Here's my definition of a Clifford algebra:
Definition: Let $B(\cdot,\cdot)$ be a symmetric bilinear form on a vector space $V$ over $\mathbb{K}$ and $Q$ its associated quadratic form. The Clifford algebra associated to the quadratic space $(V,Q)$ is a pair $\mathcal{C}l(V,Q)$ where $\mathcal{C}l(V,Q)$ is a $\mathbb{K}$ associative algebra with identity $1$ and $\varphi: V \to \mathcal{C}l(V,Q)$ is a Clifford map satisfying the following properties:
(a) $\varphi(u)\varphi(v)+\varphi(v)\varphi(u) = 2B(u,v)1$ for every $u,v \in V$.
(b) The subspace $\text{Im}\varphi$ generates the algebra $\mathcal{C}l(V,Q)$.
(c) For every Clifford map $\phi: V \to \mathcal{A}$ on $(V,Q)$ there exists a homomorphism of algebras $f: \mathcal{C}l(V,Q) \to \mathcal{A}$ such that $\phi = f\circ \varphi$.
When $B \equiv 0$, then $\mathcal{C}l(V,Q)$ becomes an exterior algebra $\bigwedge V$. How can I prove that the Clifford map $\varphi$ is injective, in this case? I've tried using the universal property of exterior algebras but got nowhere.
This seems straightforward, keeping in mind that all the maps involved are linear. Take $\phi:V\hookrightarrow \bigwedge V$ to be the natural embedding. This is a Clifford map because $$\phi(v)^2=v\wedge v=0=Q(v)1.$$ Now, by the universal property, there exists a map $f:Cl(V,Q)\to \bigwedge V$ such that $\phi=f\circ\varphi$. Since $\phi$ is injective, so is $\varphi$. (If $x\in \ker\varphi$, then $\phi(x)=f(\varphi(x))=f(0)=0$, so $x=0$ because $\phi$ is injective. Therefore $\ker\varphi=\{0\}$.)