Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.1 Exercise 3.1.7
Define $\pi : \mathbb{R}^2 \rightarrow \mathbb{R}$ by $\pi(x,y) = x + y$. Prove that $\pi$ is a surjective homomorphism and describe the fibers of $\pi$ geometrically.
Solution: First, $\pi$ is a homomorphism since \begin{align*}\pi((x_1,y_1) + (x_2,y_2)) = &\ \pi((x_1+x_2,y_1+y_2)) \\=&\ x_1+x_2+y_1+y_2\\ =&\ x_1+y_1+x_2+y_2\\ =&\ \pi(x_1,y_1) + \pi(x_2,y_2).\end{align*} Then, $\pi$ is surjective since for all $c \in \mathbb{R}$, $\pi(c,0) = c$. Clearly the fiber of $c \in \mathbb{R}$ is $\{ (x,y) \ |\ x+y = c \}$; that is, the line $x+y=c$ in $\mathbb{R}^2$.
