Let $k$ be a field. I am trying to prove the following isomorphism $$f\colon k[x_{0},x_{1},x_{2}]/I\xrightarrow{\cong}k[y_{0},y_{1}]$$
where $I=(x_{0}+x_{1}+x_{2}-1)$ and $$f(x)=\begin{cases}y_{0},&\text{if $x=x_{0}$}\\ y_{1},&\text{if $x=x_{1}$}\\1-y_{0}-y_{1},&\text{if $x=x_{2}$}\end{cases}$$
I can prove that it is well-defined and epimorphism. I am stuck at injectivity (I want to prove it rigorously). First, by the formula we have $I\subseteq \ker(f)$. Let $h(x_{0},x_1,x_2)\in\ker(f)$. Then we can write $h(x_0,x_1,x_2)=\sum_{i=0}^{n}g_{i}(x_0,x_1)x_{2}^{i}$. We have, $$\sum_{i=0}^{n}g_{i}(y_0,y_1)(1-y_{0}-y_1)^{i}=0$$ Then, we can write $$0=\sum_{i=0}^{n}g_{i}(y_0,y_1)(1-y_{0}-y_1-y_2+y_2)^{i}=\sum_{i=0}^{n}g_{i}(y_0,y_1)\sum_{k=0}^{i}\binom{i}{k}(1-y_{0}-y_1-y_2)^{i-k}y_{2}^{k}=$$ $$\sum_{i=0}^{n}g_{i}(y_0,y_1)\sum_{k=0}^{i-1}\binom{i}{k}(1-y_{0}-y_1-y_2)^{i-k}y_{2}^{k}+h(y_1,y_2,y_3)$$ and hence $h\in\ker(f)$. Is there a more efficient way to prove injectivity? Is my argument rigorous?