Nowadays, I close a new small case.
Proposition. For a surjective morphism between scheme $Xstackrel{f} o Y$, For any $Z o Y$, the base change $X imes_Y Z o Z$ is also surjective.
The diagram is as following
$$egin{array}{ccc} X imes_Y Z& o & Z\ downarrow && downarrow \ Z& o & Y\end{array}$$
In the first place, we will reduce the proposition into affine case.Since the proof involves some essential computation of tensor product, I will deal with secondly. At the end of the post, I will close the proof.
First Step (reduce to affine case). We will prove a stronger statement,
For any $zin Z$, let $yin Y$ be its image, if there exists $xin X$ such that $f(x)=y$, then exists $win X imes_Y Z$ mapsto $y$.
Take an affine set $operatorname{Spec}A, operatorname{Spec}B, operatorname{Spec}C$ of $x,y,z$ such that the image of $operatorname{Spec} A$ and $operatorname{Spec} C$ is in $operatorname{Spec} B$. So the problem reduce to the following statement.
Let $Astackrel{varphi}leftarrow Bstackrel{psi} o C$ be ring homomorphisms, and primes $mathfrak{p}, mathfrak{r}$ of $A,C$ respectively, such that $mathfrak{q}=varphi^{-1}(mathfrak{p})=psi^{-1}(mathfrak{r})$. Then there exists a prime $mathfrak{s}$ of $Aotimes_B C$, such $mathfrak{r}$ is the inverse image of $mathfrak{s}$.
$$egin{array}{ccc} Aotimes_B C& leftarrow & A\ uparrow && uparrow \ C& leftarrow & B \ end{array}qquad egin{array}{ccc} mathfrak{s}& mapsto & mathfrak{p}\ overline{downarrow} && overline{downarrow} \ mathfrak{r}& mapsto & mathfrak{q} \ end{array} $$
Second Step (some computation of tensor product). We show the following
Consider the tensor product of $k$-algebra $R_1otimes_k R_2$. For a mutiplitive subset $S$ of $R_1$, one have $$S^{-1}(R_1otimes_k R_2)=S^{-1} R_1otimes_{overline{S}^{-1}k} overline{S}^{-1} R_2$$Where $overline{S}subseteq k$ is the inverse image of $S$, and $k$ is not necessary to be a field.
The proof is nothing but check the structure of tensor product. More precisely, $S^{-1}(R_1otimes_kR_2)=S^{-1}R_1otimes_{R_1}R_1otimes_k R_2 =S^{-1}R_1 otimes_kR_2$ and $$egin{cases} frac{r_1}{s}otimes frac{r_2}{s'} = frac{r_1}{ss'}s'otimes frac{r_2}{s'}=frac{r_1}{ss'}otimes s'frac{r_2}{s'}=frac{r_1}{ss'}otimes r_2\frac{r_1}{s_1}frac{k}{s}otimes frac{r_2}{s_2}=frac{r_1}{s_1}frac{k}{s}otimes sfrac{1}{s}frac{r_2}{s_2}=frac{r_1}{s_1}kotimes frac{1}{s}frac{r_2}{s_2}=frac{r_1}{s_1}otimes frac{k}{s}frac{r_2}{s_2}end{cases}$$
Third Step (finish the proof). By the second step, we can assume $B, C$ to be local ring. Then it reduces to whether $Aotimes_B C otimes C/mathfrak{r}=0$. We have know that $Aotimes_B B/mathfrak{q} eq 0$ by the assumption on $mathfrak{q}$. One have $$Aotimes_B Cotimes_C C/mathfrak{r}=underbrace{Aotimes_B B/mathfrak{q}}_{ eq 0}otimes_{B/mathfrak{q}}otimes C/mathfrak{r}$$But now, $B/mathfrak{q}$ and $C/mathfrak{r}$ is field, thus, it is not zero either, the proof is complete.
Appendix (The fiber of $yin Y$ in the morphism $X o Y$ is $X imes_Y k(y)$). We only need to prove the affine case. Let $Bstackrel{varphi} o A$ be the associated ring homomorphism, given a prime $mathfrak{q}$ of $B$, one have $$egin{array}{rl}f^{-1}(mathfrak{q})& = { extrm{prime } mathfrak{p}subseteq A: varphi^{-1}(mathfrak{p})=mathfrak{q}} \ & ={ extrm{prime } mathfrak{p}subseteq A: varphi^{-1}(mathfrak{p})subseteq mathfrak{q}, varphi(mathfrak{q})subseteq mathfrak{p} }\ & cong { extrm{prime } mathfrak{p}subseteq A_mathfrak{q}/varphi(mathfrak{q})A_{mathfrak{q}}} \ & cong operatorname{Spec} (A_mathfrak{q}/varphi(mathfrak{q})A_{mathfrak{q}})=operatorname{Spec}( Aotimes_B B_mathfrak{q}/mathfrak{q}B_{mathfrak{q}})=operatorname{Spec} (Aotimes_B k(mathfrak{q}))end{array}$$Where $k(mathfrak{q})=operatorname{Frac} B/mathfrak{q}=B_{mathfrak{q}}/mathfrak{q}B_{mathfrak{q}}$ is the residual field of the point $mathfrak{q}$.
