Nowadays, I close a new small case.

Proposition. For a surjective morphism between scheme $X\stackrel{f}\to Y$, For any $Z\to Y$, the base change $X\times_Y Z\to Z$ is also surjective.

The diagram is as following

$$\begin{array}{ccc} X\times_Y Z& \to & Z\\ \downarrow && \downarrow \\ Z& \to & Y\\\end{array}$$

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 more stronger statement,

For any $z\in Z$, let $y\in Y$ be its image, if there exists $x\in X$ such that $f(x)=y$, then exists $w\in X\times_Y Z$ 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 $A\stackrel{\varphi}\leftarrow B\stackrel{\psi}\to C$ 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 $A\otimes_B C$, such $\mathfrak{r}$ is the inverse image of $\mathfrak{s}$.

$$\begin{array}{ccc} A\otimes_B C& \leftarrow & A\\ \uparrow && \uparrow \\ C& \leftarrow & B \\ \end{array}\qquad \begin{array}{ccc} \mathfrak{s}& \mapsto & \mathfrak{p}\\ \overline{\downarrow} && \overline{\downarrow} \\ \mathfrak{r}& \mapsto & \mathfrak{q} \\ \end{array} $$

Second Step (some computation of tensor product). We show the following

Consider the tensor product of $k$-algebra $R_1\otimes_k R_2$. For a mutiplitive subset $S$ of $R_1$, one have $$S^{-1}(R_1\otimes_k R_2)=S^{-1} R_1\otimes_{\overline{S}^{-1}k} \overline{S}^{-1} R_2$$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, $S^{-1}(R_1\otimes_kR_2)=S^{-1}R_1\otimes_{R_1}R_1\otimes_k R_2 =S^{-1}R_1 \otimes_kR_2$ and $$\begin{cases} \frac{r_1}{s}\otimes \frac{r_2}{s'} = \frac{r_1}{ss'}s'\otimes \frac{r_2}{s'}=\frac{r_1}{ss'}\otimes s'\frac{r_2}{s'}=\frac{r_1}{ss'}\otimes r_2\\\frac{r_1}{s_1}\frac{k}{s}\otimes \frac{r_2}{s_2}=\frac{r_1}{s_1}\frac{k}{s}\otimes s\frac{1}{s}\frac{r_2}{s_2}=\frac{r_1}{s_1}k\otimes \frac{1}{s}\frac{r_2}{s_2}=\frac{r_1}{s_1}\otimes \frac{k}{s}\frac{r_2}{s_2}\end{cases}$$

Third Step (finish the proof). By the second step, we can assume $B, C$ to be local ring. Then it reduces to whether $A\otimes_B C \otimes C/\mathfrak{r}=0$. We have know that $A\otimes_B B/\mathfrak{q}\neq 0$ by the assumption on $\mathfrak{q}$. One have $$A\otimes_B C\otimes_C C/\mathfrak{r}=\underbrace{A\otimes_B B/\mathfrak{q}}_{\neq 0}\otimes_{B/\mathfrak{q}}\otimes C/\mathfrak{r}$$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 $y\in Y$ in the morphism $X \to Y$ is $X\times_Y k(y)$). We only need to prove the affine case. Let $B\stackrel{\varphi}\to A$ be the associated ring homomorphism, given a prime $\mathfrak{q}$ of $B$, one have $$\begin{array}{rl}f^{-1}(\mathfrak{q})& = \{\textrm{prime } \mathfrak{p}\subseteq A: \varphi^{-1}(\mathfrak{p})=\mathfrak{q}\} \\ & =\{\textrm{prime } \mathfrak{p}\subseteq A: \varphi^{-1}(\mathfrak{p})\subseteq \mathfrak{q}, \varphi(\mathfrak{q})\subseteq \mathfrak{p} \}\\ & \cong \{\textrm{prime } \mathfrak{p}\subseteq A_\mathfrak{q}/\varphi(\mathfrak{q})A_{\mathfrak{q}}\} \\ & \cong \operatorname{Spec} (A_\mathfrak{q}/\varphi(\mathfrak{q})A_{\mathfrak{q}})=\operatorname{Spec}( A\otimes_B B_\mathfrak{q}/\mathfrak{q}B_{\mathfrak{q}})=\operatorname{Spec} (A\otimes_B k(\mathfrak{q}))\end{array}$$Where $k(\mathfrak{q})=\operatorname{Frac} B/\mathfrak{q}=B_{\mathfrak{q}}/\mathfrak{q}B_{\mathfrak{q}}$ is the residual field of the point $\mathfrak{q}$.