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This question and many resources describe that $\sqrt{z^2+1}$ does not have a branch point at infinity, but I'm having trouble making this statement rigorous. In the linked question, the argument is made that $\sqrt{z^2+1}\approx z$ about infinity (a bit more rigorously), but this involves an explicit choice of branch cut. In Ahlfors' complex analysis, "branch point" is never rigorously defined, except intuitively as a point that staples together sheets of a Riemann surface.

It seems to me that I could put in by hand a branch point at infinity: we have a manifold $\mathcal{M}$ defined by $\pm\sqrt{z^2+1}$. I can add a single point at infinity $\mathcal{M}'=\mathcal{M}\cup\{\infty\}$, add a bunch of open sets $\{\pm \sqrt{z^2+1} : |z|>R\}\cup \{\infty\}$ to my topology, and now infinity is a point that glues together the two sheets of the Riemann surface. Everything is still Hausdorff and I don't think any desirable properties are violated!

I found a document on compact Riemann surfaces which, on page six (page seven of the pdf), gives a generic compactification which I guess is different from mine, adding two points at infinity so that the sheets are not "stapled together".

My question is: is this just a matter of convention? Or am I really violating/contradicting some definitions with my compactification?

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    The two meromorphic branches at $\infty$ are $g_1(z) = z \sum_{k=0}^\infty {1/2 \choose k} z^{-2k}$ and $g_2(z) = -g_1(z)$. To make $\sqrt{z^2+1}$ a map between Riemann surfaces, the problem is similar as for $z^{1/2}$. – reuns Oct 08 '17 at 00:15
  • You need to put a complex chart around your new $\infty$ to make it into a complex manifold. You can't just glue things randomly and call it a Riemann surface. – mercio Oct 08 '17 at 08:55
  • Even simpler, you can consider $\sqrt {z^2}$ around zero. I think the point is that if you choose a path going through zero, then indeed you have an ambiguity, you can move to another sheet or stay on the same sheet. But if you choose a loop around zero, you will not be able to move from one sheet to another by the analytic continuation. Which is why it's not a branch point. – Maxim Apr 11 '18 at 18:15

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