Define $\phi: \mathbb{R}^x \mapsto \{\pm1\}$ by letting $\phi(x)$ be $x$ divided by the absolute value of $x$. Describe the fibers of $\phi$ and that $\phi$ is a homomorphism.
Need help getting a grasp of this topic. For reference, this is Dummit and Foote, 3ed., section 3.1, question 6.
Description of the fiber
The only two possible fibers from the set $\{\pm1\}$ are the fibers $X_{-1}$ and $X_1$. By definition, these are represented as follows
$X_{1} = \phi^{-1} (1) = \{x \in \mathbb{R} | \phi(x) = 1\}$
$X_{-1} = \phi^{-1} (-1) = \{x \in \mathbb{R} | \phi(x) = -1\}$
However, since the mapping $\phi$ is defined as $\phi(x) = \frac{x}{|x|}$, we can re-write the above as
$X_{1} = \phi^{-1} (1) = \{x \in \mathbb{R} | x>0\}$
$X_{-1} = \phi^{-1} (-1) = \{x \in \mathbb{R} | x<0\}$
Homomorphism(of particular concern)
Let $x,y \in \mathbb{R}^x$. Consider three cases. (Do I need to do three cases?)
If $x>0$, $y<0$, then $xy<0$.
$\phi(xy) = -1 = 1 \times -1 = \phi(x)\phi(y)$
If $x>0$, $y>0$, then $xy>0$.
$\phi(xy) = 1 = 1 \times 1 = \phi(x)\phi(y)$
$x<0$, $y<0$, then $xy>0$.
$\phi(xy) = 1 = -1 \times -1 = \phi(x)\phi(y)$
Thus, $\phi$ is a homomorphism.
Best Answer
Perfect.
You don't need to separate cases, observe only that $|xy|=|x|\,|y|$, and then it follows that $\phi=x\mapsto x/|x|$ also preserves multiplication.