Why is $C\ell$ functorial (why can we apply the appropriate universal property)?

Question

I'm struggling to understand how a map $f:(V,q)\to (V',q')$ with $q'(f(v))=q(v)$, namely $f^*q'=q$ gives a algebra homomorphism of Clifford algebras $C\ell(V,q)\to C\ell(V',q')$. I understand this is a consequence of the universal property; for any linear map $f:V\to A$ with $f(v)^2=-q(v)$, there is a unique algebra homomorphism extending $f$ to a algebra homomorphism $C\ell(V,q)\to A$. I understand that we can regard $V\subset C\ell(V,q)$ by the quotient map $\pi_q|_V:V\hookrightarrow C\ell (V,q)$, so we have a linear map $f:V\to C\ell(V',q')$ with $q'(f(v))=q(v)$. In order to apply the universal property though, we need to have $f(v)^2=-q(v)$. I can't clearly see this although it should be easy; any guidence here would be greatly appreciated.

Answer 1

You correctly noted that we have a linear map $F := \pi_{q'}|_{V'} \circ f : V \rightarrow C\ell (V',q')$. So let's do a little computation, assuming $v \in V$: \begin{equation*} F(v)^{2} = \left(\pi_{q'}|_{V'}f(v)\right)^{2} = -q'(f(v)) = -q(v), \end{equation*} which is the desired relation that shows that $F$ is a Clifford map.

Answer 2

The expression $f(v)^2$ means multiplication in the Clifford algebra $C\ell(V',q')$, right? So by the definition of that Clifford algebra it equals $-q'(f(v))$. And by hypothesis, this equals $-q(v)$.