How to show that $r\colon Sh(X_{et})\to Sh(X_{et}^f)$ is a full functor $\newcommand{\rx}{X_{\acute et}^f}$
$\newcommand{\x}{X_{\acute et}}$
$\newcommand{\F}{\mathscr F}$
$\newcommand{\G}{\mathscr G}$
$\newcommand{\a}{\alpha}$
$\newcommand{\b}{\beta}$
$\newcommand{\h}{\operatorname{Hom}}$
Let $X$ be a quasi-compact and quasi-separated scheme. Then the restricted etale site $\rx$ is defined to be the full subcategory of the etale site $\x$ consisting of finitely presented etale $X$-schemes. The inclusion $i\colon\rx\to\x$ induces a functor $i^p\colon PSh(\x)\to PSh(\rx)$ by $i^p\F(U)=\F(i(U))$ (for any $U\in\rx$). Denote by $r$ the restriction of $i^p$ to $Sh(\x)$. I am trying to finish the proof of Proposition 5.2.8 in Fu's Etale Cohomology Theory:

Proposition. The functor $r\colon Sh(\x)\to Sh(\rx)$ is an equivalence of categories.

I am stuck in showing that $r$ is a full functor. To show $$\h_{Sh(\x)}(\F,\G)\to\h_{Sh(\rx)}(r(\F),r(\G))
$$
surjective for any $\F,\G\in Sh(\x)$, we take any $\phi\in\h_{Sh(\rx)}(r(\F),r(\G))$ and wish to find a preimage $\psi\in\h_{Sh(\x)}(\F,\G)$. For any $U\in\x$, we have an etale covering $\{U_\a\to U\}_{\a\in I}$ with $U_\a\in\rx$ and $I$ finite (this is possible since $X$ is quasi-compact and quasi-separated). Then we have the commutative diagram
$$\require{AMScd} \begin{CD}
0 @>>> \F(U)@>>>\prod_{\a\in I}\F(U_\a)@>{f}>>\prod_{\a,\b\in I}\F(U_\a\times_UU_\b)\\
@.@.@V{\phi}VV @VV{\phi}V\\
0@>>> \G(U)@>>>\prod_{\a\in I}\G(U_\a)@>{g}>>\prod_{\a,\b\in I}\G(U_\a\times_UU_\b)
\end{CD}$$
Then $\F(U)\simeq\ker f$ and $\G(U)\simeq \ker g$ and thus $\phi$ induces a morphism $\F(U)\to\G(U)$.
Where I am stuck is that I cannot show that the induced morphism is independent of the choice of the etale covering, and that (--if so, we can denote the morphism by $\psi(U)$) $\psi$ really gives a morphism between $\F$ and $\G$.
Any help is appreciated.
 A: Note that the first $\phi$ in your diagram is actually $\prod_{\alpha \in I} \phi_{|U_\alpha}$. Denote by $\phi^{\{U_\alpha \to U\}}$ the induced map.
Take another cover of $U$, say $\{V_\beta \to U, \beta \in J\}$, and denote by $\phi^{\{V_\beta \to U\}}$ the induced map. We need to show that for any $\sigma \in \mathscr F (U)$, $\phi^{\{U_\alpha \to U\}}(\sigma) = \phi^{\{V_\beta \to U\}}(\sigma)$. To compare the two images lets denote by $\{W_{\alpha,\beta} = V_\beta \times_U U_\alpha \to U, \alpha \in I, \beta\in J\}$ the new cover formed by intersections. If we show that for any $(\alpha, \beta) \in I\times J$,
$$
\phi^{\{U_\alpha \to U\}}(\sigma)_{|W_{\alpha,\beta}} = \phi^{\{V_\beta \to U\}}(\sigma)_{|W_{\alpha,\beta}}
$$
we have the desired result by the sheaf property of $\mathscr G $. But that is true since by definition
$$
\phi^{\{U_\alpha \to U\}}(\sigma)_{|W_{\alpha,\beta}} = \phi_{|W_{\alpha,\beta}}(\sigma_{|W_{\alpha,\beta}}) = \phi^{\{V_\beta \to U\}}(\sigma)_{|W_{\alpha,\beta}}.
$$
