[proofplan] We use the chosen bijection $\tau:G \to Q$ to rewrite the given [isotopy](/page/Isotopy) entirely on the underlying set $G$. Composing the original isotopy maps $A,B,C:Q \to G$ with $\tau$ produces [permutations](/page/Permutation) $\alpha,\beta,\gamma$ of $G$. The transported operation is then obtained by applying the isotopy identity to $\tau(x)$ and $\tau(y)$ and solving for $x*y$ with $\gamma^{-1}$. The argument uses the existence of the isotopy identity and the relevant bijections; no separate quasigroup cancellation axiom is needed. [/proofplan] [step:Transport the isotopy maps to permutations of $G$] Define the maps $\alpha:G \to G$, $\beta:G \to G$, and $\gamma:G \to G$ by \begin{align*} \alpha(x)=A(\tau(x)), \quad \beta(x)=B(\tau(x)), \quad \gamma(x)=C(\tau(x)) \end{align*} for every $x \in G$. Since $\tau:G \to Q$ and $A,B,C:Q \to G$ are bijections, each composite $\alpha=A\circ \tau$, $\beta=B\circ \tau$, and $\gamma=C\circ \tau$ is a bijection from $G$ to $G$. Hence $\alpha,\beta,\gamma$ are permutations of $G$. [guided] The original isotopy is written using elements of $Q$, while the theorem wants a formula for the transported operation on the set $G$. The chosen bijection $\tau:G \to Q$ is exactly the device that lets us replace an element $x \in G$ by the corresponding element $\tau(x) \in Q$. Define the maps $\alpha:G \to G$, $\beta:G \to G$, and $\gamma:G \to G$ by \begin{align*} \alpha(x)=A(\tau(x)), \quad \beta(x)=B(\tau(x)), \quad \gamma(x)=C(\tau(x)) \end{align*} for every $x \in G$. These are legitimate maps from $G$ to $G$ because $\tau$ first sends an element of $G$ into $Q$, and then $A$, $B$, or $C$ sends that element of $Q$ back into $G$. We also need these maps to be permutations, not merely functions. Since $\tau$ is a bijection and $A$ is a bijection, their composite $A\circ \tau$ is a bijection from $G$ to $G$. Thus $\alpha$ is a permutation of $G$. The same composition argument applies to $B\circ \tau$ and $C\circ \tau$, so $\beta$ and $\gamma$ are also permutations of $G$. [/guided] [/step] [step:Rewrite the isotopy identity in transported coordinates] Let $x,y \in G$. By definition of the transported binary operation $*:G\times G \to G$, \begin{align*} \tau(x*y)=\tau(x)\circ \tau(y). \end{align*} Applying $C$ to both sides gives \begin{align*} C(\tau(x*y))=C(\tau(x)\circ \tau(y)). \end{align*} Using the isotopy identity with $q=\tau(x)$ and $r=\tau(y)$, and then using the definitions of $\alpha$ and $\beta$, we obtain \begin{align*} C(\tau(x*y))=\alpha(x)+\beta(y). \end{align*} Since $\gamma=C\circ \tau$, the left-hand side is $\gamma(x*y)$. Therefore \begin{align*} \gamma(x*y)=\alpha(x)+\beta(y). \end{align*} [guided] We now translate the original [isotopy](/page/Isotopy) equation from elements of $Q$ into elements of $G$. Let $x,y \in G$. The transported binary operation $*:G\times G \to G$ is defined by requiring that applying $\tau$ recovers the original product in $Q$: \begin{align*} \tau(x*y)=\tau(x)\circ \tau(y). \end{align*} This equation is valid because $\tau:G\to Q$ is a bijection, so each product $\tau(x)\circ\tau(y)\in Q$ has a unique preimage in $G$. Apply the isotopy map $C:Q\to G$ to both sides. This gives \begin{align*} C(\tau(x*y))=C(\tau(x)\circ \tau(y)). \end{align*} The isotopy identity says that for all $q,r\in Q$, \begin{align*} C(q\circ r)=A(q)+B(r). \end{align*} We may use it with $q=\tau(x)$ and $r=\tau(y)$, since $\tau(x),\tau(y)\in Q$. Hence \begin{align*} C(\tau(x)\circ \tau(y))=A(\tau(x))+B(\tau(y)). \end{align*} By the definitions $\alpha=A\circ\tau$ and $\beta=B\circ\tau$, this becomes \begin{align*} C(\tau(x*y))=\alpha(x)+\beta(y). \end{align*} Finally, by the definition $\gamma=C\circ\tau$, the left-hand side is $\gamma(x*y)$. Therefore \begin{align*} \gamma(x*y)=\alpha(x)+\beta(y). \end{align*} This is the transported form of the isotopy identity. [/guided] [/step] [step:Solve for the transported product] Because $\gamma:G \to G$ is a permutation, its inverse map $\gamma^{-1}:G \to G$ exists. Applying $\gamma^{-1}$ to the equality \begin{align*} \gamma(x*y)=\alpha(x)+\beta(y) \end{align*} yields \begin{align*} x*y=\gamma^{-1}(\alpha(x)+\beta(y)). \end{align*} This holds for every $x,y \in G$, which is the desired principal form of the transported group isotope. [guided] It remains only to isolate the transported product $x*y$. From the previous step, for arbitrary $x,y\in G$ we have \begin{align*} \gamma(x*y)=\alpha(x)+\beta(y). \end{align*} The map $\gamma:G\to G$ is a [permutation](/page/Permutation), hence it is bijective and has an inverse map $\gamma^{-1}:G\to G$. Applying this inverse to both sides gives \begin{align*} \gamma^{-1}(\gamma(x*y))=\gamma^{-1}(\alpha(x)+\beta(y)). \end{align*} Since $\gamma^{-1}\circ\gamma$ is the identity map on $G$, the left-hand side reduces to $x*y$. Therefore \begin{align*} x*y=\gamma^{-1}(\alpha(x)+\beta(y)). \end{align*} Because $x$ and $y$ were arbitrary elements of $G$, this formula holds for every pair $x,y\in G$. This is exactly the principal form of the transported group isotope. [/guided] [/step]