Let $X$ be a scheme and consider any big site structure on $X$. Is it still true that the inclusion $Qcoh(X) \hookrightarrow Mod(X)$ is exact? In particular, is it left exact? For the usual Zariski topology this is clearly true. But I doubt when the morphisms in a site have no conditions (specifically no flatness) if this is still true. I am not being able to give precise arguments to assert this. Can someone please help me out?


Thanks in advance! 



Exactness of the inclusion of quasi-coherent sheaves into the sheaves of modules on a big site
You are probably thinking about the following type of phenomenon:


**Example.** Let $X = \mathbb A\_k^1$, and consider the ringed site $((\operatorname{\underline{Sch}}/X)\_{\text{fppf}},\mathcal O)$, where $\mathcal O$ means the sheaf whose value on $T \in \operatorname{\underline{Sch}}/X$ is $\Gamma(T,\mathcal O\_T)$. The pullback functor
$$\operatorname{\underline{Qcoh}}(\mathcal O\_X) \to \operatorname{\underline{Qcoh}}((\operatorname{\underline{Sch}}/X)\_{\text{fppf}},\mathcal O)$$
is an equivalence of categories by a standard descent argument [[Tag 03DX](http://stacks.math.columbia.edu/tag/03DX)]. Thus, as you already indicated by your notation, $\operatorname{\underline{Qcoh}}(X)$ does not depend on the site in this case.


However, if we take an injection of quasi-coherent $\mathcal O\_X$-modules like $\mathcal O\_X \stackrel{x}\to \mathcal O\_X$ and pull back along the morphism $\operatorname{Spec} k = \{0\} \to \mathbb A^1\_k$, we get the map $k \stackrel 0 \to k$. This is not injective.


Since injectivity of a morphism of sheaves $\mathscr F \to \mathscr G$ on a ringed site $\mathscr C$ means that for every $U \in \mathscr C$ the map $\mathscr F(U) \to \mathscr G(U)$ is injective, we see that the map $\mathcal O\_X \stackrel x \to \mathcal O\_X$ is not injective as a map of $\mathcal O$-modules on $(\operatorname{\underline{Sch}}/X)\_{\text{fppf}}$.


