Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.6 Exercise 1.6.15
Define a map $\pi : \mathbb{R}^2 \rightarrow \mathbb{R}$ by $\pi((x,y)) = x$. Prove that $\pi$ is a homomorphism and find the kernel of $\pi$.
Solution: To show that $\pi$ is a homomorphism, let $(x_1,y_1), (x_2,y_2) \in \mathbb{R}^2$. Then $$\pi((x_1,y_1) + (x_2,y_2)) = \pi((x_1+x_2, y_1+y_2)) = x_1+x_2 = \pi((x_1,y_1)) + \pi((x_2,y_2)).$$ Now I claim that $\mathsf{ker}\ \pi = 0 \times \mathbb{R}$.
($\subseteq$) If $(x,y) \in \mathsf{ker}\ \pi$ then we have $x = \pi((x,y)) = 0$. Thus $(x,y) \in 0 \times \mathbb{R}$.
($\supseteq$) If $(x,y) \in 0 \times \mathbb{R}$, we have $x = 0$ and thus $\pi((x,y)) = 0$. Hence $(x,y) \in \mathsf{ker}\ \pi$.
