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.

  • 2
    $\begingroup$ This all looks fine. You can clean up the final part by noting that $|xy| = |x| |y|$, so that $$\varphi(xy) = \frac{xy}{|xy|} = \frac{x}{|x|}\frac{y}{|y|} = \varphi(x) \varphi(y)$$ $\endgroup$ – user61527 Mar 8 '14 at 0:03


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.


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