Let $(X, O_X)$ be a ringed space, $I$ be a sheaf of ideals on $(X, O_X)$ and let $F$ be an $O_X$-module. I have two questions.
a) Is there an example where the presheaf $P : U \mapsto I(U) \cdot F(U) \leq F(U)$ is not a sheaf on $X$ ?
b) Is it correct that we define the product $I \cdot F$ as being the image of the sheaf morphism $$I \otimes_{O_X} F \to O_X \otimes_{O_X} F \cong F,$$ and if so, do we necessarily have that $I \cdot F$ is isomorphic (as sheaf of $O_X$-module) to the sheafification of $P$ ? According to this question, this is true but the comment has no further explanation as to why this is true.
For a), if I consider a collection of pairwise compatible sections $s_U \in P(U)$, we can write $$s_U = \sum_{i=1}^{n_U} a_i^U b_i^U$$ with $a_i^U \in I(U), b_i^U \in F(U)$, but I don't see how the conditions $s_U\vert_{U \cap V} = s_V\vert_{U \cap V}$ could help to glue the various $s_U$ together.
Thank you for your help.
Let me answer b) first, because it helps with answering a): The image of the sheaf morphism $$I \otimes F \to O_X \otimes F \cong F$$ is by definition the sheafification of the presheaf $$U \mapsto Im(I(U) \otimes_{O_X(U)} F(U) \to F(U))$$ which is just your presheaf $P$. So the image is by definition the sheafification of your presheaf.
Answering a): Now I can talk about presheaves of tensor products being or not being sheaves, where an extremely important and instructive example is given by the projective line $X = \mathbb{P}^1_k$ over a field $k$. The ideal sheaf $I$ corresponding to a closed, $k$-rational point is isomorphic to $O(-1)$. Considering $F = O(1)$ we know $$I \otimes F \cong O_X,$$ but $I(X) = 0$, so the tensor product presheaf is not equal to its sheafification which satisfies $O_X(X) = k$.
Note that $F$ is a flat $O_X$-module, so the tensor product presheaf is actually isomorphic to the product, so this also gives an example for your situation.