Is the Reversion map in Geometric Algebra well-defined? I am studying from the book "Geometric Algebra for Physicists: by Chris Doran and Anthony Lasenby."  In the book they define a map $\dagger : \mathcal G \to \mathcal G$, where $\mathcal G$ is a geometric algebra, by $(a_1\ldots a_r)^\dagger=a_r\ldots a_1$ where $a_1,\ldots,a_r$ are vectors. My question is if $$a_1\ldots a_r=b_1 \ldots b_s$$ where $r$ is not necessarily equal to $s$ how does one know that $(a_1\ldots a_r)^\dagger=(b_1\ldots b_s)^\dagger$?
 A: Such a map exists and is unique by the universal property of Clifford algebras : let $\mathcal{C}=\mathcal{C}(V,Q)$ be the Clifford algebra on $V$ where $Q$ is a quadratic form on the $\Bbbk$-vector space $V$. By definition, $\mathcal{C}$ is generated as a $\Bbbk$-algebra by the vectors of $V$ and the relations $vv+Q(v)=0$ (conventions differ). It is universal in that for any $\Bbbk$-algebra $\mathcal A$ and any map of $\Bbbk$-vector spaces $\phi:V\to\mathcal{A}$ such that for all $v\in V$, $\phi(v)^2+Q(v)=0$, there exits a unique extension of $\tilde\phi$ to $\mathcal{C}\to\mathcal{A}$ that is a morphism of $\Bbbk$-algebras:
$$\begin{matrix}
V&\xrightarrow{\phi}&\mathcal A\\
\downarrow&\nearrow_{\tilde\phi}\\
\mathcal C
\end{matrix}$$
You then simply apply the universal property to the $\Bbbk$-algebra $\mathcal A=\mathcal C^{\text{op}}$ and the canonical inclusion $V\to\mathcal A=\mathcal C$ (the equality is that of vector spaces). By definition, this map reverses the order of multiplication, that is
$$\tilde\phi(e_1\cdots e_n)=\phi(e_1)\times^{\text{op}}\cdots\times^{\text{op}}\phi(e_n)=e_1\times^{\text{op}}\cdots\times^{\text{op}}e_n=e_n\times\cdots\times e_1$$
(multiplication in the two central terms is that of $\mathcal C^{\text{op}}$, and the right hand side is the translation in terms of products in $\mathcal C$.)
