I want to prove the following statement directly. However, I've got the contradictory results; could you help me?
If $C \subset \mathbb{P}_{k}^{2} \setminus \{ [0:0:1]\}$ is a curve and $\rho_{1}: \mathbb{P}_{k}^{2} \to \mathbb{P}_{k}^{1}$ is defined via $$\rho_{1}( [x_{0}:x_{1}:x_{2}]) =[x_{0}:x_{1}] $$ Then, $\rho_{1}(C)$ is closed. (We assume that affine space is algebraically closed.)
What I attempt to prove this is below; which gives that $\rho_{1}(C)$ is open.
without loss of generality, let $C$ be irreducible, defined by some $d$-degree homogeneous polynomial $f(x_{0}, x_{1},x_{2})$. Then, we can rewrite $$ f = \sum_{i=0}^{d}c_{i}(x_{0},x_{1})x_{0}^{i}$$ Then for any $[x_{0}:x_{1}:x_{2}]$ two cases hold;
First of all, if $c_{i}(x_{0},x_{1}) = 0$ for all $i\geq 1$ but $c_{0}(x_{0},x_{1}) \neq 0$, then $f(x_{0},x_{1},x_{2}) = c_{0}(x_{0},x_{1}) \neq 0$. Hence, $[x_{0}:x_{1}:x_{2}] \not\in C$.
Also note that $c_{i}(x_{0},x_{1}) = 0$ for all $i$ cannot happen since $[0:0:1] \not\in C$. ?Thus, the second case is that at least for some $i\geq 1$, $c_{i}(x_{0},x_{1})\neq0$. Then, by the fundamental theorem of algebra with respect to $x_{2}$, we have some $x_{2}$ such that $[x_{0}:x_{1}:x_{2}] \in C$.
Thus, $\rho_{1}(C) = \cup_{i=1}^{d}\{[x_{0}:x_{1}]: c_{i}(x_{0},x_{1})\neq 0 \}$, hence it is open.
I think it is suspicious; could you help me for proving this?
Your argument (or something like it) is correct, and it implies that $p_1(C)=\Bbb P^1$. You're finding that for any $[x_0:x_1]\in\Bbb P^1$, there exists some $x_2$ so that $[x_0:x_1:x_2]$ is in $C$. Thus this morphism is surjective on the level of closed points and therefore surjective, so $p_1(C)=\Bbb P^1$ and since every set is closed in itself, you are done.