My question: Let $S$ be a base scheme and let $(Sch/S)_{fppf}$ be a big fppf site. Let $U$ be a scheme over $S$. Suppose that there is a surjective morphism $\Phi_{U}:U\rightarrow U$. Then can we show that the induced morphism of sheaves $h_{U}\rightarrow h_{U}$ is locally surjective? It seems that this is false.

Note that $h_{U}={\rm{Hom}}(-,U)$ is a representable sheaf. A map of sheaves $F\rightarrow G$ on $(Sch/S)_{fppf}$ is locally surjective if for every scheme $U\in{\rm{Ob}}((Sch/S)_{fppf})$ and every $s\in G(U)$, there exists a covering $\{U_{i}\rightarrow U\}_{i\in I}$ such that for all $i$, $s|_{U_{i}}$ is in the image of $F(U_{i})\rightarrow G(U_{i})$.

Answer: Let $S:={\rm Spec}(k)$ be a field and let $U={\rm Spec}(k[t]/t^2)$.

The ring $k[t]/t^2$ is a $k$-algebra and there is a map of $k$-algebras $k[t]/t^2\to t$ sending $t$ to $0$ so we get morphisms $U\to{\rm Spec}(k)$ and ${\rm Spec}(k)\to U$.

Let $a:U\to U$ be the composition of these two morphisms. This is surjective.

Now if the answer to your question is positive, there is a faithfully flat morphism $b:V\to U$ and an element $\rho\in{\rm Hom}(V,U)$ such that $a\circ\rho=b$ (applying local surjectivity to ${\rm Id}_U\in{\rm Hom}(U,U)$). In other words, the morphism $a_V:U\times_U V\to V$ has a section. However, if the morphism $a_V$ has a section then so has the morphism $S\times_U V\to V$ obtained by base-change from the morphism $S={\rm Spec}(k)\to U$. The morphism is $S\times_U V\to V$ is a closed immersion, so it is an isomorphism, since it has a section. Isomorphisms descend along faithfully flat morphisms, so we deduce that ${\rm Spec}(k)\to U$ is an isomorphism, which is false. So this provides a counterexample.