Question about showing $R^{*}/N\cong R^{**}$ For the following question: 
Let $R^{*}$ be the multiplicative group of nonzero real numbers and let $N$ be the subgroup $\{1,-1\}$.  Prove that $R^{*}/N$ is isomorphic to the multiplicative group $R^{**}$ of positive numbers.
I need to write an explicit map from $R^{*}/N$ to $R^{**}$.  $\textbf{Can I let}$ $f(Nx)=\lvert x\rvert$ from the beginning instead of $f(\lvert x\rvert)=Nx$. The student solutions manual first write the function:  $f(x)=Nx$, then the absolute value of x is used when it justifies that $f$ is surjective:

For any $x\in R^{*},$ the absolute value $\lvert x\rvert$ is in $R^{**}$.  Also $\lvert x\rvert\in Nx$ so that $f(\lvert x\rvert)=Nx.$

Thank you in advance
 A: 
$\textbf{Can I let}$ $f(Nx)=\lvert x\rvert$ from the beginning instead of $f(\lvert x\rvert)=Nx$.

Yes, as long as you check the due things, accordingly. Moreover, everything can be done even prior to letting kernel concept setting in (let alone the homomorphism theorems), which, IMHO, tends here to hastily sell off the basics.
So, if you address the (tentative) map $f\colon \Bbb R^*/N\longrightarrow\Bbb R^*_{>0}$, defined by $f(Nx):=|x|$, then you first have to check whether this is well-defined, as $f$'s definition depends on the choice of coset's representative. But $Nx=\{x,-x\}$, so $f(N(-x))=\left|-x\right|=|x|=f(Nx)$, and you are done. Next, $f(NxNy)=f(Nxy)=|xy|=|x||y|=f(Nx)f(Ny)$ (operation-preserving). Surjectivity is plain, because every positive real number $y$ is the modulus of itself, and hence $\Bbb R^*_{>0}\ni y=|y|=f(Ny)$. Injectivity: $f(Ny)=f(Nx)\Longrightarrow |y|=|x|\Longrightarrow y=\pm x\Longrightarrow y\in Nx\Longrightarrow Ny=Nx$.
On the other hand, if you address the (tentative) map $\hat f\colon \Bbb R^*_{>0}\longrightarrow \Bbb R^*/N$, defined by $f(x):=Nx$, then good-definiteness is not in the agenda, as there's no arbitrary choice in the definition of $\hat f$. Next, $\hat f(xy)=Nxy=NxNy=\hat f(x)\hat f(y)$ (operation-preserving). Surjectivity: since $Nx=N(-x)$, either $Nx=\hat f(x)$ or $Nx=\hat f(-x)$, according to which one is positive between $x$ and $-x$. Injectivity: $\hat f(y)=\hat f(x)\Longrightarrow Ny=Nx\Longrightarrow y=\pm x\Longrightarrow y=x$ (as $x,y$ are both positive).
Therefore, $\Bbb R^*/N\stackrel{f}{\cong}\Bbb R^*_{>0}$ or, equivalently, $\Bbb R^*_{>0} \stackrel{\hat f}{\cong}\Bbb R^*/N$.
