[proofplan] The proof uses only the defining torsor property of an [affine space](/page/Affine%20Space): for a fixed origin $o$, every point of $A$ is reached from $o$ by a unique vector in $V$. This gives surjectivity by sending $v$ to $o+v$, and injectivity by uniqueness of the vector carrying $o$ to a point. The compatibility formula follows from associativity of the affine action, namely $(p+u)+v=p+(u+v)$. [/proofplan] [step:Use the affine difference to define the coordinate map] Because $A$ is an affine space modeled on $V$, the action of $V$ on $A$ is free and transitive. Therefore, for each $p\in A$, there exists a unique vector $w\in V$ such that \begin{align*} o+w=p. \end{align*} By definition of affine difference, this unique vector is $p-o$. Hence the assignment \begin{align*} \phi_o:A\to V,\qquad p\mapsto p-o \end{align*} is well-defined. [/step] [step:Show that every model vector is attained] Let $v\in V$. The point $o+v$ belongs to $A$ by the affine action. By definition, $\phi_o(o+v)$ is the unique vector $w\in V$ such that \begin{align*} o+w=o+v. \end{align*} Freeness of the affine action gives $w=v$. Thus \begin{align*} \phi_o(o+v)=v. \end{align*} Since $v\in V$ was arbitrary, $\phi_o$ is surjective. [guided] We want to prove surjectivity, so we start with an arbitrary vector $v\in V$ and look for a point of $A$ that maps to it. The natural candidate is the point obtained by translating the chosen origin $o$ by $v$, namely $o+v\in A$. By definition, $\phi_o(o+v)$ is the unique vector $w\in V$ satisfying \begin{align*} o+w=o+v. \end{align*} The affine action is free: if two vectors translate the same point to the same point, then the vectors are equal. Applying freeness at the point $o$ gives $w=v$. Therefore \begin{align*} \phi_o(o+v)=v. \end{align*} Since this construction works for every $v\in V$, every vector in $V$ lies in the image of $\phi_o$, so $\phi_o$ is surjective. [/guided] [/step] [step:Show that two points with the same coordinate vector are equal] Let $p,q\in A$ and suppose that \begin{align*} \phi_o(p)=\phi_o(q). \end{align*} By the definition of $\phi_o$, this means \begin{align*} p-o=q-o. \end{align*} The defining property of affine difference gives \begin{align*} o+(p-o)=p \end{align*} and \begin{align*} o+(q-o)=q. \end{align*} Substituting $p-o=q-o$ into these identities yields $p=q$. Hence $\phi_o$ is injective. [/step] [step:Verify compatibility with the affine action] Let $p\in A$ and $v\in V$. Set $u:=\phi_o(p)=p-o\in V$. Then \begin{align*} o+u=p. \end{align*} Using associativity of the affine action, we obtain \begin{align*} p+v=(o+u)+v=o+(u+v). \end{align*} Thus the unique vector carrying $o$ to $p+v$ is $u+v$. Therefore \begin{align*} \phi_o(p+v)=u+v=\phi_o(p)+v. \end{align*} The previous steps show that $\phi_o$ is a bijection, and this identity proves the asserted compatibility with the affine action. [/step]