Show that $f(x, y)=(-1)^y x$ is an isomorphism from $\mathbb{R}^{+} \times \mathbb{Z}_2$ to $\mathbb{R}^*$. (REMARK: To combine elements of $\mathbb{R}^{+}\times \mathbb{Z}_2$, one multiplies first components, adds second components.) Conclude that $\mathbb{R}^* \cong \mathbb{R}^+ \times \mathbb {Z}_2$.
I am stuck how to prove that the function is a bijection.
I can show it is a surjection. Showing that it is an injection gives me an equation of $x$ and $y$. I think fact that $y$ only takes the values of 0 and 1 is essential here.
Should I start with $y_1=y_2=0$ then 1 then show that $x_1$ = $x_2$ ?
It's a homomorphism due to index rules.
You have shown it's surjective.
Suppose $f(a,b)=f(x,y)$. Then $(-1)^ba=(-1)^yx$. Then, since $a,x>0$, $a=x$. It follows that $b=y$ since $b,y\in\{0,1\}$ (which is how I presume you conceptualise $\Bbb Z_2$). Thus $(a,b)=(x,y)$. Thus $f$ is injective.