**Proof.**
To see part (1) let $y$ be an object of $\mathcal{Y}$ which lies over a scheme $V$ such that the corresponding morphism $y : V \to \mathcal{Y}$ is flat. Then $g(y) : V \to \mathcal{Y} \to \mathcal{X}$ is flat as a composition of flat morphisms (see Morphisms of Stacks, Lemma 101.25.2) hence $\mathcal{F}(g(y))$ is zero by assumption. Since $g^*\mathcal{F} = g^{-1}\mathcal{F}(y) = \mathcal{F}(g(y))$ we conclude $g^*\mathcal{F}$ is parasitic.

To see part (2)(a) note that if $\{ x_ i \to x\} $ is a $\tau $-covering of $\mathcal{X}$, then each of the morphisms $x_ i \to x$ lies over a flat morphism of schemes. Hence if $x$ lies over a scheme $U$ such that $x : U \to \mathcal{X}$ is flat, so do all of the objects $x_ i$. Hence the presheaf $\mathcal{F}^+$ (see Sites, Section 7.10) is parasitic if the presheaf $\mathcal{F}$ is parasitic. This proves (2)(a) as the sheafification of $\mathcal{F}$ is $(\mathcal{F}^+)^+$.

Let $\mathcal{F}$ be a parasitic $\tau $-module. It is immediate from the definitions that any submodule of $\mathcal{F}$ is parasitic. On the other hand, if $\mathcal{F}' \subset \mathcal{F}$ is a submodule, then it is equally clear that the presheaf $x \mapsto \mathcal{F}(x)/\mathcal{F}'(x)$ is parasitic. Hence the quotient $\mathcal{F}/\mathcal{F}'$ is a parasitic module by (2)(a). Finally, we have to show that given a short exact sequence $0 \to \mathcal{F}_1 \to \mathcal{F}_2 \to \mathcal{F}_3 \to 0$ with $\mathcal{F}_1$ and $\mathcal{F}_3$ parasitic, then $\mathcal{F}_2$ is parasitic. This follows immediately on evaluating on $x$ lying over a scheme flat over $\mathcal{X}$. This proves (2)(b), see Homology, Lemma 12.10.2.

Let $f_ i : \mathcal{X}_ i \to \mathcal{X}$ be a jointly surjective family of smooth morphisms of algebraic stacks and assume each $f_ i^*\mathcal{F}$ is parasitic. Let $x$ be an object of $\mathcal{X}$ which lies over a scheme $U$ such that $x : U \to \mathcal{X}$ is flat. Consider a surjective smooth covering $W_ i \to U \times _{x, \mathcal{X}} \mathcal{X}_ i$. Denote $y_ i : W_ i \to \mathcal{X}_ i$ the projection. It follows that $\{ f_ i(y_ i) \to x\} $ is a covering for the smooth topology on $\mathcal{X}$. Since a composition of flat morphisms is flat we see that $f_ i^*\mathcal{F}(y_ i) = 0$. On the other hand, as we saw in the proof of (1), we have $f_ i^*\mathcal{F}(y_ i) = \mathcal{F}(f_ i(y_ i))$. Hence we see that for some smooth covering $\{ x_ i \to x\} _{i \in I}$ in $\mathcal{X}$ we have $\mathcal{F}(x_ i) = 0$. This implies $\mathcal{F}(x) = 0$ because the smooth topology is the same as the étale topology, see More on Morphisms, Lemma 37.38.7. Namely, $\{ x_ i \to x\} _{i \in I}$ lies over a smooth covering $\{ U_ i \to U\} _{i \in I}$ of schemes. By the lemma just referenced there exists an étale covering $\{ V_ j \to U\} _{j \in J}$ which refines $\{ U_ i \to U\} _{i \in I}$. Denote $x'_ j = x|_{V_ j}$. Then $\{ x'_ j \to x\} $ is an étale covering in $\mathcal{X}$ refining $\{ x_ i \to x\} _{i \in I}$. This means the map $\mathcal{F}(x) \to \prod _{j \in J} \mathcal{F}(x'_ j)$, which is injective as $\mathcal{F}$ is a sheaf in the étale topology, factors through $\mathcal{F}(x) \to \prod _{i \in I} \mathcal{F}(x_ i)$ which is zero. Hence $\mathcal{F}(x) = 0$ as desired.

Proof of (4): omitted. Hint: similar, but simpler, than the proof of (3).
$\square$

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