Exercise 3.1.6 (Homomorphism $x \mapsto x/|x|$)

Question

Define $\varphi : \mathbb{R}^	imes 	o \{\pm 1\}$ by letting $\varphi(x)$ be $x$ divided by the absolute value of $x$. Describe the fibers of $\varphi$ and prove that $\varphi$ is a homomorphism.

Richard Ganaye · Accepted Answer

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

\begin{proof} 
By definition,
$$
\varphi\ 
\left\{
\begin{array}{ccl}
 \mathbb{R}^	imes &	o& \{-1, 1\}\
  x & \mapsto &\frac{x}{|x|}
\end{array}
ight..
$$
Note that for all $x \in \R^	imes$,
$$\varphi(x) = 1 \iff x = |x| \iff x>0.$$
Hence the two fibers are
\begin{align*}
\varphi^{-1}(1) &= \{x \in \R \mid x >0\} = \R^	imes_+,\
\varphi^{-1}(-1) &= \{x \in \R \mid x <0\} = \R^	imes_-.
\end{align*}
Moreover, for all $x, y \in \R^	imes$,
$$\varphi(xy) = \frac{xy}{|xy|} = \frac{x}{|x|} \frac{y}{|y|} = \varphi(x) \varphi(y).$$
So $\varphi$ is a homomorphism.
\end{proof}

Exercise 3.1.6 (Homomorphism $x \mapsto x/|x|$)

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Exercise 3.1.6 (Homomorphism $x \mapsto x/|x|$)

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