= Solution
The original PDF shows the chain $A\to C\to B$, not a collider. Its <Bayesian network> factorization is $p(a,b,c)=p(a)p(c\mid a)p(b\mid c)$. For any $c$ with positive marginal <probability>,
$$
p(a,b\mid c)=\frac{p(a)p(c\mid a)p(b\mid c)}{p(c)}=p(a\mid c)p(b\mid c).
$$
Integrating or summing over $a$ also gives the same conditional marginal $p(b\mid c)$. Thus
$$
\boxed{A\mathrel{\perp\!\!\!\perp}B\mid C.}
$$
This is <conditional independence>; the graph does not generally imply marginal <independence>, since summing $p(a)p(c\mid a)p(b\mid c)$ over $c$ can transmit dependence from $A$ to $B$. Special <statistical parameter> choices can make marginal <independence> hold as well, so the claim is that only the conditional statement is guaranteed by the graph. <Conditional distributions> on null values of $C$ are immaterial to this assertion.
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