Exercise 3.1.8 (Image and fibers of $\varphi : \mathbb{R}^ imes o \mathbb{R}^ imes$ defined by $\varphi(x) = |x|$)

Question

Let $\varphi : \mathbb{R}^	imes 	o \mathbb{R}^	imes$ be the map sending $x$ to the absolute value of $x$. Prove that $\varphi$ is a homomorphism and find the image of $\varphi$. Describe the kernel and the fibers of $\varphi$.

Richard Ganaye · Accepted Answer

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

\begin{proof} 
For all $x, y \in \R$,
$$\varphi(xy) = |xy| = |x| |y| = \varphi(x) \varphi(y),$$
so $\varphi$ is a homomorphism.

For all $x 
e 0$, $\varphi(x) = |x| >0$, so $\mathrm{im}(\varphi) \subseteq \R^	imes_+$.

Conversely, if $x \in R^	imes_+$, then $x = |x| = \varphi(x)$, so $x \in \mathrm{im}(\varphi)$. This shows the inclusion $R^	imes_+ \subseteq \mathrm{im}(\varphi)$, so
$$\mathrm{im}(\varphi) =  \R^	imes_+ = \{x \in \R \mid x>0\}.$$
For all $x \in \R^	imes$,
$$\varphi(x) = 1 \iff |x| = 1 \iff x \in \{-1,1\},$$
so
$$\ker(\varphi) = \{-1, 1 \}.$$
Similarly,
$$\varphi(x) = a \iff |x| = a \iff x \in \{-a,a\},$$
so the fibers of $\varphi$ are the sets
$$\varphi^{-1}(a) = \{-a, a\}.$$
\end{proof}

Exercise 3.1.8 (Image and fibers of $\varphi : \mathbb{R}^\times \to \mathbb{R}^\times$ defined by $\varphi(x) = |x|$)

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Exercise 3.1.8 (Image and fibers of $\varphi : \mathbb{R}^\times \to \mathbb{R}^\times$ defined by $\varphi(x) = |x|$)

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