Construction of $C\ell(V,q)$ for a quadratic space $(V,q)$ and proving an injection $V\hookrightarrow C\ell(V,q)$ exists. I've been reading Lawson & Michelsohn's Spin Geometry, and struggled particularly with their (apparently wrong) proof that a quadratic space embeds into its Clifford algebra. I did a bit of reading and came to the following understanding. Could anyone confirm to me that this works (or if it doesn't, why, and how can I fix it)? Many thanks.
Here all algebras are unital and associative. Let $(V,q)$ be a quadratic (finite-dimensional $k$-vector) space, and define the polarisation of $q$ to be $2q(a,b)=q(a+b)-q(a)-q(b)$.

Definition: Form a category $A_k(V,q)$ by taking objects to be linear maps $f:M\to A$ for some algebra $A$, with the property that for each $a\in V$, $f(a)^2=q(a)1_A$. Take a morphism between objects $f:V\to A$ and $g:V\to B$ to be a $k$-algebra homomorphism $u:A\to B$ with $u\circ f=g$.

I claim that $A_k(V,q)$ has an initial universal object, namely a map $\varphi:V\to C$ (with $\text{char}\neq 2$ ) for some $k$-algebra $C$ so that for any map $f:V\to A$ there is a morphism $\overline{f}:C\to A$ which has $f=\overline{f}\circ \varphi$. First, notice that if such a $\varphi$ was to exist,
\begin{equation}
\varphi(a+b)^2=\varphi(a^2+ab+ba+b^2)=q(a)1+\varphi(ab+ba)+q(b)1
\end{equation}
But also $\varphi(a+b)^2=q(a+b)1$, so $2q(a,b)=ab+ba$ for any $a,b\in V$. Write $C\ell(V,q)$ for $C$.

Proposition: Let $\mathscr{T}(V)$ be the associated tensor algebra, and let $\mathscr{I}_q(V)\subset\mathscr{T}(V)$ be the ideal generated by $v\otimes v -q(v)1$ for $v\in V$. Define $C\ell(V,q)$ to be the quotient $\mathscr{T}(V)/\mathscr{I}_q(V)$. We claim the canonical map
$$\varphi=\pi_q\circ i:V\hookrightarrow \mathscr{T}(V)\to \mathscr{T}(V)/\mathscr{I}_q(V)=C\ell (V,q)$$
is the initial universal object in $A_k(V,q)$.

Proof: Let $f:V\to A$ be an object of $A_k(V,q)$. Notice that by the universal property of $\mathscr{T}(V)$, the map $f$ extends uniquely to an algebra map $\overline{f}:\mathscr{T}(V)\to A$. Since $f(v)^2=q(v)1$, $\overline{f}$ vanishes on $\mathscr{I}_q(V)$ so by the universal property of quotients, $\overline{f}$ ascends uniquely to an algebra map $\widetilde{f}:C\ell(V,q)\to A$. We know $\widetilde{f}\circ \pi_q= \overline{f}$. Descending to $V\cong T^1(V)$, we see $\widetilde{f}\circ\varphi=f$.



Proposition: The space $V$ embeds into $C\ell(V,q)$, namely the canonical map $\varphi:V\hookrightarrow C\ell(V,q)$ is an injection.

Proof: First suppose the polarisation of $q$ is nondegenerate. Then fix a nonzero $a\in V$ and take (by nondegeneracy) some $b\in V$ which has $q(a,b)\neq 0$. From the previous remark, we see $\varphi(a)\varphi(b)+\varphi(b)\varphi(a)=2q(a,b)$, so it must be the case that $\varphi(a)\neq 0$. That is, the kernel of $\varphi$ is trivial. Now, let $q$ be totally degenerate on some subspace $W\subset V$, and without loss of generality, suppose $q$ is nondegenerate on another subspace $Z$, so that $V=Z\oplus W$. For each $a\in W$, $q(a,b)=0$ for every $b\in V$, so $C\ell(V,q)=\bigwedge(V)$, and $\varphi:V\hookrightarrow \bigwedge(V)$ is injective. Then since $C\ell(V,q)=C\ell(Z, q)\oplus C\ell(W,0)$ and $\varphi$ is injective on both summands, the result follows.


 A: It so happens that
$$C\ell(W\bot Z)\color{red}{\not\cong} C\ell(W)\oplus C\ell(Z)$$ (by $\bot$ I mean the orthogonal sum of spaces), but rather
$$C\ell(W\bot Z) \cong C\ell(W) \hat \otimes C\ell(Z)$$
where $\hat \otimes$ is the super tensor product (which is the coproduct in the category of commutative superalgebras, much like the usual tensor product of not necessarily commutative algebras is the coproduct when restricted to the category of commutative algebras). The simplest way to prove this is to show that $C\ell(W) \hat \otimes C\ell(Z)$ satisfies the universal property of $C\ell(W\bot Z)$.

To finish your proof you might go this way: let $q$ be a degenerate form, and let $W=\operatorname{rad} q$ the radical of the form (elements $a\in V$ such that $\forall b\in V\, q(a,b)=0$). Then, $V=W\bot Z$ for some non-degenerate subspace $Z$ (you can either show this directly or use Witt's theorem). Thus,
$$C\ell(V)=C\ell(W\bot Z) \cong C\ell(W) \hat \otimes C\ell(Z)=\Lambda(W)\hat\otimes C\ell(Z).$$
We have the following commutative diagram, thanks to functoriality of Clifford algebras:
$$\begin{matrix}
W & \color{red}\rightarrow & V \\
\color{blue}\downarrow && \color{red}\downarrow \\
\Lambda(W) & \color{blue}\rightarrow & C\ell(V)\cong\Lambda(W)\hat\otimes C\ell(Z)
\end{matrix}$$
where all maps are inclusions. We are interested in whether the red map is injective; commutativity of the diagram means the red map equals the blue map. Now, we take the injectivity of $W \hookrightarrow \Lambda(W)$ as a known result, and injectivity of the bottom blue map comes from the properties of super tensor products of superalgebras.
