How can we evaluate a homomorphism in $\alpha$? Some confusion in homomorphism . My confusion given below marked in red line

My question is that how  can we evaluate a homomorphism in $\alpha$?
My attempt :  I was  thinking $f : L \rightarrow k[x]$ and $g : L \rightarrow k[x]$
How can we show that $f=g$ ?

They've chosen an unnecessarily awkward way to phrase it, but for any polynomial ring $k[x]$ and any element $\alpha\in k$, there is a homomorphism (not homeomorphism) $E_\alpha\colon k[x]\to k$ given by $E_\alpha(f(x)) = f(\alpha)$ for any $f(x)\in k[x]$. (It's a good exercise to verify that this is a ring homomorphism.)
It would be easier to understand if they'd just said "evaluate both sides at $x=\alpha$".
