Point-motion boundary map is a homomorphism

Independence of the evaluation lift. Let $\tilde{\alpha },{\tilde{\alpha }}^{\mathrm{\prime }}$ be lifts of the same based loop $\alpha$ with $\tilde{\alpha }(0)={\tilde{\alpha }}^{\mathrm{\prime }}(0)=\mathrm{id}$, and put $g_t:=\tilde{\alpha }(t{)}^{-1}\circ {\tilde{\alpha }}^{\mathrm{\prime }}(t)\in {\mathrm{Homeo}}^{+}(D^2,\mathrm{\partial }{D}^2)$; this is a continuous path by [L3] with $g_0=\mathrm{id}$. For each $t$, the tuples $x:=\tilde{\alpha }(t)(Q_n)$ and $y:={\tilde{\alpha }}^{\mathrm{\prime }}(t)(Q_n)$ satisfy $[x]=[y]=\alpha (t)$, so $y=\sigma \cdot x$ for a permutation $\sigma \in S_n$; applying the homeomorphism $\tilde{\alpha }(t{)}^{-1}$ coordinatewise gives $\tilde{\alpha }(t{)}^{-1}(y)=\sigma \cdot Q_n$, because $\tilde{\alpha }(t{)}^{-1}(\tilde{\alpha }(t)(Q_n))=Q_n$. Hence $\mathrm{ev}(g_t)=[\sigma \cdot Q_n]=[Q_n]$ for every $t$, so $t\mapsto g_t$ is a path in the fibre $F$ from $\mathrm{id}$ to $g_1=\tilde{\alpha }(1{)}^{-1}\circ {\tilde{\alpha }}^{\mathrm{\prime }}(1)$. Therefore $[g_1]=[\mathrm{id}]$ in ${\pi }_0(F)$, that is $[\tilde{\alpha }(1{)}^{-1}][{\tilde{\alpha }}^{\mathrm{\prime }}(1)]=[\mathrm{id}]$ by the product rule of [L3]; multiplying by the inverse of $[{\tilde{\alpha }}^{\mathrm{\prime }}(1)]=[{\tilde{\alpha }}^{\mathrm{\prime }}(1{)}^{-1}{]}^{-1}$ gives $[\tilde{\alpha }(1{)}^{-1}]=[{\tilde{\alpha }}^{\mathrm{\prime }}(1{)}^{-1}]$, and by [L2] the value $\delta ([\alpha ])$ does not depend on the chosen lift.

Independence of the representative. Let $H:I\times I\to B$ be a path homotopy relative to $\{0,1\}$ from $\alpha$ to a second based loop ${\alpha }^{\mathrm{\prime }}$ at $[Q_n]$, so $H(0,t)=\alpha (t)$, $H(1,t)={\alpha }^{\mathrm{\prime }}(t)$ and $H(s,0)=H(s,1)=[Q_n]$. Let $\tilde{\alpha }$ be a lift of $\alpha$ with $\tilde{\alpha }(0)=\mathrm{id}$, prescribe the constant lift $\mathrm{id}$ on the bottom edge $I\times \{0\}$ and $\tilde{\alpha }$ on the left edge $\{0\}\times I$ of the square: the two prescriptions agree at the corner $(0,0)$ because $\tilde{\alpha }(0)=\mathrm{id}$. By the relative lifting clause of [L4] there is $\tilde{H}:I\times I\to {\mathrm{Homeo}}^{+}(D^2,\mathrm{\partial }{D}^2)$ with $\mathrm{ev}\circ \tilde{H}=H$ agreeing with these values; in particular the right edge $s\mapsto \tilde{H}(1,s)$ is a lift of ${\alpha }^{\mathrm{\prime }}$ starting at $\tilde{H}(1,0)=\mathrm{id}$, and the top edge $s\mapsto \tilde{H}(s,1)$ has image under $\mathrm{ev}$ constantly equal to $H(s,1)=[Q_n]$, hence lies in $F$ and joins $\tilde{H}(0,1)=\tilde{\alpha }(1)$ to $\tilde{H}(1,1)$. Thus $[\tilde{\alpha }(1)]=[\tilde{H}(1,1)]$ in ${\pi }_0(F)$, and applying the continuous inversion of [L3] gives $[\tilde{\alpha }(1{)}^{-1}]=[\tilde{H}(1,1{)}^{-1}]$; by [L2] and step 1.1, $\delta ([\alpha ])=\delta ([{\alpha }^{\mathrm{\prime }}])$.

Multiplicativity. Let $\alpha ,\beta$ be based loops at $[Q_n]$ with lifts $a,b:I\to {\mathrm{Homeo}}^{+}(D^2,\mathrm{\partial }{D}^2)$ from $\mathrm{id}$, and write $a_1:=a(1)$, $b_1:=b(1)$, both in $F$ by [L1]. Define $c:I\to {\mathrm{Homeo}}^{+}(D^2,\mathrm{\partial }{D}^2)$ by $c(t):=a(2t)$ for $t\le \frac{1}{2}$ and $c(t):=b(2t-1)\circ a_1$ for $t\ge \frac{1}{2}$. The two formulas agree at $t=\frac{1}{2}$ because $a(1)=a_1=b(0)\circ a_1$, so $c$ is continuous by [L3]; also $c(0)=a(0)=\mathrm{id}$. For $t\le \frac{1}{2}$ one has $\mathrm{ev}(c(t))=\alpha (2t)$, and for $t\ge \frac{1}{2}$ one has $\mathrm{ev}(c(t))=[b(2t-1)(a_1(Q_n))]=[b(2t-1)(Q_n)]=\beta (2t-1)$, where the middle equality uses $a_1\in F$, so $a_1(Q_n)=Q_n$ as sets; hence $\mathrm{ev}\circ c=\alpha \ast \beta$. Therefore $c$ is a lift of $\alpha \ast \beta$ from $\mathrm{id}$ with terminal value $c(1)=b_1\circ a_1$, and [L2] together with the identity $(b_1{a}_1{)}^{-1}=a_1^{-1}{b}_1^{-1}$ in the group $F$ gives $\delta ([\alpha ][\beta ])=\delta ([\alpha \ast \beta ])=[(b_1{a}_1{)}^{-1}]=[a_1^{-1}{b}_1^{-1}]=[a_1^{-1}][b_1^{-1}]=\delta ([\alpha ])\delta ([\beta ]),$ the third equality being the product rule for ${\pi }_0(F)$ from [L3] and the first the product convention [L5].

Conclusion, and agreement with the exact sequence. Steps 1.1 and 2.1 show that $\delta$ is well defined on ${\pi }_1(B,[Q_n])$, and step 2.2 shows that it preserves products; by [L7] it is a group homomorphism. For a lift $g$ of $\gamma$ from the identity, write $a=g(1)\in F$. The path $k(t):=g(1-t)\circ a^{-1}$ starts at the identity, ends at $a^{-1}$, and evaluates to $\gamma (1-t)$ because $a^{-1}(Q_n)=Q_n$ setwise. Hence the endpoint-component action of [L6] on the inverse loop class gives ${\mathrm{\partial }}_p[\gamma ]=[\mathrm{id}]\cdot [\gamma {]}^{-1}=[a^{-1}]=\delta ([\gamma ])$ by [L2]. When $n=0$, $B$ is a singleton and the constant path at the identity is one lift of its unique loop. Step 1.1 shows that every other lift gives the same identity component; indeed it is already a path in $F=E$. By [L8], ${\pi }_0(F)$ is trivial as well; when $n=1$ there is no collision condition and the displayed square and concatenation arguments apply verbatim.