Conjugating loop classes by a path is an isomorphism of fundamental groups

Concatenation respects path homotopy rel endpoints. Let $\alpha \simeq {\alpha }^{\mathrm{\prime }}$ rel endpoints by $H$ and $\beta \simeq {\beta }^{\mathrm{\prime }}$ rel endpoints by $K$, where $\alpha (1)=\beta (0)$ and ${\alpha }^{\mathrm{\prime }}(1)={\beta }^{\mathrm{\prime }}(0)$ so that both concatenations are defined. Put $G(s,t):=H(2s,t)$ for $0\le s\le \frac{1}{2}$ and $G(s,t):=K(2s-1,t)$ for $\frac{1}{2}\le s\le 1$. At $s=\frac{1}{2}$ the two formulas give $H(1,t)=\alpha (1)$ and $K(0,t)=\beta (0)$ by the rel-endpoints condition, and these agree because the middle endpoints agree; hence $G$ is a well-defined function on $I\times I$. The two closed sets $[0,\frac{1}{2}]\times I$ and $[\frac{1}{2},1]\times I$ cover $I\times I$, and on each of them $G$ is a composite of $H$ or $K$ with the continuous affine map $(s,t)\mapsto (2s,t)$ or $(s,t)\mapsto (2s-1,t)$, so [L4] makes $G$ continuous. Finally $G(s,0)=\alpha \ast \beta (s)$, $G(s,1)={\alpha }^{\mathrm{\prime }}\ast {\beta }^{\mathrm{\prime }}(s)$, $G(0,t)=\alpha (0)={\alpha }^{\mathrm{\prime }}(0)$ and $G(1,t)=\beta (1)={\beta }^{\mathrm{\prime }}(1)$, so $G$ is a path homotopy $\alpha \ast \beta \simeq {\alpha }^{\mathrm{\prime }}\ast {\beta }^{\mathrm{\prime }}$ rel endpoints. A constant homotopy on one factor is the case $\beta ={\beta }^{\mathrm{\prime }}$, $K(s,t):=\beta (s)$, so the same statement applies when only one of the two factors is deformed.

Reparametrisation does not change the class. Let $\lambda :I\to X$ be a path and let $\phi :I\to I$ be continuous with $\phi (0)=0$ and $\phi (1)=1$. Then $H(s,t):=\lambda ((1-t)\phi (s)+ts)$ is continuous because the argument is obtained from the continuous maps $s\mapsto \phi (s)$, $s\mapsto s$ and $t\mapsto t$ by products, sums and the continuous inclusion of $I$ in $\mathbb{R}$, and it satisfies $H(s,0)=\lambda (\phi (s))$, $H(s,1)=\lambda (s)$, $H(0,t)=\lambda (0)$ and $H(1,t)=\lambda (1)$: the last two because $\phi (0)=0$ and $\phi (1)=1$. So $\lambda \circ \phi \simeq \lambda$ rel endpoints. Consequently the two bracketings of a triple concatenation of composable paths are reparametrisations of one another, so they are path-homotopic rel endpoints, and for a path $\lambda$ from $x$ to $y$ the concatenations $\lambda \ast c_y$ and $c_x\ast \lambda$ with the constant paths at the endpoints are reparametrisations of $\lambda$, so both are path-homotopic to $\lambda$ rel endpoints. Hence constant factors may be inserted and deleted inside a larger concatenation up to path homotopy rel endpoints.

A path cancels its reversal. Let $\lambda :I\to X$ be a path from $x$ to $y$ and put $H(s,t):=\lambda (2s(1-t))$ for $0\le s\le \frac{1}{2}$ and $H(s,t):=\lambda (2(1-s)(1-t))$ for $\frac{1}{2}\le s\le 1$. At $s=\frac{1}{2}$ both formulas give $\lambda (1-t)$, and the two closed pieces cover $I\times I$, so [L4] makes $H$ continuous. One has $H(s,0)=\lambda \ast \bar{\lambda }(s)$, $H(s,1)=\lambda (0)=x$, and $H(0,t)=\lambda (0)=x=H(1,t)$, so $H$ is a path homotopy $\lambda \ast \bar{\lambda }\simeq c_x$ rel endpoints, where $c_x$ is the constant path at the initial point. Applying the same statement to the reversed path $\bar{\lambda }$, whose reversal is $\lambda$, gives $\bar{\lambda }\ast \lambda \simeq c_y$ rel endpoints.

Well-definedness of ${\phi }_c$. By [F1] the path $\bar{c}$ joins $x_1$ to $x_0$ and the concatenation $(\bar{c}\ast \alpha )\ast c$ is a loop at $x_1$ for every based loop $\alpha$ at $x_0$, so the formula of the statement defines a function on the set of based loops. Let $\alpha \simeq {\alpha }^{\mathrm{\prime }}$ rel endpoints. Step 1.1 applied to the pair $\alpha \simeq {\alpha }^{\mathrm{\prime }}$ and the constant homotopy of $\bar{c}$ gives $\bar{c}\ast \alpha \simeq \bar{c}\ast {\alpha }^{\mathrm{\prime }}$ rel endpoints, and step 1.1 applied again to that homotopy and the constant homotopy of $c$ gives $(\bar{c}\ast \alpha )\ast c\simeq (\bar{c}\ast {\alpha }^{\mathrm{\prime }})\ast c$ rel endpoints. Both are loops at $x_1$, so their classes in ${\pi }_1(X,x_1)$ coincide by [F2], and ${\phi }_c$ is independent of the representative of $[\alpha ]$.

${\phi }_c$ is a homomorphism. Let $\alpha ,\beta$ be based loops at $x_0$. Then, using the product formula of [F2] and the bracketing freedom of step 1.2, ${\phi }_c([\alpha ]){\phi }_c([\beta ])=[(\bar{c}\ast \alpha \ast c)\ast (\bar{c}\ast \beta \ast c)]\simeq [(\bar{c}\ast \alpha )\ast (c\ast \bar{c})\ast (\beta \ast c)]\simeq [(\bar{c}\ast \alpha )\ast (\beta \ast c)]\simeq [\bar{c}\ast (\alpha \ast \beta )\ast c]={\phi }_c([\alpha ][\beta ]),$ where the second reduction replaces the loop $c\ast \bar{c}$ at $x_0$ by a constant path using step 1.3 and deletes that constant factor using step 1.2, and where each replacement is licensed inside the ambient concatenation by step 1.1. Hence ${\phi }_c$ is a group homomorphism.

${\phi }_{\bar{c}}$ is a two-sided inverse. The assignment ${\phi }_{\bar{c}}([\beta ]):=[(c\ast \beta )\ast \bar{c}]$ is well defined by the argument of step 2.1 with $c$ replaced by $\bar{c}$, and it maps ${\pi }_1(X,x_1)$ to ${\pi }_1(X,x_0)$. For a based loop $\alpha$ at $x_0$ one has ${\phi }_{\bar{c}}({\phi }_c([\alpha ]))=[c\ast ((\bar{c}\ast \alpha )\ast c)\ast \bar{c}]$; reassociating by step 1.2 and applying step 1.1 to insert the pairs, this class equals $[(c\ast \bar{c})\ast \alpha \ast (c\ast \bar{c})]$, and since $c\ast \bar{c}$ is homotopic to the constant path at $x_0$ by step 1.3, deleting both constant factors with step 1.2 gives $[\alpha ]$. Symmetrically, for a based loop $\beta$ at $x_1$ one has ${\phi }_c({\phi }_{\bar{c}}([\beta ]))=[\bar{c}\ast ((c\ast \beta )\ast \bar{c})\ast c]\simeq [(\bar{c}\ast c)\ast \beta \ast (\bar{c}\ast c)]=[\beta ]$ by the same two steps, because $\bar{c}\ast c$ is homotopic to the constant path at $x_1$ by step 1.3. So the two composites are the identities.

Conclusion. Steps 2.1, 2.2 and 3.1 exhibit ${\phi }_c$ as a well-defined group homomorphism with a two-sided inverse, hence a bijection, and [L5] makes it a group isomorphism. The final assertion follows because a path from $x_0$ to $x_1$ exists exactly when the two points lie in the same path component of $X$ by [F1].