The Clifford quotient has its canonical parity grading

In the tensor-length parity grading of [F2], each $r_v=j(v{)}^2-Q(v)1$ is even, since its two terms have tensor lengths $2$ and $0$. Let $I$ be the subspace of finite sums of elements $a{r}_{v}b$ with $a,b\in T_k(V)$ and $v\in V$. It is a two-sided ideal containing every $r_v$, and every two-sided ideal containing all $r_v$ contains $I$; hence [F1] gives $I=J_Q$.

Write $a=a_0+a_1$ and $b=b_0+b_1$ according to tensor-length parity. Then $a{r}_{v}b={\sum }_{i,j\in \{0,1\}}{a}_i{r}_v{b}_j$, and each summand has parity $i+j(\mathrm{m}\mathrm{o}\mathrm{d}2)$ and belongs to $I$. Thus every homogeneous component of every generator $a{r}_{v}b$ lies in $I$; by linearity the same holds for every element of $I=J_Q$. Therefore $J_Q$ is a graded ideal.

Since $J_Q$ is graded, its quotient decomposes as the direct sum of the images of the two parity components of $T_k(V)$, and the quotient multiplication preserves their parities. The tensor-algebra unit has even length, while each $j(v)$ has odd length; their quotient images therefore satisfy $1\in {\mathrm{C}\mathrm{l}}^0(V,Q)$ and $\bar{v}\in {\mathrm{C}\mathrm{l}}^1(V,Q)$. This is the claimed canonical grading, including when $Q$ is degenerate.