Show that $\mathbb{R}^* / \{-1, 1\} \cong \mathbb{R}^+$

Question

Let $f$ be a function such that $f: \mathbb{R}^* \rightarrow \mathbb{R}^+$ and $f(x) = x^2$. Now, we know that the kernel of the function is $\{-1, 1\}$ because: \begin{align*} f(x) &= 1 \\ x^2 &= 1 \\ x &= \pm 1 \end{align*} Now, we know that $\mathbb{R}^+$ is the image of the function because $x^2$ only gives out positive numbers. So, according to the first isomorphism theorem, if $f: R^* \rightarrow R^+$ is a homomorphism, then $R^*/ker(f) \cong R^+$.

But, "we know that $\mathbb{R}^+$ is the image of the function because $x^2$ only gives out positive numbers" only says that image of f is a subset of $R^*$. How do I approach this problem?

Answer 1

You just need to show that $f$ is a homomorphism. But $f(ab)=(ab)^2=a^2b^2=f(a)f(b)$.

The image of $f$ is indeed $\Bbb R^+$, as required, as every positive real number has a nonzero real square root (actually it has two, a positive one and a negative one.)

Note that $f$ is $2-1$, as the kernel has order $2$.

Answer 2

Assuming you do not have to use the function provided, the answer below seems to work:

First, define $\phi$ by:

$$\phi:\mathbb{R}^*\rightarrow\mathbb{R}_{>0}$$

$$\phi(x)=|x|$$

This is simple to check that it's a homomorphism:

$$\phi(xy)=|xy|=|x||y|=\phi(x)\phi(y)$$

If $a\in\mathbb{R}_{>0}$, then:

$$\phi(a)=|a|=a$$

Therefore, $\phi$ is surjective, i.e. $\text{Im}(\phi)=\mathbb{R}_{>0}$. It's also clear that $1,-1$ are the only things sent to the identity element in $\mathbb{R}_{>0}$ which is $1$, so it's clear that $(1,-1)$ is the kernel of $\phi$. So, by the Homomorphism Theorem, we have that:

$$\mathbb{R}^*/\{1,-1\}\cong\mathbb{R}_{>0}$$