Possible images of $\mathbb P^2$ Let $X$ be the blow up of $\mathbb P^2$ at one point.  Does there exist a nonconstant map $\mathbb P^2 \to X$?
 A: I don't think there can be such a map. Suppose $\phi:\mathbb{P}^2\rightarrow X$ is a map. We divide our work into two cases.
Case 1: $\deg(\phi)>0$. In this case, $\phi$ is surjective. Let $E\in X$ be the exceptional divisor and $L\in X$ be a line disjoint from $E$. Then, $\phi^{*}E$ and $\phi^{*}L$ are divisors on $\mathbb{P}^2$ such that their intersection number is 0. 
This means either $\phi^{*}E$ or $\phi^{*}L$ is trivial by Bezout's theorem. However, this is impossible since they are nontrivial divisors (this is where we use $\phi$ is surjective). 
Case 2: $\deg(\phi)=0$. In this case, if $\phi$ is nonconstant, its image is a curve $C\subset X$. Since $\mathbb{P}^2$ is irreducible, $C$ is irreducible. Since $\mathbb{P}^2$ is reduced, the map $\mathbb{P}^2\rightarrow X$ factors through $C$, so we have a surjective map $\mathbb{P}^2\rightarrow C$ onto an irreducible curve.
Since $\mathbb{P}^2$ is normal, this map factors through the normalization $\tilde{C}\rightarrow C$, so we have a map $\mathbb{P}^2\rightarrow \tilde{C}$ onto a smooth curve. Any smooth curve admits a map to $\mathbb{P}^1$, so we have a surjective map $\mathbb{P}^2\rightarrow \mathbb{P}^1$, which is impossible. 
