I have seen that a question on this specific problem has been already asked and answered here A map is injective if it is nonzero at the generic point?
Just to give a context we have a finite separable morphism of non-singular curves $f: X \to Y$ and we want to prove the injectivity of $f^* \Omega_Y \to \Omega_X$. Hartshorne claims that it is sufficient to show that the map is nonzero at the generic point, since both $f^* \Omega_Y$ and $\Omega_X$ are invertible sheaves on $X$.
Now, in the question I linked, the answer (in my opinion) explains why it is sufficient to show that the map is non zero but doesn't cite anywhere why it is sufficient to look $\textbf{only}$ at the generic point. The answer underlines that it is sufficient to check the injectivity of $(f^* \Omega_Y)_x \to \Omega_{X,x}$ $\forall x$. Maybe it's trivial or maybe I don't understand where it is written between the lines. Anyway, I tried to give my self the following esxplaination:
$f^* \Omega_Y$ and $\Omega_X$ are invertible sheaves on $X$, so are isomorphic to $\mathcal{O}_X$. Moreover, since $X$ is integral, $\forall U \subset X$ we have an injective map $\mathcal{O}_X(U) \to \mathcal{O}_{X,\xi}$ and so $\mathcal{O}_{X,x} \to \mathcal{O}_{X,\xi}$. From here, checking the injectivity of the map $(f^* \Omega_Y)_\xi \to \Omega_{X,\xi}$ is easy and since the composition of two injective maps is itself injective, we're done. Is my explaination correct? Even if correct, is it useless?
Question: "Now, in the question I linked, the answer (in my opinion) explains why it is sufficient to show that the map is non zero but doesn't cite anywhere why it is sufficient to look only at the generic point."
Answer: If $S:=Spec(A)$ is a noetherian integral domain and if $g:E \rightarrow F$ is a map of finite rank projective $A$-modules it follows $E$ is torsion free hence the kernel $C:=ker(g)$ is torsion free. Let $\eta \in S$ be the generic point. It follows $C=0$ iff $C_{\eta}=0$: Assume $C_{\eta}=0$. $C$ has a finite set of generators $x_1,..,x_n$ and since $C_{\eta}=0$ it follows there is a non zero element $s \in A$ with $sx_i=0$ and since $C$ is torsion free it follows $x_i=0$ hence $C=0$.
If you choose an open affine set $U:=Spec(B) \subseteq X$ and consider the induced map $df_U: E:=f^*(\Omega^1_Y)(U) \rightarrow \Omega^1_X(U):=F$ and let $C(U):=ker(df_U)$. It follows $B$ is a noetherian integral domain and $E$ is a torsion free $B$-module, hence if $C(U)_{\eta}=0$ it follows $C(U)=0$. Hence if $df_{\eta}=0$ it follows $df_U=0$. The open set $U$ is arbitrary hence if $df_{\eta}$ is injective it follows $df$ is injective.
Claim: $f^* \Omega_Y$ and $\Omega_X$ are invertible sheaves on $X$, so are isomorphic to $\mathcal{O}_X$.
Comment: This is not correct. If $X:=\mathbb{P}^1$ it follows $\Omega^1_X \cong \mathcal{O}(-2)$.