Restriction of an injection of sheaves Let $X$ be a smooth projective variety over $\mathbb{C}$. Let $D$ be an effective divisor on $X$ (we may assume that $D$ is smooth if needed). Suppose $F$ and $G$ are coherent sheaves on $X$ such that we have an injective map of sheaves
$$ 0\rightarrow F\rightarrow G.$$
Assume further that the quotient $G/F$ is torsion-free. Then is the following restriction map still injective $$F|_D\rightarrow G|_D?$$
 A: Let me elaborate on the comment above.
The question is clearly local, so we can reduce to the affine case, i.e. we have an exact sequence of $R$-modules $$0 \to F \to G \to C \to 0$$ with $C$ torsion free.
Now we tensor this sequence with $R/f$, where $f$ is a regular element of $R$ and we want to see that the sequence is still exact.
We surely get an exact sequence
$$\operatorname{Tor}^R_1(C,R/f) \to F \otimes R/f \to G \otimes R/f \to C \otimes R/f \to 0$$
, hence we are left to show that $\operatorname{Tor}^R_1(C,R/f)=0$ holds.
To do that, we take the exact sequence $$0 \to R \xrightarrow{\cdot f}R \to R/f \to 0$$ and after tensoring with $C$, we get the long exact sequence
$$\operatorname{Tor}^R_1(C,R) \to \operatorname{Tor}^R_1(C,R/f) \to C \xrightarrow{\cdot f} C \to C/fC \to 0$$
$C$ is torsion free, hence multiplication by $f$ is injective. Thus the map $\operatorname{Tor}^R_1(C,R/f) \to C$ is the zero map, but it is also injective because $\operatorname{Tor}^R_1(C,R)=0$. This implies the desired $\operatorname{Tor}^R_1(C,R/f)=0$.
