A path between basepoints induces an isomorphism of fundamental groups

Concatenation respects path homotopy. Let $\alpha\simeq\alpha'$ rel endpoints by $H$, and let $\beta\simeq\beta'$ rel endpoints by $K$, with the terminal point of $\alpha$ equal to the initial point of $\beta$. Setting $G(s,t)=H(2s,t)$ for $s\le\tfrac12$ and $G(s,t)=K(2s-1,t)$ for $s\ge\tfrac12$ gives a map on the two closed sets $[0,\tfrac12]\times I$ and $[\tfrac12,1]\times I$, which cover $I\times I$ and meet where $H(1,t)$ and $K(0,t)$ are both the shared endpoint. Each piece is a composite of $H$ or $K$ with an affine map of $I\times I$, so [L4] makes $G$ continuous, and it is a path homotopy $\alpha*\beta\simeq\alpha'*\beta'$ rel endpoints.

Reparametrisation does not change the class. Let $\varphi:I\to I$ be continuous with $\varphi(0)=0$ and $\varphi(1)=1$, and let $\lambda$ be a path. Then $H(s,t)=\lambda\bigl((1-t)\varphi(s)+ts\bigr)$ is continuous, starts at $\lambda\circ\varphi$, ends at $\lambda$, and is constant at each endpoint, so $\lambda\circ\varphi\simeq\lambda$ rel endpoints. Both bracketings of a triple concatenation, and each concatenation of a path with a constant path at its own endpoint, differ from the path itself precisely by such a $\varphi$. Hence concatenation is associative and the constant paths act as identities, in both cases up to path homotopy rel endpoints.

A path cancels its reversal. For a path $\lambda$ from $x$ to $y$, put $H(s,t)=\lambda(2s(1-t))$ for $s\le\tfrac12$ and $H(s,t)=\lambda(2(1-s)(1-t))$ for $s\ge\tfrac12$. The two closed pieces agree at $s=\tfrac12$, where both give $\lambda(1-t)$, so [L4] makes $H$ continuous. At $t=0$ it is $\lambda*\bar\lambda$ and at $t=1$ it is the constant path at $x$, and $H(0,t)=H(1,t)=x$ throughout. Thus $\lambda*\bar\lambda\simeq c_x$ rel endpoints, and applying this to $\bar\lambda$ gives $\bar\lambda*\lambda\simeq c_y$.

Define $\gamma_\#:\pi_1(X,x_0)\to\pi_1(X,x_1)$ by $\gamma_\#([\alpha])=[\bar\gamma*\alpha*\gamma]$, bracketed as $(\bar\gamma*\alpha)*\gamma$. The path $\bar\gamma$ runs from $x_1$ to $x_0$ and $\gamma$ from $x_0$ to $x_1$, so this is a loop at $x_1$. If $\alpha\simeq\alpha'$ rel endpoints, step 1.1 applied twice gives $\bar\gamma*\alpha*\gamma\simeq\bar\gamma*\alpha'*\gamma$, so $\gamma_\#$ is independent of the representative.

Homomorphism. For loops $\alpha,\beta$ at $x_0$, reassociating by step 1.2 turns $(\bar\gamma*\alpha*\gamma)*(\bar\gamma*\beta*\gamma)$ into $\bar\gamma*\alpha*(\gamma*\bar\gamma)*\beta*\gamma$; step 1.3 replaces the middle $\gamma*\bar\gamma$ by the constant path at $x_0$, step 1.2 deletes that constant factor, and each replacement is licensed inside the larger concatenation by step 1.1. The result is $\bar\gamma*(\alpha*\beta)*\gamma$, so $\gamma_\#([\alpha])\gamma_\#([\beta])=\gamma_\#([\alpha][\beta])$ by [L1].

Two-sided inverse. The same construction applied to $\bar\gamma$, whose reversal is $\gamma$, gives $\bar\gamma_\#:\pi_1(X,x_1)\to\pi_1(X,x_0)$. Composing, $\bar\gamma_\#(\gamma_\#([\alpha]))=[\gamma*(\bar\gamma*\alpha*\gamma)*\bar\gamma]$, and steps 1.2 and 1.3 reduce the two inserted pairs $\gamma*\bar\gamma$ and $\gamma*\bar\gamma$ to constant paths, leaving $[\alpha]$. The other composite is identical with the roles exchanged.

Thus $\gamma_\#$ is a homomorphism with a two-sided inverse, hence bijective, and [L5] makes it a group isomorphism $\pi_1(X,x_0)\cong\pi_1(X,x_1)$.