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I am having trouble understanding one step in Hatcher's construction of generalized lens spaces (p 144 of Hatcher's Algebraic Topology

For the generalized lens space $L_m(\ell_1,\ldots,\ell_n)$ he shows by induction that he can build Lens spaces with one cell for every dimension.

He considers the unit circle, $C$, in the last component of $\mathbb{C}^n$. Then he joins the $m$th roots of unit in this circle by arcs of great circles to the subsphere $S^{2n-3}$. He calls these arcs $B_j^{2n-2}$.

Up until here I have followed. However, at this point he says "Similarly joing the $j$th edge of $C$ to $S^{2n-3}$ gives a ball $B_j^{2n-1}$ bounded by $B_j^{2n-2}$ and $B_{j+1}^{2n-2}$ ". I am confused what he means by "joining" in this context. From my understanding, when the quotient is taken to pass to the Lens space, this $C$ will not be identified with $S^{2n-3}$. Why is an extra $1$-cell not required here?

After this step, I understand the rest of the proof as he argues that these $B_j^{2n-1}$ and $B_j^{2n-2}$ are just sent to each other under the rotations under which identifications are made. However, I am lost on that one point about relating to $C$.

Thank you for any help.

Atticus Christensen
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    I believe I have solved my own question. What Hatcher means by joining the $j$th edge of $C$ to $S^{2n-3}$ is to make $B^{2n-1}j$ equal to $(z_1,\ldots,z{n-1},z_n)$, where $z_n$ is between the $j$th and $j+1$st root of unity in $C$. In this description, it is clear the interior of this is homeomorphic to a $2n-1$ disc and the boundaries are $B_j^{2n-2}$ and $B_{j+1}^{2n-2}$. – Atticus Christensen Apr 28 '14 at 15:14
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    If you've solved your own question, you should answer it and accept your own answer to take this question off the unanswered list. –  Apr 29 '14 at 06:47

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