Exercise 1.6.15 (Kernel of a projection)

Question

Define the map $\pi : \mathbb{R}^2 	o \mathbb{R}$ by $\pi((x,y)) = x$. Prove that $\pi$ is a homomorphism and find the kernel of $\pi$ (cd. Exercise 14).

Richard Ganaye · Accepted Answer

ewcommand{\R}{\mathbb{R}}

\begin{proof} Let $(x,y)$ and $(z,t)$ be any elements of $\R^2$. Then
$$\pi((x,y) + (z,t))  = \pi((x + z, y+ t)) = x+ z =\pi((x,y)) + \pi(z,t).$$
This shows that $\pi$ is a group homomorphism.

Moreover
$$(x,y) \in \ker(\pi)\iff \pi((x,y)) = 0  \iff x = 0,$$
so
$$\ker(\pi) = \{0\} 	imes \R.$$

\end{proof}

Exercise 1.6.15 (Kernel of a projection)

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Exercise 1.6.15 (Kernel of a projection)

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