3
$\begingroup$

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.

$\endgroup$
  • 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
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.