The non-source projection theorem

Proof. (N1) If $s_0=s_1$ then $a_Q(s_0)=a_Q(s_1)$, contradicting (13); so $s_0\neq s_1$. These two distinct points have the same image under $p$, so $p$ is not injective. Since $p$ is surjective (Lemma 2.3) but not injective, $p$ is not bijective; hence the readout does not identify each source state uniquely , the states $s_0$ and $s_1$ share the readout value $p(s_0)=p(s_1)$. We emphasize that this argument is direct and invokes no separating hypothesis: Theorem 3.4 is unavailable here, since Assumption 3.3 is not among the hypotheses of the present theorem, and it is not needed , the witness (13) exhibits the noninjectivity outright.

(N2) This is Lemma 2.3, which used only $\simeq_{\mathrm{phys}}$-invariance and surjectivity of $\Pi_{T_1}$, and is untouched by (13). The point deserves emphasis: exactness of the readout and failure of source completeness are perfectly compatible. Losing the fiber’s internal structure is not a defect of exactness; it is what exactness of a noninjective quotient means.

(N3) Suppose such $\bar a$ existed. By Lemma 3.2((1)$\Rightarrow$(2)), $a_Q$ would be constant on $p$-fibers. But $p(s_0)=p(s_1)$ places $s_0,s_1$ in one fiber while $a_Q(s_0)\neq a_Q(s_1)$; contradiction.

(N4) $u_Q$ is surjective by Lemma 2.7. Set $t:=p(s_0)=p(s_1)$ and consider $$e_Q(s_0)=\bigl(t,\,a_Q(s_0)\bigr),\qquad e_Q(s_1)=\bigl(t,\,a_Q(s_1)\bigr).$$ Both lie in $E_Q$; they are distinct, because their second coordinates differ by (13); and $u_Q$ sends both to $t$. Hence $u_Q$ is not injective, i.e. some fiber of $u_Q$ has at least two points, which is what “$E_Q$ properly refines $T_1$” asserts.

(N5) Proved as Theorem 4.6 below.

(N6) By construction (1), $S=\Omega/\!\simeq_{\mathrm{phys}}$, so any two raw realizations related by a declared redundancy have already been identified. Since $s_0\neq s_1$ in $S$, no such redundancy relates their representatives; a fortiori no relabelling in the declared class maps one to the other. Cf. (2). ◻

Proof. Existence. Since $e$ is surjective, by Lemma 3.2 it suffices to check that $e_Q$ is constant on the fibers of $e$. Let $e(s)=e(s')$. Then:

• $p(s)=\pi(e(s))=\pi(e(s'))=p(s')$, using (10);

• $a_Q(s)=\hat a(e(s))=\hat a(e(s'))=a_Q(s')$, using (11).

Hence $e_Q(s)=(p(s),a_Q(s))=(p(s'),a_Q(s'))=e_Q(s')$. Lemma 3.2 now supplies a unique $\phi\colon E\to E_Q$ with $\phi\circ e=e_Q$, which is also the asserted uniqueness.

Surjectivity. $e_Q$ is surjective onto $E_Q$ (Lemma 2.7) and factors as $\phi\circ e$; hence $\mathop{\mathrm{Im}}\phi\supseteq\mathop{\mathrm{Im}}(\phi\circ e)=E_Q$.

Compatibility. For $s\in S$: $u_Q(\phi(e(s)))=u_Q(e_Q(s))=p(s)=\pi(e(s))$, and since $e$ is surjective, $u_Q\circ\phi=\pi$ on all of $E$. Likewise $\rho_Q(\phi(e(s)))=\rho_Q(e_Q(s))=a_Q(s)=\hat a(e(s))$, giving $\rho_Q\circ\phi=\hat a$. ◻

Proof. The hypothesis says exactly $a_Q(s_0)=\theta(r(s_0))\neq\theta(r(s_1))=a_Q(s_1)$ with $p(s_0)=p(s_1)$, i.e. (13); apply Theorem 4.5. For the last claim, $r(s_0)\neq r(s_1)$ (else $\theta$ would take equal values), so $e_{\{r\}}(s_0)\neq e_{\{r\}}(s_1)$ while $u_{\{r\}}$ identifies them, exactly as in the proof of (N4). ◻

Proof. $K(\Psi+n)=K\Psi+Kn=K\Psi$, so $p(\Psi+n)=p(\Psi)$; with (14) this is the witness (13) with $s_0=\Psi$, $s_1=\Psi+n$. (Note $n\neq0$ follows from (14).) Apply Theorem 4.5 and Theorem 4.6. That every sufficient extension separates the two cosets: if $e(\Psi)=e(\Psi+n)$, sufficiency (11) would force $a_Q(\Psi)=a_Q(\Psi+n)$, contradiction. ◻

Proof. Suppose $\sigma$ were such a selector and let $t$ satisfy (15). Since $p(\sigma(t))=t$, we have $\sigma(t)\in p^{-1}(t)$. Let $g\in G_t$. Then, using equivariance of $\sigma$ and $g\cdot t=t$, $$g\cdot\sigma(t)=\sigma(g\cdot t)=\sigma(t).$$ As $g\in G_t$ was arbitrary, $\sigma(t)$ is fixed by every element of $G_t$, i.e. $\sigma(t)\in\mathop{\mathrm{Fix}}_{G_t}(p^{-1}(t))$, contradicting (15). ◻