I tried different books and lecture notes to understand sheafification, but for instance in Hartshore or Shafarevich's book, but I found it hard to understand.
The following is the approach my professor used. But also here I have some questions. But first here is what he wrote:
Let $\mathscr{F}$ be a presheaf.
$$\mathscr{F}^g(U):= \left\{\begin{array}{rl} \{f_P\}_{P\in U}| & f_P\in\mathscr{F}_P, \forall P\in U\textrm{ }\exists V\ni P, V\subset U\textrm{ and a }f\in\mathscr{F}(V) \\ | & \textrm{ with }f\in\mathscr{F}(V)\to\mathscr{F}_Q, f\mapsto f_Q \end{array}\right\}$$
First see that $\mathscr{F}^g$ is a presheaf:
Let $U'\subset U$. $f=\{f_P\}_{P\in U}\in\mathscr{F}^g(U)\mapsto \bar{f}:=\{f_P\}_{P\in U'}\in\mathscr{F}^g(U')$.
So $\forall P\in U'$ $\exists V\subset U'$ such that $V'=V\cap U'$ and $\mathscr{F}(U)\to\mathscr{F}_Q$ with $f\mapsto f_Q$.
$$\begin{array}{ccc} \mathscr{F}(V) & & \\ \downarrow{\varphi} & \searrow & & \\ \mathscr{F}_Q & \xleftarrow{} & \mathscr{F}(V')& \end{array}$$
For all $Q\in V'$. and $\varphi(\bar{f})$
Let $U=\bigcup\limits_{i\in I}U_i$.
$\mathscr{F}^g(U)\overset{\beta}{\to}\prod\limits_{i\in I}\mathscr{F}^g(U_i)$
$\prod\limits_{i\in I}\mathscr{F}^g(U_i)\overset{\alpha}{\to}\prod\limits_{i,j\in I}\mathscr{F}^g(U_{ij})$, ($U_{ij}=U_i\cap U_j$) with $\{f_i\}_{i\in I}\mapsto \{f_{ij}\}_{ij}$,
and $f_{ij} = \rho_{U_i U_{ij}}f_i - \rho_{U_j U_{ij}}f_j$ and $f_i = \{f_{iP}\}_{P\in U_i}\in\mathscr{F}^g(U_i)$
$(*)$ $Im(\beta)=ker(\alpha)$. If $\beta(f)=0$ $\Rightarrow f_P=0$ $\forall P\in U$ $\Rightarrow f = \{f_P\}_{P\in U}=0$.
$(**)$ If $P\in U_{ij}$, then $\{f_i\}_{i\in I}\in ker(\alpha)$.
Show $f_{ip}=f_{jp}$: Let $U_i\supseteq V_i\ni f\in V_j\subseteq U_j$
$\mathscr{F}(V_i)\to \mathscr{F}_{iP}$ with $\tilde{f}_i\mapsto f_{iP}$ and $\mathscr{F}(V_j)\to \mathscr{F}_{jP}$ with $\tilde{f}_j\mapsto f_{jP}$.
$(***)$ $\rho_{U_i U_{ij}}f_i = \rho_{U_j U_{ij}}f_j\Rightarrow f_{ip}=f_{jp}$ $\forall P\in U_{ij}$.
$\tilde{f}=\{f_{P}\}_{P\in U}$. $f_P=f_{iP}$ for $i$ with $P\in U_i$.
$\tilde{f}\in\mathscr{F}^g(U)$ fulfils the condition, and so $\mathscr{F}^g$ is a sheaf.
Our definition for a sheaf: Let $\mathscr{F}: ouv(X)\to \textrm{(abelian groups)}$ be a presheaf.
Then $\mathscr{F}$ is a sheaf if $\forall U\subset X$ open:
$U=\bigcup\limits_{i\in I}U_i$ open couver $\Rightarrow \mathscr{F}(U)\overset{\beta}{\to}\prod\limits_{i\in I}\mathscr{F}(U_i)$ is injective and $Im(\beta)=\{(f_i)_{i\in I}|\rho_{U_i U_{ij}} f_i = \rho_{U_j U_{ij}} f_j\}$.
Now my questions:
To $(*)$: Here we want to show that $Im(\beta)=ker(\alpha)$ holds right? But how did he show it? I don't understand what he does to afterwards.
To $(**)$: Why do we have ''If $P\in U_{ij}$, then $\{f_i\}_{i\in I}\in ker(\alpha)$.''?
To $(***)$: I understand why ''$f_{ip}=f_{jp}$'' follows, but not that ''$\rho_{U_i U_{ij}}f_i = \rho_{U_j U_{ij}}f_j$'' holds.
Thanks and all the best!
Clearly $\mathcal{F}^g$ is a presheaf.
In order to show that it is a sheaf we need to show that for ervery open set $U$ and every covering $\{U_i\}$ then
is an equalizer. The two maps $$ \prod_i \mathcal{F} ^g (U_i) \longrightarrow \prod_{i,j} \mathcal{F} ^g (U_{i,j}) $$ are induced by the two different restrictions into $\mathcal{F} ^g (U_{i,j})$ i.e. $\mathcal{F} ^g (U_{i}) \longrightarrow \mathcal{F} ^g (U_{i,j})$ and $\mathcal{F} ^g (U_{j}) \longrightarrow \mathcal{F} ^g (U_{i,j})$.
If $\mathcal{F}$ is a sheaf of abelian groups or rings or some other "nice" algebraic object this is the same as showing that
$$ 0 \longrightarrow \mathcal{F}^g(U) \overset{\beta}{\longrightarrow} \prod_i \mathcal{F} ^g (U_i) \overset{\alpha}{\longrightarrow} \prod_{i,j} \mathcal{F} ^g (U_{i,j}) $$ is exact with the $\alpha$ and $\beta$ you defined.
I guess that $(*)$ should be $Ker(\beta)=0$ and this is in fact what is shown.
Next we want to show that $Im(\beta) = Ker(\alpha)$.
First $Im(\beta) \subset Ker(\alpha)$. This is $(***)$: starting with $f \in \mathcal{F}^g(U)$ we can restrict it to $U_{i,j}$ in two different ways, first to $U_i$ and then to $U_{i,j}$ or first to $U_{j}$ and then to $U_{i,j}$. However both are the same since anyway we look at it it is just restricting from $U$ to $U_{i,j}$.
Next $Im(\beta) \supset Ker(\alpha)$.
This is $(**)$. Let $\{f_i\} \in Ker(\alpha)$. That is $\rho_{U_i U_{i,j}}f_i - \rho_{U_j U_{i,j}}f_j=0$ for every $i,j$. For every point $P \in U_{i,j}$ we have an equality $(f_i)_P=(f_j)_P$ in the stalk $\mathcal{F}^g_P=\mathcal{F}_P$. So there is a neighborhood of $P \subset V_{i,j,P} \subset U_{i,j}$ and an element $\tilde{f}_{i,j,P}$ such that for every point $Q \in V_{i,j,P}$ the restriction to the stalk is $(\tilde{f}_{i,j,P})_Q=(f_i)_Q=(f_j)_Q$.
Now we can define an element of $\mathcal{F} ^g (U)$ as $\tilde{f}=\{(f_i)_P\}$. By construction $\tilde{f} \in Im(\beta)$.