The even Clifford norm detects the identity component in mixed signature

For $A\in SO(p,q)$, $\det K_{+}\det K_{-}=\det K=1$, so the two block determinants have the same sign. The polar path in [F1] joins $A$ to $K\in S(O(p)\times O(q))$. For each $m\ge 1$, $SO(m)$ is path connected: $SO(1)=\{I\}$; for $m\ge 2$, rotate $A{e}_m$ to $e_m$ in their plane, using a half-turn in a plane containing $e_m$ if $A{e}_m=-e_m$, and then apply induction to the restriction to $e_m^{\perp }$. Each determinant component of $O(m)$ is a translate of $SO(m)$. Thus the two common determinant signs label exactly the two components of $SO(p,q)$, and the sign $+1$ component is $S{O}_0(p,q)$.

Along any path in $SO(p,q)$, a chosen lift exists by [F2]. Indeed, cover the compact parameter interval by evenly covered neighborhoods, choose a finite subdivision subordinate to this cover, and continue the lift over each subinterval in the sheet agreeing at its initial endpoint. By [F3], its norm is a continuous function with values in $\{1,-1\}$, so it is constant along the lifted path.

If the form is definite and $\dim V>0$, every normalized vector has the same value of $Q$, either $1$ or $-1$. Every full-even Spin element is a product of an even number of such vectors, so [F3] gives norm $1$; hence the full-even and norm-one subgroups coincide. If $V=0$, the definition gives full Spin $=\{1,-1\}$ and both elements have norm $1$. The rotation argument in step 1.1 also shows that definite $SO(V)$ is connected, including $SO(0)=\{I\}$.

Factor $K_{+}$ and $K_{-}$ into reflections using [F4]. Their combined number is even because $\det K=1$, so normalized vector lifts give a full-even Spin lift of $K$. A reflection in the positive subspace contributes norm $-1$, and one in the negative subspace contributes norm $+1$, by [F3]. If $k_{+}$ is the number of positive-subspace reflections, the lift therefore has norm $(-1{)}^{k_{+}}=\det K_{+}$. The two lifts differ by $-1$ by [F2], and $N(-1)=1$, so this norm is independent of the chosen lift. By step 2.1 it is the same for every lift of $A$ joined to $K$. Since [F2] gives a full-even lift for every $A$, norm one occurs exactly when $\det K_{+}=1$. Step 1.1 identifies these matrices with $S{O}_0(p,q)$, proving the image equality.

In signature $(1,1)$, put $E=e_{+}{e}_{-}$ with $e_{+}^2=1$ and $e_{-}^2=-1$. Then $E^2=1$ and $\bar{E}=-E$. For $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in SO(1,1)$ and $J=\mathrm{diag}(1,-1)$, orthogonality gives $A^{-1}=J{A}^{T}J$, while determinant one gives $A^{-1}=\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)$. Equating entries yields $d=a$ and $c=b$; the $(1,1)$ entry of $A^{T}JA=J$ then gives $a^2-b^2=1$. Thus $S{O}_0(1,1)$ is the branch $a>0$, parameterized by $a=\cosh t$, $b=\sinh t$. Set $w(t)=\cosh (t\mathrm{/}2)e_{+}-\sinh (t\mathrm{/}2)e_{-}$; then $Q(w(t))=1$ and $x(t)=e_{+}w(t)=\cosh (t\mathrm{/}2)-E\sinh (t\mathrm{/}2)$ is a product of two normalized vectors, hence a full-even Spin lift. Since $\bar{E}=-E$, direct multiplication gives $N(x(t))=1$. Also $[E,e_{+}]=-2e_{-}$ and $[E,e_{-}]=-2e_{+}$, so ${\mathrm{ad}}_E^2=4I$ on the span of $e_{+},e_{-}$. Since $x(t)$ is even, its twisted action is conjugation; from $x(t)=e^{-tE\mathrm{/}2}$ and the exponential series, $e^{-t{\mathrm{ad}}_E\mathrm{/}2}=\cosh (t)I-\frac{1}{2}\sinh (t){\mathrm{ad}}_E$. Applying this formula to $e_{+}$ and $e_{-}$ gives ${\rho }_{x(t)}(e_{+})=\cosh (t)e_{+}+\sinh (t)e_{-}$ and ${\rho }_{x(t)}(e_{-})=\sinh (t)e_{+}+\cosh (t)e_{-}$, which is the matrix in the identity branch. By [F2], every fiber has exactly the two lifts $\pm x(t)$, so the full preimage of $S{O}_0(1,1)$ is $\{\pm (\cosh s+E\sinh s):s\in \mathbb{R}\}$. Each sign branch is connected, and the scalar coordinate is respectively positive or negative, so these branches are separated. In particular $1$ and $-1$ lie in different connected components of the norm-one preimage, which is therefore not the identity component of full-even Spin.