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Consider the space $X$ which is $S^2$ with $x\sim -x$ for $x$ on the equator $S^1$. I was reading this answer here.

When putting a cell structure on $X$, it says the two hemispheres are wound twice around the equator, and in opposite directions.

Why is that? How do we know they wind around twice, and why must it be in opposite directions?

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Geovanna Anthony
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    The map $p\colon S^1\to S^1$ given by $p(x)=[x]$ is a 2-fold covering map because $p^{-1}([x])={x,-x}$. The 'opposite directions' bit really just comes from your frame of reference - it's the same direction if you consider one disk to be 'upside down' or it's different directions if you think they should both be right side up and we instead turn the circle 'upside down'. – Dan Rust May 26 '14 at 13:14

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Consider the adjunction space $Z=(D^2_1\sqcup D^2_2)\cup_f S^1$, where $f:\partial D^2_1⊔∂D^2_2\to S^1$ is the map sending $s\in∂D^2_1$ to $s^2$ and $t\in∂D^2_2$ to $t^{-2}$. That's the CW complex which we construct from two disks glued to the circle by gluing maps winding around the circle twice and in opposite directions. Note that it's also a quoteint space of the two disks, by the relation generated by $\pm s\in ∂D_1^2\sim \pm s^{-1}\in ∂D^2_2$

On the other hand, we have $W=S^2/\sim$ which, since $S^2$ itself is a quotient of two disks, namely by $s\in D_1^2\sim s^{-1}\in D_2^2$, is a quotient of $D^2_1⊔D^2_2$ where we identify a point in the boundary with its inverse in the other boundary and then with their respective negations.

The opposite direction in the attaching of the lower hemisphere isn't really necessary here. It only depends on how you look at the lower disk, either from below (and from outside of the sphere) or from above (which is from the inside of the sphere), in the latter case you don't reverse the direction.

Stefan Hamcke
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