In the book Elliptic Curves: Number Theory and Cryptography by Lawrence C. Washington, it says the following (see picture given below):
"Since $\text{E}(\overline{K})$ is infinite and the kernel of $\alpha$ is finite, only finitely many points of $\text{E}(\overline{K})$ can map to a point with a given $x$-coordinate."
I don't see why it is true that only finitely many such points exists. I tried to come up with a contradiction to no avail. Any help is appreciated.
This answer is based on Adrien's comment. Fix $(x,y)$ in $E(\overline{K})$, suppose there exists $P$ in $E(\overline{K})$ such that $\alpha(P)=(x,y)$ (if not, then the statement is trivial), consider the following map: $$f:\{Q\in E(\overline{K}) : \alpha(Q)=(x,y)\}\to \text{ker}(\alpha), \; Q \mapsto Q - P.$$ Clearly $f$ maps to $\text{ker}(\alpha)$. Moreover, $f$ is surjective; pick any element $Z$ in $\text{ker}(\alpha)$, then consider $Z+P$ in $E(\overline{K})$. Note that $\alpha(Z+P)=\alpha(Z)+\alpha(P)=\infty+(x,y)=(x,y)$ (using that $\alpha$ is a homomorphism), hence $Z+P \in \{Q\in E(\overline{K}) : \alpha(Q)=(x,y)\}$. Evidently, $f(Z+P)=Z$. So $f$ is surjective. Now suppose that $f(V)=f(W)$ for some $V,W$ in the domain. Then $V-P=W-P$ implying that $V=W$, hence $f$ is injective. So $f$ is bijective and since $\text{ker}(\alpha)$ is finite it follows that only finitely many elements map to a point with a given $x$-coordinate.