Riemann Surface Comprehension I have a question about an example in following script: https://math.berkeley.edu/~teleman/math/Riemann.pdf
Here the excerpt:

Why there don't exist a continuous choice of $w$ near $\pm1$ and $\pm k$? Why it does lead to the opposite choice upon return? What do they mean with return?
Above I have drawn a (in my opinion) possible choice for $w$. Why it is wrong?
 A: The problem with your diagram is that it shows only real values of $z$ near $-1$. If you let $z$ move in a small circle in the complex plane, centered at $-1$, then the value of $\sqrt{(z^2-1)(z^2-k^2)}$ will continuously change from positive to negative on returning to $z=-1+\epsilon$ ($\epsilon>0$).
For the problem you're having, it's enough to look at the simpler surface $\sqrt{z^2-1}$. Here's a picture I created of part of it around $\pm1$:

The x-axis runs from the bottom left to the top right. On the left, I've tried to show the "cross-cut" where the surface appears to intersect itself. (It doesn't really intersect itself, that's just an artifact of trying to draw the surface in regular Euclidean space. Like the circle of "self-intersection" in drawings of a Klein bottle.)
On the right hand side, I've sliced the picture along the y-axis, and display only the part with $y>0$. (When I say "x-axis" and "y-axis", I'm assuming $z=x+iy$.) In my drawing, I focussed on the situation near $1$, so let's work with that; the story near $-1$ is basically the same.
Now consider a small circle in the $z$-plane centered at $1$. Let $z$ go around that circle, starting $w$ on the upper surface. As the point passes passes through the cross-cut, it moves to the lower surface. When $z$ returns to its original position, $w$ is now minus its original value.
You can sort of see this algebraically. In $w=\sqrt{z^2-1}=\sqrt{(z-1)(z+1)}$, for $z$ close to 1, $z+1\approx 2$. So $\sqrt{(z-1)(z+1)}\approx\sqrt{2(z-1)}$. So near $1$, the surface should look a lot like the surface of the square root, except with the branch point at $1$ instead of $0$. (There are good pictures of the Riemann surface of $\sqrt{z}$ all over the web.)
There is a reason Telemann is looking at the more complicated surface $\sqrt{(z^2-1)(z^2-k^2)}$. (Spoiler warning!) I've shown a portion of the surface of $\sqrt{z^2-1}$ double-covering a part of the $z$ plane near the origin. The double covering can be extended to a double covering of the entire Riemann sphere. (The branch points need special attention, but I'll leave that to the textbooks.) The surface $\sqrt{z^2-1}$ is topologically less interesting than the surface of $\sqrt{(z^2-1)(z^2-k^2)}$.
Finally, a recommendation: the first chapter of George Springer's Introduction to Riemann Surfaces has lovely pictures and a fine discussion. It's an old book, but well worth digging up.
