Why $\mathcal{I}_x \longrightarrow I_x$ will NOT be injective?

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Let X be a scheme, $\mathfrak{C}$ be the category of $\mathcal{O}_X$-module sheaves. I want to prove $\mathfrak{C}$ has enough injectives. I found a description like as follows in the book "Algebraic Geometry and Commutative Algebra"(Siegfried Bosch, 2011):

Fix an $\mathcal{O}_{X, x}$-module $I_x$ for every $x \in X$and construct an $\mathcal{O}_X$-module sheaf $I$ on $X$ by looking at the functor th at on open subsets $U \subset X$ is given by $$\mathcal{I}(U) = \prod_{x\in U}I_x$$ with canonical restriction morphisms. One might think that the stalk $\mathcal{I}_x$ of $\mathcal{I}$ at any point $x \in X$ will coincide with $I_x$. But a careful analysis shows that there is just a canonical map $\mathcal{I}_x \longrightarrow I_x$ and that the latter will not be injective in general.

I can't understand this claim. Why the canonical map $\mathcal{I}_x \longrightarrow I_x $ will not be injective? Do I have some mistake?