Exercise 3.1.7 (Fibers of $\pi : \mathbb{R^2} o \mathbb{R}$ defined by $\pi((x,y)) = x + y$)

Question

Define $\pi : \mathbb{R}^2 	o \mathbb{R}$ by $\pi((x,y)) = x + y$. Prove that $\pi$ is a surjective homomorphism and describe the kernel and fibers of $\pi$ geometrically.

Richard Ganaye · Accepted Answer

ewcommand{\R}{\mathbb{R}}
\begin{proof} 
Consider the map
$$
\pi\ 
\left\{
\begin{array}{ccl}
\R^2 & 	o & \R\
(x,y) & \mapsto & x + y
\end{array}
ight..
$$
Let $u = (x,y) \in \R^2$ and $ v = (z,t) \in R^2$. Then
\begin{align*}
\varphi(u + v) &= \varphi((x,y) + (z,t))\
&=\varphi(x + z, y + t)\
&= (x+z) + (y + t)\
&= (x + y) + (z + t)\
&= \varphi((x,y)) + \varphi((z,t))\
&= \varphi(u) + \varphi(v).
\end{align*}
So $\varphi$ is a homomorphism.

Moreover, let $z$ be any real number. Then $\varphi((z,0)) = z$, so there exists $ u \in \R^2$ (i.e. $u = (z,0)$), such that $\varphi(u) = z$. This shows that $\varphi$ is a surjective homomorphism.

The kernel of $\varphi$ is the line of equation $x +y = 0$, that is the line 
$$\ker(\varphi) = D = \{\lambda \cdot (1,-1)\mid  \lambda \in \R\}.$$

The fibers are the translate of this line: for any real $a$,
$$\varphi^{-1}(a) = \{(x,y) \in \R^2 \mid x + y = a\} =(a,0)  + D$$
is a line parallel to $D$.
\end{proof}

Exercise 3.1.7 (Fibers of $\pi : \mathbb{R^2} \to \mathbb{R}$ defined by $\pi((x,y)) = x + y$)

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Exercise 3.1.7 (Fibers of $\pi : \mathbb{R^2} \to \mathbb{R}$ defined by $\pi((x,y)) = x + y$)

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