Let $X$ be a scheme. A $\mathcal O_X$-module $\mathcal L$ is called invertible if, for every point $x\in X$, there is an open neighborhood $U$ of $x$ and an isomorphism of $U$-modules $\mathcal O_X|_U \cong \mathcal L|_U$.
I am struggling with the basics of sheaf theory, and I have the following questions.
$1.$ Why is the tensor product of two invertible sheaves invertible?
$2$. Why is the pullback of an invertible sheaf by a morphism an invertible sheaf?
$3.$ Why is the twist $\mathcal O_X(k)$ on $X=\operatorname{Proj} B$ for a graded ring $B$ invertible, where $k$ is any integer, possibly negative?
I have a few jumbled thoughts.
For $1$, recall that the tensor product of two sheaves takes the presheaf that assigns to an open set $U$ the module $ \mathcal L_1(U) \otimes_{\mathcal O_{X}|_U} \mathcal L_2(U)$ and sheafifies it. If we pick open sets $U_i$ such that the sheaf $\mathcal L_i$ is trivialized on $U_i$, then on $U_1\cap U_2$ they both have trivializations, and their tensor product on this open set is isomorphic to $\mathcal O_X(U) \otimes_{$\mathcal O_X(U)} \mathcal O_X(U)=\mathcal O_X(U)$, which gives the desired result on the presheaf. I do not see why this still holds after sheafificaiton.
For $2$, I tried using the local tensor product definition of the pullback sheaf, but I could not simply it.
For $3$, it suffices to show it for $k=\pm 1$ and use $1$, because $\mathcal O_X(n) \otimes_{\mathcal O_X} \mathcal O_X(m)=\mathcal O_X(n+m)$.
It is not enough to argue with sections. You have to consider the whole sheaf! Tensor products commute with restrictions to open subsets. Therefore it is enough to remark that $\mathcal{O}_X \otimes \mathcal{O}_X = \mathcal{O}_X$.
Pullbacks commute with restrictions to open subsets. Therefore it is enough to remark that $f^* \mathcal{O}_X = \mathcal{O}_Y$ for a morphism $f : Y \to X$.
In general we have $\mathcal{O}(n+m) \cong \mathcal{O}(n) \otimes \mathcal{O}(m)$. You can find this in every introduction to algebraic geometry (of course the same remark applies to 1. and 2.). In particular, $\mathcal{O}(n) \otimes \mathcal{O}(-n) \cong \mathcal{O}$. Alternatively, $\mathcal{O}(n)$ is freely generated by $f^n$ on $D_+(f)$, where $f \in B_1$.