Homomorphism and kernel proof help

Question

Consider the map $\phi: \mathbb R^* \to \mathbb R^*$ (under multiplication) defined by $\phi(x) = |x|$. Prove that $\phi$ is a homomorphism and find $\ker \phi$.

Proof:
Now $\phi(xy) = |xy| = |x||y| = \phi(x) \phi(y)$.
Hence, $\phi$ is a homomorphism.

This is the form my professor wants the kernel proof in.

proof:
Now $x$ is in the kernel of $\phi$ if and only if $\phi(x) =$ identity in $\mathbb R^*$.
If and only if $|x| = 1$.
If and only if $x = 1$.
Hence, the $\ker \phi = \{1\}$.

The kernel of phi is trivial since 1 is the identity of the domain. Is that why it is 1 and not 0?

Answer 1

Is $\mathbb{R}^*$ supposed to be $\mathbb{R}\setminus\{0\}$? If so, you're forgetting something.. $\lvert-1\rvert = 1$...

Answer 2

The group $\mathbb R^\times$ consists of all real numbers except for $0$. Are you sure that $+1$ is the only real number with absolute value $1$?