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Consider the CW-complex structure of X: Delete the interiors of two disjoint subdiscs in the interior of $D^2$, and then identify all three resulting boundary circles via homeomorphisms preserving clockwise orientation of these circles.

I saw the structure below with boundary of 2-cell: $aba^{−1}b^{−1}ca^{−1}$, and I am confused: (1) why do we have the 1-cells b and c?(why can't we just have one 0-cell:x and one 1-cell a and a 2-cell u?) (2) Can I say the boundary is $ababca$? I am stuck on them, can you give me a explanation? Thank you!

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First, it looks to me like you left off the final term in the attaching map of the boundary of the 2-cell: $a b a^{-1} b^{-1} c a^{-1} c^{-1}$.

To get to your questions, you cannot leave out $b$ and $c$ because then $u$ is not homeomorphic to the interior of $D^2$, as is required for a 2-cell. With the 1-cells $b$ and $c$ present, $u$ is indeed homeomorphic to the interior of $D^2$. Remember that in a 2-dimensional CW complex, each component of the complement of the 1-skeleton must be homeomorphic to the interior of $D^2$.

To describe the attaching map of the boundary of the 2-cell, it's best to have a good picture, which is too hard for me to reproduce here, but I'll try to describe how you can produce the required picture. Imagine that the interior of $u$ is filled out by concentric topological circles; of course, those circles will be distorted in the picture since the interior of $u$ is not literally round in the given picture, so this is going to take some visualization skill. Draw one of those circles which tracks close to the $1$-skeleton. Trace exactly one time around that circle, writing down the concatenation of 1-cells which that circle tracks close to. For example, if you start near $x$ going in the clockwise direction around the outer $a$ circle, but staying just inside that circle, and continue close to the 1-skeleton but never touching the 1-skeleton, then you will get $$aba^{-1}b^{-1}ca^{-1}c^{-1} $$ If you start at a different point you will get a cyclic permutation (e.g. $b a^{-1} b^{-1} c a^{-1} c^{-1} a$). If you go in the opposite direciton you will get the inverse (meaning $c a c^{-1} b c b^{-1} a^{-1}$), or a cyclic permutation of the inverse (e.g. $a c^{-1} b c b^{-1} a^{-1} c$).

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  • $\begingroup$ Thank you, that makes sense. $\endgroup$ – Katherine Apr 18 '16 at 2:40

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