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Sorry if the question is simple.

Why does not exist a surjective morphism (as morphism of schemes) from $\mathbb{A}^1$ onto a variety $Y$ with only two points?

I know that there does not exist such a morphism but I don't know why.

J.Harris says that in "Algebraic Geometry (A first course) in Lecture 10, Quotiens (pg 124)"

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    The image of a connected set is connected. – 2017-01-18
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    Thank you, is simple and elegant. – 2017-01-18
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    You're welcome ! – 2017-01-18

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Here $k$ is a field. Such a morphism is induced by a morphism of algebras $f:k\oplus k\rightarrow K[X]$,

the image of $f((1,1)=1$. you also have $f(1,0)f(0,1)=f(0,0)=0$. You can suppose $f(1,0)=0$, without restricting the generality. This implies $f(0,1)=1$. If $P$ is an irreducible polynomial, $f^{-1}((P))=(1,0)$ since $f(0,1)=1$ is not in $P$. and $(1,0)$ is the image of the map $Spec(k[X])\rightarrow Spec(k\oplus k)$ induced by $f$.