Understanding branch cuts for functions with multiple branch points My question was poorly worded and thus confusing. I'm going to edit it to make it clearer, and then I'm going to give a brief answer.

Take, for example, the function $$f(z) = \sqrt{1-z^{2}}= \sqrt{(1+z)(1-z)} = \sqrt{|1+z|e^{i \arg(1+z)} |1-z|e^{i \arg(1-z)}}.$$
If we restrict $\arg(1+z)$ to $-\pi < \arg(1+z) \le \pi$, then the half-line $[-\infty,-1]$ needs to be omitted.
But if we restrict $\arg(1-z)$ to $0 < \arg(1-z) \le 2 \pi$, why does the half-line $(-\infty, 1]$ need to be omitted and not the half-line $[1, \infty)$?
And if we define $f(z)$ in such a way, how do we show that $f(z)$ is continuous across $(-\infty,-1)$?
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The answer to the first question is $(1-z)$ is real and positive for $z \in (-\infty,1)$.
And with regard to the second question, to the left of $z=x=-1$ and just above the real axis,
$$f(x) = \sqrt{(-1-x)e^{i (\pi)} (1-x)e^{i (2 \pi)}} = e^{3 \pi i /2} \sqrt{x^{2}-1} = -i \sqrt{x^{2}-1} .$$ 
While to the left of $z=x=-1$ and just below the real axis, 
$$f(x) = \sqrt{(-1-x)e^{i (-\pi)} (1-x)e^{i (0)}} = e^{-i \pi /2} \sqrt{x^{2}-1} = -i \sqrt{x^{2}-1} .$$

There are no "natural" or "canonical" "cuts" associated to a "multivalued function", despite much tradition that may accidentally give this impression. Rather, what we are doing in telling a "cut" is restricting the function(s) to a slightly smaller domain on which there are well-defined holomorphic (and continuous) "single-valued" versions of the function.
For explicit functions such as $\log(z)$ and $\sqrt{1-z^2}$, we understand the "bad points" (in these cases $0,\infty$ and $\pm 1$, respectively). In the case of $\log$ we have to prevent ourselves making a loop around $0$ that returns to the same point. In the case of $\sqrt{1-z^2}$, there are perhaps-surprising options. Namely, following a loop around either $\pm 1$ flips the sign of the square root. This is "bad". How to prevent this? Well, we can prevent all loops around both $\pm 1$ by cuts $(-\infty,-1]$ and $[1,\infty)$. Or cuts vertically to $+i\infty$ from both $\pm 1$, for that matter.
Or we can simultaneously prevent such sign-flips by slitting along some (non-self-intersecting) path from $-1$ to $+1$, for example, $[-1,+1]$. Or an arc of the unit circle going from $-1$ to $+1$. Edit: to be clearer! ... the point is that if we constrain ourselves to go around either none, or both the bad points, the net sign flip is ... null. 
(Edited...) A sillier example, to amplify the point, is about $\sqrt{(1-z^2)(4-z^2)}$. Here there are 4 bad points, $\pm 1,\pm 2$. Any combination of slits connecting one of these to another... such as $1$-to-$2$ and $-1$-to-$-2$, or $1$-to-$-2$ and $-1$-to-$2$, causes the sign-flips created by travelling loops to be "doubled", that is, no net sign flip, thus, a well-defined function.
The issue is not about supposed relations between args, really, but, rather, to kill off enough of the homology of the domain so that under the Monodromy Map all that's left maps to $\{1\}$ in the permutation of function elements under analytic continuation... 
