Transition functions of sheaf tensor product

Question

Suppose $\mathcal{L},\mathcal{M}$ are invertible sheafs on a scheme $X$. I've seen an abstract construction of $\mathcal{L}\otimes_X \mathcal{M}$, but I'm having trouble connecting this with a more down-to-earth construction. Specifically, suppose that $\{U_i\}$ is an open cover of $X$ and that $\mathcal{L},\mathcal{M}$ have transition maps $l_{ij},m_{ij}\colon U_i\to U_j$ between local trivializations. How does it follow from the universal property of $\mathcal{L}\otimes_X \mathcal{M}$ that this sheaf has transition functions $l_{ij}m_{ij}$?

Answer 1

We have an open cover $X = \bigcup_i U_i$ with $\mathcal L_{|U_i} = \mathcal O_{X|U_i}\langle l_i\rangle$ and transition functions $l_{ij} \in \mathcal O^*_X(U_i \cap U_j)$, which means that $l_{ij}l_j=l_i$ holds.

We also have the same notion with $\mathcal M, m_i, m_{ij}$.

Now the open cover is also a trivialization for $\mathcal L \otimes \mathcal M$, because on $U_i$ we have the generator $l_i \otimes m_i$ (It is clearly a generator of the tensor product-presheaf and since sheafification is exact, it is a generator of the tensor product itself).

So all we have to check is $l_{ij}m_{ij}(l_j \otimes m_j) = l_i \otimes m_i$. But this is obvious:

$$l_{ij}m_{ij}(l_j \otimes m_j) = (l_{ij}l_j) \otimes (m_{ij}m_j) = l_i \otimes m_i$$