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

Theorem
Let $X$ be a path-connected space.

For $x \in X$ let $\pi_1(X,x)$ denote the fundamental group of closed paths with initial point $x$.

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 is a path $f$ connecting $y$ and $x$.

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

Since $\phi_f([gh])=[f^{-1}ghf]=[f^{-1}gff^{-1}hf]=[f^{-1}gf][f^{-1}hf]=\phi_f([g])\phi_f([h])$, $\phi$ is an homomorphism of groups, by definition.

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 $\phi_{f^{-1}} ([g])=[fgf^{-1}]$, is an 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.