I'm trying to understand how to define the square root of a complex function "globally". Let's say we have some function from some set $X$ onto $\mathbb{C} - \{0\}$: $$ f:X\to\mathbb{C}-\{0\} $$ and we define some number by $N(x) := \frac{f(x)}{\sqrt{f(x)^2}}$.
It is clear that this number is always equal to plus or minus one: $ N(x)\in\{1,-1\}$.
However, what is unclear to me, is how to distinguish between these two cases, given a (complex) numerical value for $f(x)$.
I know that the square root of a complex number is only defined up to a sign: $\sqrt{z^2}=\pm z$ due to the multi-valuedness of the argfunction and the definition $$ z^n := \exp(\log(z^n)) = \exp(\log(|z|^n)+i \arg(z^n)) = |z|^n \exp(i \arg(z^n)) = |z|^n \exp(i n\varphi + in2\pi k) \forall k\in\mathbb{Z}$$ where $\varphi\in[0,2\pi)$ is some angle we choose to represent $z$ and when $n=\frac{1}{2}$ we have only two distinct values of $k$, namely $k\in\{0,1\}$ so that $\sqrt{z} \in \{\sqrt{|z|} \exp(i\frac{\varphi}{2}), \sqrt{|z|} \exp(i(\frac{\varphi}{2}+\pi))\}$. where, again, $\varphi\in[0,2\pi)$ is some angle we choose to represent $z$.
The choice of which one of these possibilities we choose defines a branch, with $[0,\infty)$ being the branch cut, through which $\sqrt{}$ is not continuous.
What I don't understand is as follows:
I read in some paper (unrelated, physics) that "because $f$ is never zero globally and is continuous, we may define a global branch for the square root and it is uniquely defined in each point."
OK, so let's say we choose globally the first branch, in which $\sqrt{z} = \sqrt{|z|} \exp(i\frac{\varphi}{2})$. Then how will we ever get $-1$ for $N(x)$? It seems to me like it will always be $+1$: $$ N(x) \equiv \frac{f(x)}{\sqrt{f(x)^2}} = \frac{|f(x)| \exp (i\varphi)}{\sqrt{|f(x)|^2 \exp (i 2 \varphi)}} = 1$$
How do you ever get -1???
Naively there must be some mistake in what I did because we know that if $f(x) = -2$ then $\sqrt{f(x)^2} = 2$ and so we get -1 indeed. But I cannot think of a general rule.
EDIT: Following a request, I am adding additional context and information about the actual problem at hand.
We assume there is a continuous map $f:\mathbb{T}^2\to\mathbb{C}^\ast$ where $\mathbb{T}^2$ is the 2-torus, and we are given the value of the function at four points on the torus: $\{f((0,0)), f((0,\pi)), f((\pi,0)), f((\pi,\pi))\}$, but not the actual map $f$.
How is it possible to then determine the sign of the following product: $$ \frac{f((0,0))}{\sqrt{f((0,0))^2}}\frac{f((0,\pi))}{\sqrt{f((0,\pi))^2}}\frac{f((\pi,0))}{\sqrt{f((\pi,0))^2}}\frac{f((\pi,\pi))}{\sqrt{f((\pi,\pi))^2}} $$?
The paper you read means the following: The squaring map
$$sq: \mathbb C^* \longrightarrow \mathbb C^*, sq(z):=z^2,$$
is a covering map in the sense of algebraic topology.
The set $\mathbb C$ is simply connected. Hence it has vanishing fundamental group $\pi_1(\mathbb C,*) = 0$. Therefore $f: \mathbb C \longrightarrow \mathbb C^*$ lifts to any covering of $\mathbb C^*$, in particular to the domain of $sq$: A continous map $g: \mathbb C \longrightarrow \mathbb C^*$ exists with $sq \circ g = f$, i.e. $g^2 = f$. A second lift differs by the sign, because you may choose one point $z_0 \in \mathbb C$ and prescribe one value
$$g(z_0) \in sq^{-1}(f(z_0))=\{ +\sqrt {f(z_0)}, -\sqrt {f(z_0) }\},$$
q.e.d.
Note. Concerning the lifting property see the lifting theorem, e.g., in "Spanier, Edwin: Algebraic Topology. Chap. 2, sect. 4, Theor. 5". The lifting theorem is a simple but powerful tool.