Suppose you two locally free sheaves $L,M$ on an integral scheme $X$ over $k$. Is it true that the map
$L(X) \otimes M(X) \to ( L \otimes M)(X)$
Is injective? The map is formally induced by the general map from a presheaf to its sheafification, recalling that the tensor of two sheaves is the sheafification of the presheaf which is "openwise" $U \mapsto L(U)\otimes M(U)$.
Concretely, is just thinking a product of sections in the tensor sheaf.
My try:
Let U be a open such that both L,M are free. Then the tensor sheaf coincides with the tensor presheaf, because for a ring $A$, $f \in A$, and $r,s$ natural holds $(A^{rs})_f \simeq A^r_f \otimes A^s_f $ By the exactness of localisation.
Suppose $\sum s_i \otimes t_i $ goes to zero. Then it's stalks are everywhere null, and in particular by 1 there exist an open set $U$ such that $\sum s_i \otimes t_i | U = 0$. Recall that $X$ is integral, thus $U$ is dense.
I would like now to conclude the restriction map to $U$ is injective, because is the tensor of two injective maps over $k$. Is is true?
Best, Andrea
This is not injective in general. Consider $X = \Bbb P^1_k$, $L = M = O_X(1)$. Then $L(X) = M(X) \cong k^2$ so that $L(X) \otimes_k M(X) \cong k^4$, but $L \otimes_{O_X} M \cong O_X(2)$ has a $3$-dimensional space of global sections.
A presheaf $F$ is said to be separated when the sheafification map $F \to F^+$ is injective. The example above is a non-separated presheaf $U \mapsto L(U) \otimes_{O_X(U)} M(U)$. See also here.
Let me call $f$ the natural map $L(X) \otimes_{O_X(X)} M(X) \to (L \otimes_{O_X} M)(X)$. In your second step, when you are writing $\sum s_i \otimes t_i | U = 0$, I guess that you mean $$\mathrm{Res}_U^X \left[ f\left(\sum_i s_i \otimes t_i \right)\right] = 0$$ But even if the restriction map $\mathrm{Res}_U^X$ was injective (as you mention in your 3rd point), you've proven nothing, since you don't know if $f$ is injective. This is precisely what you are trying to prove. Unless I misunderstood your reasoning.