I am trying to understand the proof of Stacks Project of the fact that the Picard group of a UFD is trivial. I think I understand all the steps in the proof. What I don't understand is how they give the conclusion. I'll explain: what this proof actually shows is that if $L$ is an arbitrary invertible $R$-module, where $R$ is a UFD, then, for any $\varphi\in\operatorname{Hom}_R(L,R)$, it holds that $\varphi(L)\subset R$ is a principal ideal. They implicitly say that this is sufficient to conclude that $L\cong R$ (i.e., that $\operatorname{Pic}(R)=0$).
But I have sincerely no idea why is this last claim is true. I think the reasoning might have to do with the fact that if $L$ is invertible then it is finitely generated and torsion-free. Since this information is mentioned at the beginning of the proof but on no other place of the proof the finiteness and torsion-freeness of $L$ are used in any way.
The point is that if $\varphi\in\operatorname{Hom}_R(L,R)$ is nonzero, then it is injective so it gives an isomorphism from $L$ to $\varphi(L)$, and $\varphi(L)\cong R$ since it is a principal ideal.
Here is one way you can see that $\varphi$ is injective. Note that if $K$ is the fraction field of $R$ then $K\otimes L\cong K$ (since $L$ is locally free of rank $1$) and $\varphi$ induces a homomorphism $\varphi\otimes K:L\otimes K\to R\otimes K$. Since $L$ is torsion free, it embeds in $L\otimes K$, so if $\varphi$ is nonzero then so is $\varphi\otimes K$. But since $L\otimes K$ and $R\otimes K$ are both $1$-dimensional vector spaces, this means $\varphi\otimes K$ is injective, and hence so is $\varphi$.