there an explicit map or should I consider some properties of path components? $\endgroup$ – Kevin Sheng Nov 19 '15 at 20:39
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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.

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