Let $f:F\rightarrow G$ be a morphism of presheaves, then there is a unique induced map of $f^+:F^+\rightarrow G^+$, where $F^+$ and $G^+$ are the sheafification of the sheaves. This follows because if $\theta$ is the sheafification morphism, there is a map $\theta\circ f:F\rightarrow G^+$, so by the universal property of sheafification we have a unique map $f^+:F^+\rightarrow G^+$ satisfying $f^+\circ \theta=\theta\circ f$.
Now, by the universal property of the direct limit, there exists a map $f^+_q:F^+_q\rightarrow G^+_q$ such that $f^+_q\circ \theta_q=\theta_q\circ f_q$. I can show that for any presheave $\theta_q$ is an isomorphism, however, the problem I am working on claims that $(f^+)_q=f_q$, and to me this seems to only be true if $\theta_q=\text{Id}$, however this makes no sense to me.
I get there is an isomorphism $F^+_q\cong F_q$, however, this isomorphism does not seem to be the identity. Indeed, the isomorphism is given by sending $s_q\in F_q$ to $(s_p)_q$ where $s$ is any section of $F(U)$, $U$ any open, such that the equivalence class $[U,s]$ is equal to $s_q$, and $(s_p)$ is the image of this section under the map $F(U)\rightarrow F(U)^+$. This is clearly well defined, and it is easy to see it is a bijection, and thus an isomorphism of the underlying rings, but how is it the identity, when the stalks aren't as sets on the nose the same?