Whats the relationship between a presheaf and its sheafification?

Question

Given a presheaf $F'$ of rings, say, we have a morphism $F' \rightarrow F$ where $F$ is the sheafification of $F'$. I see no reason this morphism should be injective or surjective. Is that right?

Answer 1

Here is an example where the map $F' \to F$ is not injective : take a ring $R$, $X = \{0,1\}$ with discrete topology (for example) and $F'(U) = 0$ if $U \neq X$ and $F'(X) = R$ with zero restrictions morphisms. Then, $F$ is the zero sheaf and in particular the canonical map $F' \to F$ is not injective.

For surjectivity, take $F'$ the presheaf of bounded functions on $\Bbb R$. Then, $F$ is the sheaf of locally bounded functions, in particular $\text{id}_{\Bbb R} \in F(X)$ but is not in $F'(X)$. So the map $F' \to F$ is not surjective.

Answer 2

If you notice on page 81 of the PDF I linked in my comment, there isn't necessarily an injectivity or surjectivity of sheaves in the universal mapping property of a sheafification. However, do note that the sheafification of a sheaf returns the sheaf itself. The definition of sheafification in some sense (if I remember right), makes the smallest sheaf which somehow contains the original presheaf. (I hope that doesn't make things worse.)