Part (1) we saw in Proposition \ref{proposition-equivalence-quasi-coherent}. \medskip\noindent We have seen in Schemes, Section \ref{schemes-section-quasi-coherent} that a colimit of quasi-coherent sheaves on a scheme is a quasi-coherent sheaf. Moreover, in Remark \ref{remark-change-topologies-ringed-sites} we saw that $\mathcal{F} \mapsto \mathcal{F}^a$ is the pullback functor for a morphism of ringed sites, hence commutes with all colimits, see Modules on Sites, Lemma \ref{sites-modules-lemma-exactness-pushforward-pullback}. Thus (3) and its special case (3) hold. \medskip\noindent This also shows that the functor is right exact (i.e., commutes with finite colimits), hence (4). \medskip\noindent The functor $\QCoh(\mathcal{O}_S) \to \QCoh(S_\etale, \mathcal{O})$, $\mathcal{F} \mapsto \mathcal{F}^a$ is left exact because an \'etale morphism is flat, see Morphisms, Lemma \ref{morphisms-lemma-etale-flat}. This proves (5). \medskip\noindent To see (6), suppose that $S = \Spec(\mathbf{Z})$. Then $2 : \mathcal{O}_S \to \mathcal{O}_S$ is injective but the associated map of $\mathcal{O}$-modules on $(\Sch/S)_\tau$ isn't injective because $2 : \mathbf{F}_2 \to \mathbf{F}_2$ isn't injective and $\Spec(\mathbf{F}_2)$ is an object of $(\Sch/S)_\tau$. \medskip\noindent We omit the proofs of (7) and (8). \medskip\noindent Let $0 \to \mathcal{F}_1^a \to \mathcal{E} \to \mathcal{F}_2^a \to 0$ be a short exact sequence of $\mathcal{O}$-modules with $\mathcal{F}_1$ and $\mathcal{F}_2$ quasi-coherent on $S$. Consider the restriction $$ 0 \to \mathcal{F}_1 \to \mathcal{E}|_{S_{Zar}} \to \mathcal{F}_2 $$ to $S_{Zar}$. By Proposition \ref{proposition-same-cohomology-quasi-coherent} we see that on any affine $U \subset S$ we have $H^1(U, \mathcal{F}_1^a) = H^1(U, \mathcal{F}_1) = 0$. Hence the sequence above is also exact on the right. By Schemes, Section \ref{schemes-section-quasi-coherent} we conclude that $\mathcal{F} = \mathcal{E}|_{S_{Zar}}$ is quasi-coherent. Thus we obtain a commutative diagram $$ \xymatrix{ & \mathcal{F}_1^a \ar[r] \ar[d] & \mathcal{F}^a \ar[r] \ar[d] & \mathcal{F}_2^a \ar[r] \ar[d] & 0 \\ 0 \ar[r] & \mathcal{F}_1^a \ar[r] & \mathcal{E} \ar[r] & \mathcal{F}_2^a \ar[r] & 0 } $$ To finish the proof it suffices to show that the top row is also right exact. To do this, denote once more $U = \Spec(A) \subset S$ an affine open of $S$. We have seen above that $0 \to \mathcal{F}_1(U) \to \mathcal{E}(U) \to \mathcal{F}_2(U) \to 0$ is exact. For any affine scheme $V/U$, $V = \Spec(B)$ the map $\mathcal{F}_1^a(V) \to \mathcal{E}(V)$ is injective. We have $\mathcal{F}_1^a(V) = \mathcal{F}_1(U) \otimes_A B$ by definition. The injection $\mathcal{F}_1^a(V) \to \mathcal{E}(V)$ factors as $$ \mathcal{F}_1(U) \otimes_A B \to \mathcal{E}(U) \otimes_A B \to \mathcal{E}(U) $$ Considering $A$-algebras $B$ of the form $B = A \oplus M$ we see that $\mathcal{F}_1(U) \to \mathcal{E}(U)$ is universally injective (see Algebra, Definition \ref{algebra-definition-universally-injective}). Since $\mathcal{E}(U) = \mathcal{F}(U)$ we conclude that $\mathcal{F}_1 \to \mathcal{F}$ remains injective after any base change, or equivalently that $\mathcal{F}_1^a \to \mathcal{F}^a$ is injective.