Fundamental Group is Independent of Base Point for Path-Connected Space

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Theorem

Let $X$ be a path-connected space.

For $x \in X$ let $\pi_1(X,x)$ denote the fundamental group.

For $x, y \in X$, there is an isomorphism:

$\phi: \pi_1 \left({X, x}\right) \to \pi_1 \left({X, y}\right)$


Proof

Since $X$ is path-connected there exists a path $f$ connecting $y$ and $x$.

We define $\phi_f: \pi_1 \left({X, x}\right) \to \pi_1 \left({X, y}\right)$ by $\phi_f \left({ \left[{g}\right] }\right) =\left[{ f^{-1} g f }\right]$.

$\phi$ is a homomorphism of groups, as is seen from:

$\displaystyle \phi_f \left({ \left[{g h}\right] }\right) = \left[{ f^{-1} g h f }\right] = \left[{ f^{-1} g f f^{-1} h f }\right] = \left[{ f^{-1} g f }\right] \left[{ f^{-1} h f }\right] = \phi_f \left({ \left[{g}\right] }\right) \phi_f \left({ \left[{h}\right] }\right)$


Also, $f^{-1}$ is a path from $x$ to $y$. Then by the same argument as before, $\phi_{f^{-1}}: \pi_1 \left({X, y}\right) \to \pi_1 \left({X, x}\right)$ where $\phi_{f^{-1}} \left({ \left[{g}\right] }\right) = \left[{ f g f^{-1} }\right]$, is a homomorphism of groups.

Trivially, $\phi_f \circ \phi_{f^{-1}} = I_{\pi_1 \left({X, y}\right)}$ and $\phi_{f^{-1}} \circ \phi_f=I_{\pi_1 \left({X, x}\right)}$.

Then $\phi_f^{-1} = \phi_{f^{-1}}$, so $\phi_f$ is bijective and thus an isomorphism.

$\blacksquare$