From 51b64cc427c1b877e34387057feb6a0fe25fafdc Mon Sep 17 00:00:00 2001 From: partowp Date: Mon, 3 Aug 2026 20:29:50 +0100 Subject: [PATCH] minor --- draft/draft.tex | 18 +++++++++--------- 1 file changed, 9 insertions(+), 9 deletions(-) diff --git a/draft/draft.tex b/draft/draft.tex index 861be01..e1f0014 100644 --- a/draft/draft.tex +++ b/draft/draft.tex @@ -577,24 +577,24 @@ The definition derives the ordering on morphisms of each hom-set $\Hom(X,\sub Y) \end{cases} \end{gather*} 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*} \sum_{x \in X} \mu'(x)&\\ - =& \sum_{x \in X} \frac{\nu(g(x))}{Sg(\mu)(g(x))} \comp \mu(x)\\ - =& \sum_{y \in Y} \sum_{x \in g^{\mone}(y)} \frac{\nu(y)}{Sg(\mu)(y)} \comp \mu(x)\\ - =& \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 Sg(\mu)(y)\\ + =& \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)&(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)&(Sg(\mu)(g(x))\neq 0)\\ + =& \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)\\ \leq& 1 \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: \begin{align*} Sg(\mu')(y)&\\ = &\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)\\ - = &\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)\\ + = &\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 Sg(\mu)(y)&(Sg(\mu)(y)\neq 0)\\ = &\mu(y) \end{align*} (If $Sg(\mu)(y) = 0$, then $\nu(y) = 0$, and the sum is $0$ as well, so the equality still holds.) \\