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Let $k$ be a field. Let $\Omega$ be an algebraically closed extension of $k$. Let $X$ be a scheme of finite type over $k$. We denote $Hom_k(Spec(\Omega), X)$ by $X(\Omega)$. Let $\eta$ be the unique point of $Spec(\Omega)$. Let $\alpha,\beta \in X(\Omega)$. If $\alpha(\eta) = \beta(\eta)$, we write $\alpha \sim \beta$.

Let $f\colon X \rightarrow Y$ be a morphism of schemes of finite type over a field $k$. $f$ induces a map $\bar f\colon X(\Omega) \rightarrow Y(\Omega)$. Suppose $\bar f(\alpha) = \beta$. Suppose $\beta \sim \beta'$. Does there exist $\alpha'$ such that $\alpha \sim \alpha'$ and $\bar f(\alpha') = \beta'$?

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Yes. Note that to give a geometric point $\alpha\in X(\Omega)$ is equivalent to give a point $x\in X$ ($x=\alpha(\eta)$) and a homomorphism $k(x)\to \Omega$ of $k$-algebras.

So let $x=\alpha(\eta)$ and $y=f(x)=\beta(\eta)$. Let $\phi: k(y)\to k(x)$ be the canonical $k$-homomorphism induced by $f$. The geometric point $\beta'\in Y(\Omega)$ has image $y$ and corresponds to a $k$-homomorphism $k(y)\to\Omega$: $$\begin{matrix} k(y) & \stackrel{\beta'}{\longrightarrow} & \Omega\\ {\phi} \downarrow && \\ k(x)&\stackrel{\alpha}{\rightarrow}& \Omega \end{matrix} $$ We want to find a $k$-homomorphism $\alpha' : k(x)\to \Omega$ such that $\alpha'\circ \phi=\beta'$. Let $(t_i)_i$ be a basis of transcendance of $k(x)$ over $\phi(k(y))$. Define $\gamma': k(y)(t_i)_i\to\Omega$ by $\gamma'=\beta'$ on $k(y)$ and $\gamma'(t_i)=\alpha(t_i)$.
$$\begin{matrix} k(y) & \stackrel{\beta'}{\longrightarrow} & \Omega\\ \downarrow && ||\\ k(y)(t_i)_i & \stackrel{\gamma'}{\longrightarrow} & \Omega \\ {\phi'} \downarrow && \\ k(x) \end{matrix} $$ Now as the extension $k(x)/k(y)(t_i)_i$ is algebraic and $\Omega$ is algebraically closed, $\gamma'$ extends to a $k$-homomorphism $\alpha': k(x)\to\Omega$. By "extends" I mean $\alpha'\circ \phi'=\gamma'$, where $\phi': k(y)(t_i)_i\to k(x)$ is the natural extension of $\phi$. The map $\alpha'$ and $x$ form a geometric point $\in X(\Omega)$ as desired.

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