This is a problem from Hatcher's ALgebraic Topology
"Compute the homology of the space obtained from $D^2$ by first deleting the interiors of two disjoint subdisks in the interior of $D^2$ and then identifying all three resulting boundary circles together via homeomorphisms preserving clockwise orientations of these circles."
I found a solution here https://web.stanford.edu/class/math215b/Sol4.pdf. From the photo you can see that the solution uses a CW structure and said the $2$-cell $U$ attaches to the word $aba^{-1}b^{-1}ca^{-1}c^{-1}$. My question is: Why is that?
It seems more rational to me to attach $U$ to $abab^{-1}cac^{-1}$ since we want all 3 circles to be clockwise. I can vaguely understand the procedure: we start at $x$, then we go around $a$, now we go through $b$ to reach the inner circle from the outer one, and then we go around $a$ again, then we do the same for $c$. But why we go counterclockwise when we reach inside?
Think of yourself sitting inside $U$ and think of how the boundary wraps around. You start at the top $x$ then take a clockwise trip round the outer circle ($a$), then a walk along segment $b$ (now you have done $ab$) then an anticlockwise walk along the left inner circle ($aba^{-1}$), then back along $b$ ($aba^{-1}b^{-1}$) etc.
The point is that when you're inside $U$ the walks along the inner circles are anticlockwise; opposite in direction to the walk around the outer circle. Just remember that the interior of $U$ is always on the same side as one walks along its boundary.