A path between basepoints induces an isomorphism of fundamental groups

Concatenation respects path homotopy. Let $\alpha \simeq {\alpha }^{\mathrm{\prime }}$ rel endpoints by $H$, and let $\beta \simeq {\beta }^{\mathrm{\prime }}$ 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 $s\le \frac{1}{2}$ and $G(s,t)=K(2s-1,t)$ for $s\ge \frac{1}{2}$ gives a map on the two closed sets $[0,\frac{1}{2}]\times I$ and $[\frac{1}{2},1]\times I$, 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 $\alpha \ast \beta \simeq {\alpha }^{\mathrm{\prime }}\ast {\beta }^{\mathrm{\prime }}$ rel endpoints.

Reparametrisation does not change the class. Let $\phi :I\to I$ be continuous with $\phi (0)=0$ and $\phi (1)=1$, and let $\lambda$ be a path. Then $H(s,t)=\lambda ((1-t)\phi (s)+ts)$ is continuous, starts at $\lambda \circ \phi$, ends at $\lambda$, and is constant at each endpoint, so $\lambda \circ \phi \simeq \lambda$ 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 $\phi$. 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 $s\le \frac{1}{2}$ and $H(s,t)=\lambda (2(1-s)(1-t))$ for $s\ge \frac{1}{2}$. The two closed pieces agree at $s=\frac{1}{2}$, where both give $\lambda (1-t)$, so [L4] makes $H$ continuous. At $t=0$ it is $\lambda \ast \bar{\lambda }$ and at $t=1$ it is the constant path at $x$, and $H(0,t)=H(1,t)=x$ throughout. Thus $\lambda \ast \bar{\lambda }\simeq c_x$ rel endpoints, and applying this to $\bar{\lambda }$ gives $\bar{\lambda }\ast \lambda \simeq c_y$.

Define ${\gamma }_{\mathrm{\#}}:{\pi }_1(X,x_0)\to {\pi }_1(X,x_1)$ by ${\gamma }_{\mathrm{\#}}([\alpha ])=[\bar{\gamma }\ast \alpha \ast \gamma ]$, bracketed as $(\bar{\gamma }\ast \alpha )\ast \gamma$. 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 $\alpha \simeq {\alpha }^{\mathrm{\prime }}$ rel endpoints, step 1.1 applied twice gives $\bar{\gamma }\ast \alpha \ast \gamma \simeq \bar{\gamma }\ast {\alpha }^{\mathrm{\prime }}\ast \gamma$, so ${\gamma }_{\mathrm{\#}}$ is independent of the representative.

Homomorphism. For loops $\alpha ,\beta$ at $x_0$, reassociating by step 1.2 turns $(\bar{\gamma }\ast \alpha \ast \gamma )\ast (\bar{\gamma }\ast \beta \ast \gamma )$ into $\bar{\gamma }\ast \alpha \ast (\gamma \ast \bar{\gamma })\ast \beta \ast \gamma$; step 1.3 replaces the middle $\gamma \ast \bar{\gamma }$ 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 $\bar{\gamma }\ast (\alpha \ast \beta )\ast \gamma$, so ${\gamma }_{\mathrm{\#}}([\alpha ]){\gamma }_{\mathrm{\#}}([\beta ])={\gamma }_{\mathrm{\#}}([\alpha ][\beta ])$ by [L1].

Two-sided inverse. The same construction applied to $\bar{\gamma }$, whose reversal is $\gamma$, gives ${\bar{\gamma }}_{\mathrm{\#}}:{\pi }_1(X,x_1)\to {\pi }_1(X,x_0)$. Composing, ${\bar{\gamma }}_{\mathrm{\#}}({\gamma }_{\mathrm{\#}}([\alpha ]))=[\gamma \ast (\bar{\gamma }\ast \alpha \ast \gamma )\ast \bar{\gamma }]$, and steps 1.2 and 1.3 reduce the two inserted pairs $\gamma \ast \bar{\gamma }$ and $\gamma \ast \bar{\gamma }$ to constant paths, leaving $[\alpha ]$. The other composite is identical with the roles exchanged.

Thus ${\gamma }_{\mathrm{\#}}$ is a homomorphism with a two-sided inverse, hence bijective, and [L5] makes it a group isomorphism ${\pi }_1(X,x_0)\cong {\pi }_1(X,x_1)$.