I am currently studying a proof of the fact that every sheaf of abelian groups can be injected in an injective sheaf, and it seems like I'm missing something. In the proof, the idea is pretty straight-forward: We take a sheaf $E$ and for each $x \in X$ (X the base topological space), we choose an injective abelian group $I_{x}$ and a monomorphism $\psi_{x}: E_x \to I_x$ and then, with some verifications involved, we construct the preasheaf $I: U\subset X \mapsto I(U) := \prod_{z \in U} I_z$ and then it turns out to be an injective sheaf. My question is: Is the stalk of $I$ at a point $x \in X$ the abelian group $I_x$?
Thank you in advance for all the help :)
Not necessarily. For example, let $X = \mathbb{R}$ and $I_x = \mathbb{Q}$ for all $x \in \mathbb{R}$. Note that the global sections of $\prod_{x \in \mathbb{R}} I_x$ are just functions $\mathbb{R} \to \mathbb{Q}$.
Suppose for contradiction that there exists an isomorphism $\mathbb{Q} \to (\prod_{x \in \mathbb{R}} I_x)_0$. Composing with the map $(\prod_{x \in \mathbb{R}} I_x)_0 \to \mathbb{Q}$ induced by the projection $\prod_{x \in \mathbb{R}} I_x \to I_0$, we get a surjective homomorphism $\mathbb{Q} \to \mathbb{Q}$, which must therefore be an isomorphism. This means that the germ of a local section in a neighborhood of $0$ is determined by its component in $I_0$, which is false. For example, take the two global sections $f, g$ given by the functions
$$f(x) = 0$$ $$g(x) = \begin{cases} 0 &: x = 0 \\ 1 &: x \neq 0 \end{cases}$$