This is a follow-up question to: https://mathoverflow.net/questions/14024/degree-2-branched-map-from-the-torus-to-the-sphere

This is a silly computation, but for whatever reason this is taking me much, much longer than it should. So hopefully more geometrically-oriented people can answer this easily.

Let's say we have a 2-cover of S<sup>2</sup> branched at 4 points. We can visualize this, as stankewicz aptly put it, as having a torus, putting it on a skewer (that meets it in four points), and quotienting by the action of turning the torus by a 180 degrees around that skewer. If you prefer to think of the torus as the plane quotiented by a lattice, it's the same as identifying vectors with their minuses (the ramification points being the 2-torsion).

As we know $\pi_1$(Torus - 4 points,basept)$\cong$< a,b,c,d,e,f| [a,b]cdef=1>, and $\pi_1$(S<sup>2</sup> - 4 points,basept)$\cong$< g,h,x,w| ghxw=1>. This 2-cover corresponds to an identification, therefore, of < a,b,c,d,e,f| [a,b]cdef=1> with an index 2 subgroup of < g,h,x,w| ghxw=1>. For the life of me, I can't figure out how this identification would go! I've been drawing tori to no avail for two days now.