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partowp
2026-08-03 20:29:50 +01:00
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@@ -577,24 +577,24 @@ The definition derives the ordering on morphisms of each hom-set $\Hom(X,\sub Y)
\end{cases} \end{cases}
\end{gather*} \end{gather*}
First, we need to prove that $\mu'$ is a subdistribution. First, we need to prove that $\mu'$ is a subdistribution.
For every $x \in X$, $\mu'(x) \ge 0$. The total mass of $\mu'$ is For every $x \in X$, $\mu'(x) \ge 0$. We have:
\begin{align*} \begin{align*}
\sum_{x \in X} \mu'(x)&\\ \sum_{x \in X} \mu'(x)&\\
=& \sum_{x \in X} \frac{\nu(g(x))}{Sg(\mu)(g(x))} \comp \mu(x)\\ =& \sum_{x \in X} \frac{\nu(g(x))}{Sg(\mu)(g(x))} \comp \mu(x)&(Sg(\mu)(g(x))\neq 0)\\
=& \sum_{y \in Y} \sum_{x \in g^{\mone}(y)} \frac{\nu(y)}{Sg(\mu)(y)} \comp \mu(x)\\ =& \sum_{y \in Y} \sum_{x \in g^{\mone}(y)} \frac{\nu(y)}{Sg(\mu)(y)} \comp \mu(x)&(Sg(\mu)(g(x))\neq 0)\\
=& \sum_{y \in Y} \frac{\nu(y)}{Sg(\mu)(y)} \comp \sum_{x \in g^{\mone}(y)} \mu(x)\\ =& \sum_{y \in Y} \frac{\nu(y)}{Sg(\mu)(y)} \comp \sum_{x \in g^{\mone}(y)} \mu(x)&(Sg(\mu)(g(x))\neq 0)\\
=& \sum_{y \in Y} \frac{\nu(y)}{Sg(\mu)(y)} \comp Sg(\mu)(y)\\ =& \sum_{y \in Y} \frac{\nu(y)}{Sg(\mu)(y)} \comp Sg(\mu)(y)&(Sg(\mu)(g(x))\neq 0)\\
=& \sum_{y \in Y} \nu(y)\\ =& \sum_{y \in Y} \nu(y)\\
\leq& 1 \leq& 1
\end{align*} \end{align*}
Thus $\mu'$ is a subdistribution. Now, we prove $Sg(\mu') = \nu$. As for every $x\in X$ that $Sg(\mu)(g(x))=0$, $\mu'(x)=0$, it does not have effect the inequality. Thus $\mu'$ is a subdistribution. Now, we prove $Sg(\mu') = \nu$.
For any $y \in Y$, we have: For any $y \in Y$, we have:
\begin{align*} \begin{align*}
Sg(\mu')(y)&\\ Sg(\mu')(y)&\\
= &\sum_{x \in g^{\mone}(y)} \mu'(x)\\ = &\sum_{x \in g^{\mone}(y)} \mu'(x)\\
= &\sum_{x \in g^{\mone}(y)} \frac{\nu(y)}{Sg(\mu)(y)} \comp \mu(x)\;\;(Sg(\mu)(y)\neq 0)\\ = &\sum_{x \in g^{\mone}(y)} \frac{\nu(y)}{Sg(\mu)(y)} \comp \mu(x)&(Sg(\mu)(y)\neq 0)\\
= &\frac{\nu(y)}{Sg(\mu)(y)} \comp \sum_{x \in g^{\mone}(y)} \mu(x)\;\;(Sg(\mu)(y)\neq 0)\\ = &\frac{\nu(y)}{Sg(\mu)(y)} \comp \sum_{x \in g^{\mone}(y)} \mu(x)&(Sg(\mu)(y)\neq 0)\\
= &\frac{\nu(y)}{Sg(\mu)(y)} \comp Sg(\mu)(y)\;\;(Sg(\mu)(y)\neq 0)\\ = &\frac{\nu(y)}{Sg(\mu)(y)} \comp Sg(\mu)(y)&(Sg(\mu)(y)\neq 0)\\
= &\mu(y) = &\mu(y)
\end{align*} \end{align*}
(If $Sg(\mu)(y) = 0$, then $\nu(y) = 0$, and the sum is $0$ as well, so the equality still holds.) \\ (If $Sg(\mu)(y) = 0$, then $\nu(y) = 0$, and the sum is $0$ as well, so the equality still holds.) \\