I am currently studying Algebraic Geometry, and I just discover that the a presheaf $\mathscr F$ is isomorphic to its sheafification $\mathscr F^{sh}$... Something is surely going wrong but I can not figure it out, therefore I am looking for some help!
Throughout this question, I am talking about (pre)sheaf on topological space $X$ of sets and $U$ is an open subset of $X$.
There are $5$ steps toward the conclusion "a presheaf $\mathscr F$ is isomorphic to its sheafification $\mathscr F^{sh}$".
Step $1$: A section of a sheaf is determined by its germs, i.e. the natural map $\phi:\mathscr F(U) \rightarrow \prod \limits_{p \in U} \mathscr F_p$ is injective.
Step $2$: Any choice of compatible germs for a sheaf $\mathscr F$ over $U$ is the image of $\phi$ of a section of $\mathscr F$ over $U$.
Step $3$(a controversial step): Any image of $\phi$ of a section of $\mathscr F$ over $U$ consists of compatible germs.
Proof of step $3$: for any $f \in \mathscr F(U)$, $\phi(f)$ is $(\mathscr f_p)_{p \in U}$ such that $[(f,U)]=\mathscr f_p$.([] marks the representative of the equivalence class in $\mathscr F_p$.) Then for all $p \in U$, the representative $[(f,U)]$ is what we want. Because $f_p=[(f,U)]$ for all $p \in U$.
Step $4$: $\mathscr F(U)$ is isomorphic to the set {$ \prod \limits_{p \in U} \mathscr f_p \mid \prod \limits_{p \in U} \mathscr f_p \in \prod \limits_{p \in U} \mathscr F_p$ consists of compatible germs}.
Proof of step $4$: Combining Step $2$ and Step $3$, we have the image of $\phi$ equals to the set {$ \prod \limits_{p \in U} \mathscr f_p \mid \prod \limits_{p \in U} \mathscr f_p \in \prod \limits_{p \in U} \mathscr F_p$ consists of compatible germs}. And $\phi$ is injective, so $\mathscr F(U)$ is isomorphic to the image of $\phi$, thus the set {$ \prod \limits_{p \in U} \mathscr f_p \mid \prod \limits_{p \in U} \mathscr f_p \in \prod \limits_{p \in U} \mathscr F_p$ consists of compatible germs}.
Step $5$(a controversial step): Presheaf $\mathscr F$ is isomorphic to its sheafification $\mathscr F^{sh}$.
Proof of step $5$:Look closely at the construction of sheafification,$\mathscr F^{sh}(U)=$ {$(f_p \in \mathscr F_p)_{p \in U}:$ for all $p \in U$, there exists an open neighborhood $V$ of $p$, contained in $U$ and $s \in \mathscr F(V)$ with $s_q=f_q$ for all $q \in V$}. It 's basically saying that $\mathscr F^{sh}(U)=$ {$ \prod \limits_{p \in U} \mathscr f_p \mid \prod \limits_{p \in U} \mathscr f_p \in \prod \limits_{p \in U} \mathscr F_p$ consists of compatible germs}. Hence $\mathscr F^{sh}(U)$ is isomorphic to $\mathscr F(U)$ due to the Step $4$, that is, $\mathscr F^{sh}$ is isomorphic to $\mathscr F$ .
Which steps are not correct?